Chapter 17 IB Maths HL Cambridge Textbook
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Transcript of Chapter 17 IB Maths HL Cambridge Textbook
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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17 Basic integration and its applications 569
17
As in many areas of mathematics as soon as we learn a newprocess we must then learn how to undo it However it turnsout that undoing the process of differentiation opens up thepossibility of answering a seemingly unconnected problemwhat is the area under a curve
17A Reversing differentiationWe saw in the last chapter how differentiation gives us thegradient of a curve or the rate of change of one quantity withanother What then if we already know the function describinga curversquos gradient or the expression for a rate of change andwish to 1047297nd the original function Our only way of proceedingis to lsquoundorsquo the differentiation that has already taken place andthis process of reverse differentiation is known as integration
Basic
integrationand itsapplications
Introductory problem
Te amount of charge stored in a capacitor is given by thearea under the graph of current (I ) against time (t ) Whenit contains alternating current the relationship between I and t is given by When it contains direct currentthe relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whetheralternating or direct current is used
In this chapter youwill learn
to reverse the processbullof differentiation (this
process is calledintegration)
to find the equationbullof a curve given itsderivative and a pointon the curve
to integrate sinbull x cos x and tan x
to integratebull ex and 1
x to find the areabullbetween a curve andthe x - or y -axis
to find the areabullenclosed between twocurves
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570 Topic 6 Calculus
Let us look at two particular cases to get a feel for this process
Each time we will be given and need to answer the question
lsquoWhat was differentiated to give thisrsquo
If then the original function y must have contained
x 2 as we know that differentiation decreases the power by 1Differentiating x 2 gives exactly 2x so we have found that ifd y
x x = then 2
If12 then the original function y must have contained x
3
2
Differentiating32 will give
12 and we do not
want the3
However if we multiply the3
2 by then when we
differentiate the coeffi cient cancels to 1 so if x 1
2
then2 3
2
Writing out lsquoifd
d
y
x
1
2 then y x 32 is descriptive but rather
laborious and so the notation used for integration is
1 32
Here the dx simply states that the integration is taking place
with respect to the variable x in exactly the same way that in y
it states that the differentiation is taking place with respect to x
We could equally well write for example int 1 3
2
The integration symbolcomes from the oldEnglish way ofwriting the letter lsquoSrsquoOriginally it stood for theword lsquoSumrsquo (or rather m)
As you will see in latersections the integral doesindeed represent a sum ofinfinitesimally small quantities
Exercise 17A1 Find a possible expression for y in terms of x
(a) (i) 2 (ii) 4
(b) (i) = minus1
2 (ii)
x x = minus
5
You may have heardof the term lsquodifferentialequationrsquo These arethe simplest types ofdifferential equation
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17 Basic integration and its applications 571
(c) (i)d
d
y
x x
1 (ii)
d
d
y
x x
12
(d) (i)x
10 4 (ii) 1 2
17B Constant of integration
We have seen how to integrate some functions of the form x n byreversing the process of differentiation but the process as carriedout above was not complete
Let us consider again the 1047297rst example where we stated that
int 2 2
Were there any other possible answers here
We could have given int 2 or
int = minus3
2x x x
Both of these are just as valid as our original answer we know
that when we differentiate the constant ( +1 or minus ) we just
get 0 We could therefore have given any constant withoutfurther information we cannot know what this constant on the
original function was before it was differentiatedHence our complete answers to the integrals considered inSection 17A are
int =2
x c1 32
3
where the c is an unknown constant of integration
We will see later that given further information we can 1047297nd
this constant
Exercise 17B
1 Give three possible functions which when differentiatedwith respect to x give the following
(a) 3
(b) 0
We will see how to 1047297nd the constant
of integration in
Section 17F
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572 Topic 6 Calculus
2 Find the integrals
(a) (i) 7 (ii) int 1
x x
(b) (i)1
2 (ii)
3
17C Rules of integration
o 1047297nd integrals so far we have used the idea of reversingdifferentiation for each speci1047297c function Let us now thinkabout applying the reverse process to the general rule ofdifferentiation
We know that for n n nminus1 or in words
o differentiate x n multiply by the old power then decrease thepower by 1
We can express the reverse of this process as follows
o integrate x n increase the power by 1 then divide by thenew power
Using integral notation
KEY POINT 171
Te general rule for integrating x n for any rational powern ne minus1 is
x c+int 1
1
Note the condition n ne minus1 which ensures that we are notdividing by zero
It is worth remembering the formula below for integrating aconstant = +k x c which is a special case of the rule inKey point 171
1
In Key point 163 we saw that if we differentiate kf )x we getkf prime( )x
we can reverse this logic to show that
KEY POINT 172
o integrate multiples of functions
We will see how
to integrate x ndash1 inSection 17D
T he + c is a par t o f
t he ans wer and you
mus t wri te i t e ver y
time
e x a m h i n t
T his ru le on l y wor ks
i f k is a cons tan t
e x a m h i n t
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17 Basic integration and its applications 573
Since we can differentiate term by term (also in Key point 164)then we can also split up integrals of sums
KEY POINT 173
For the sum of integrals
int int +
By combining Key points 172 and 173 with k = minus1 we canalso show that the integral of a difference is the difference of theintegrals of the separate parts
Tese ideas are demonstrated in the following examples
Be warned You
canno t in tegra te
produc ts or quo tien ts
b y in tegra ting eac h
par t separa te l y
e x a m h i n t
Worked example 171
Find (a) int x x (b) int minus
( )+4
minus x d
Add one to the power anddivide by this new power
(a) 6xint minus
minus c x = +xminus
Tidy up
= minus
22
x
= minus +minus2
x
Go through term by termadding one to the powerof x and dividing by this
new powerRemember the rule forintegrating a constant
(b) 3 3 8
4
1x x+x minus +x
minus +minus minus+ +
int c
Tidy up
=minus
+minus
5 1xminus c
= + +
minus 1
3x c
Just as for differentiation it may be necessary to change termsinto the form n before integrating
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574 Topic 6 Calculus
Exercise 17C
1 Find the following integrals
(a) (i) int 9x x (ii) int 12x x
(b) (i) int (ii) int (c) (i) 9 (ii) int
1dx
(d) (i) (ii)
(e) (i) int x x (ii) int 33 x x
(f) (i)2
(ii)3
2 Find the following integrals
(a) (i) int 3 (ii) int 7 z
(b) (i) int q (ii) int r r
In t he in tegra l do no t
forge t t he d x or t he
equi va len t We wi l l
ma ke more use o f i t
la ter T he func tion
you are in tegra ting
is ac tua l l y being
mu l tip lied b y d x
so you cou ld
wri te ques tion
1( f )(ii ) as int2 x
e x a m h i n t
Worked example 172
Find (a) 3
(b) int
( )minusx
2
d
Write the cube root as a powerand use rules of exponents
(a)
int xx
= int x x
Dividing by10
3
7
31+ is the
same as multiplying by3
10
Expand the brackets first thenuse rules of exponents
(b) int +
x
x minus
Dividing by a fraction isthe same as multiplying by
its reciprocal
times10
10
3x c = +10
= minusminus xx 2
= + +5
95 3
minus times c
= ++5 3 1
minus c
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17 Basic integration and its applications 575
(c) (i) 13
(ii) int 57
(d) (i) int 2h (ii) int
d p
p4
3 Find the following integrals (a) (i) int +x minus x 2 d (ii) int x minus x + d
(b) (i) int +1 1
4d (ii) int times minus times
1 15
v d
(c) (i) int x x x d (ii) int 3
(d) (i) int ( 3 (ii) int )2
4 Find int 1
x [4 marks]
17D Integrating x ndash1 and ex
When integrating int +n
1
1 we were careful to exclude
the case n = minus1
In Key point 168 we saw that ( )n Reversing this gives
KEY POINT 174
int minusx c=x
In Key point 167 we saw that x e( We can use this to
integrate the exponential function
KEY POINT 175
e
We will modify
this rule in Section17H
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576 Topic 6 Calculus
Exercise 17D 1 Find the following integrals
(a) (i) (ii)
(b) (i) int 1
x (ii) int
1
x
(c) (i) int minus x 2 1
d (ii) int + x 3
d
(d) (i) int +
2x (ii) int
x x minusx 2
2 Find the following integrals
(a) (i) int ex (ii) int ex x
(b) (i)x
(ii)7 x
(c) (i) int + x
(ii) int +
17E Integrating trigonometric functions
We can expand the set of functions that we can integrate bycontinuing to refer back to work covered in chapter 16
We saw in Key point 166 that (s x which means
that
Similarly as ( then int si minus
KEY POINT 176
Te integrals of trigonometric functions
int s minus s
=
We do not have a function whose derivative is ta and so
have no way (yet) of 1047297nding ta We will meet a method that
enables us to establish this in chapter 19 but for completeness
the result is given here
KEY POINT 176a
tan
See Exercise 19B
for establishing this
result
T he in tegra l o f
tan x is no t gi ven
in t he Formu la
boo k le t and is wor t h
remem bering
e x a m h i n t
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17 Basic integration and its applications 577
Exercise 17E
1 Find the following integrals
(a) (i) int s n x cminus os x d (ii) x s n x d
(b) (i) int t n (ii)
sin
2 3
(c) (i)7
(ii)
(d) (i) int minus x s n x d (ii) cos minus x sminus ncos x
2 Find int s n
cos
x cos
x x d [5 marks]
3 Find int cos x sminus n [5 marks]
17F Finding the equation of a curve
We have seen how we can integrate the function y
to 1047297nd the
equation of the original curve except for the unknown constant
of integration Tis is because the gradientd y
x determines the
shape of the curve but not exactly where it is However if weare also given the coordinates of a point on the curve we can
then determine the constant and hence specify the originalfunction precisely
If we again consider x 2 which we met at the start of this
chapter we know that the original function must have equation y = x 2 + c for some constant value c
If we are also told that the curve passes through the point(1 minus1) we can 1047297nd c and specify which of the family of curvesour function must be
(1minus1)
c = minus6
c = minus4
c = minus2
c = 4
c = 2
c = 0
y
x0
Look back to Worked
example 162 where given the gradient
we could draw many
different curves bychanging the starting
point
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578 Topic 6 Calculus
Te above example illustrates the general procedure for 1047297ndingthe equation of a curve from its gradient function
KEY POINT 177
o 1047297nd the equation for y given the gradientx
and onepoint ( p q) on the curve
1 Integrate remembering +c
2 Find the constant of integration by substitutingx = p y = q
Exercise 17F
1 Find the equation of the original curve if
(a) (i) and the curve passes through (ndash2 7)
(ii) x = 2 and the curve passes through (0 5)
(b) (i)d
d
y
x x
1 and the curve passes through (4 8)
(ii)1
2 and the curve passes through (1 3)
(c) (i) y
= + and the curve passes through (1 1)
(ii)d y
x = e and the curve passes through (ln5 0)
Worked example 173
Te gradient of a curve is given by = + and the curve passes through the point
(1 ndash4) Find the equation of the curve
To find y from ddy x
we need tointegrate
Donrsquot forget + c
= + 5xminus d
= +minus +
The coordinates of the givenpoint must satisfy this
equation so we can find c
When x = = minus so
minus ) + ) += 53 2
c
rArr minus minus + == 4 + 6c rArr
there4 = minusx minus x5+ 6
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17 Basic integration and its applications 579
(d) (i)+
and the curve passes through (e e)
(ii)1
and the curve passes through (e2 5)
(e) (i)d y
x x = x and the curve passesthrough (π 1)
(ii) x 3ta and the curve passes through (0 4)
2 Te derivative of the curve y f )x is1
(a) Find an expression for all possible functions f(x)
(b) If the curve passes through the point (2 7) 1047297nd theequation of the curve [5 marks]
3 Te gradient of a curve is found to be minus2
(a) Find the x -coordinate of the maximum point justifyingthat it is a maximum
(b) Given that the curve passes through the point (0 2)show that the y -coordinate of the maximum point
is minus7 [5 marks]
4 Te gradient of the normal to a curve at any point isequal to the x -coordinate at that point If the curve passesthrough the point (e2 3) 1047297nd the equation of the curvein the form ( ) where is a rationalfunction x gt 0 [6 marks]
17G Definite integration
Until now we have been carrying out a process known as
inde1047297nite integration inde1047297nite in the sense that we have anunknown constant each time for example
1
However there is also a process called de1047297nite integration which yields a numerical answer without the involvementof the constant of integration for example
bb
3 3 3
3 3 minus
int
Here a and b are known as the limits of integration a is the
lower limit and b the upper limit
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580 Topic 6 Calculus
Te square bracket notation means that the integration hastaken place but the limits have not yet been applied o do thiswe simply evaluate the integrated expression at the upper limitand subtract the integrated expression evaluated at the lowerlimit
You may be wondering where the constant of integration hasgone We could write it in as before but we quickly realise thatthis is unnecessary as it will just cancel out at the upper andlower limit each time
b c
bb
3
3
1
1
1
Te value of x is a dummy variable it does not come intothe answer But both a and b can vary and affect the resultChanging x to a different variable does not change the answerFor example
u a x bb
31 1
int
Worked example 174
Find the exact value of1
1+
Integrate and write in squarebrackets x
+ ]xint
Evaluate the integratedexpression at the upper
and lower limits andsubtract the lower from
the upper
=(In (e) + 4 (e)) minus (In (1) + 4 (1))
= + minusminus =
Ma ke sure you
kno w ho w to e va lua te
de fini te in tegra ls on your
ca lcu la tor as e xp lained
on Ca lcu la tor s ki l ls s hee t
1 0 on t he C D - R OM
I t can sa ve you time
and you can e va lua te
in tegra ls you donrsquo t kno w
ho w to do a lge braica l l y
E ven w hen you are
as ked to find t he e xac t
va lue o f t he in tegra l you
can c hec k your ans wer
on t he ca lcu la tor
e x a m h i n t
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17 Basic integration and its applications 581
Exercise 17G
1 Evaluate the following de1047297nite integrals giving exact answers
(a) (i) x x 2
(ii)1
4
1
(b) (i)2π
int (ii) sπ
π2
(c) (i) ex 1
int (ii) 31
1
eminusint
2 Evaluate correct to three signi1047297cant 1047297gures
(a) (i)3
1 4
int (ii)1
int
(b) (i) e1
int (ii) ne
1int 13 Find the exact value of the integral e x int
π [5 marks]
4 Show that the value of the integral12
x k
k
is independentof k [4 marks]
5 If x ) 73
evaluate 13
[4 marks]
6 Solve the equation 1 [5 marks]
y
x
y = f (x)
a b
A
17H Geometrical significance of definiteintegration
Now we have a method that gives a numerical value for anintegral the natural question to ask is what does this numbermean
On Fill-in proof sheet 20 on the CD-ROM Te fundamentaltheorem of calculus we show that as long as f )x is positivethe de1047297nite integral of f )x between the limits a and b is thearea enclosed between the curve the x -axis and the lines x = a and x = b
KEY POINT 178
rea )x b
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582 Topic 6 Calculus
If you are sketching the graph on the calculator you can get it toshade and evaluate the required area see Calculator skillssheet 10 on the CD-ROM You need to show the sketch as a partof your working if it is not already shown in the question
Worked example 175
Find the exact area enclosed between the x -axis the curve y = sin x and the lines x = 0 and x = π
Sketch the graph and identifythe area required = in
3
Integrate and write in squarebrackets
A = ]= minus
0
3ππ
Evaluate the integratedexpression at the upper andlower limit and subtract the
lower from the upper
= minus )cos cminus minus os
= minus minus ( )minus =
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17 Basic integration and its applications 583
The Ancient Greekshad developed ideasof limiting processessimilar to those used
in calculus but it took nearly2000 years for these ideas
to be formalised This wasdone almost simultaneouslyby Isaac Newton andGottried Leibniz in the 17thCentury Is this a coincidenceor is it often the case that along period of slow progressis often necessary to get tothe stage of majorbreakthroughsSupplementary sheet 10
looks at some other peoplewho can claim to haveinvented calculus
In the 17th Century integration was defined as thearea under a curve The area was broken downinto small rectangles each with a height f (x ) and a
width of a small bit of x called ∆x The total areawas approximately the sum of all of these rectangles
x a
x
f x x sum )∆
Isaac Newton one of the pioneers of calculus was also abig fan of writing in English rather than Greek So sigmabecame the English letter lsquoSrsquo and delta became the Englishletter d so that when the limit is taken as the width of therectangles become vanishingly small then the expressionbecomes
x a
)x dint This illustrates another very important interpretation of
integration ndash the infinite sum of infinitesimally small parts
Worked example 176
Find the area A in this graphy
x
y = x(xminus1)(2minusx)
1 2A
0
Write down the integral to beevaluated then use calculator
x x xminusint 1
x = minus y D
The area must be positive there4 =
When the curve is entirely below the x -axis the integral will givea negative value Te modulus of this value is the area
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584 Topic 6 Calculus
Unfortunately the relationship between integrals and areas isnot so simple when there are parts of the curve above and belowthe axis Tose bits above the axis contribute positively to thearea but bits below the axis contribute negatively to the areaWe must separate out the sections above the axis and belowthe axis
Worked example 177
(a) Find x x 1
4
x int 1 d
(b) Find the area enclosed between the x -axis the curve y minus x + and the lines= 1 and 4
Apply standard integration (a) x x
4
xminus1
= ( ( ) ( ( )minus + minus +
= minus =3
4
The value found above canrsquot bethe correct area for (b)
Sketch the curve to see exactlywhich area we are being asked
to find
(b) y
2 minus + 3
1 3
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17 Basic integration and its applications 585
Te fact that the integral was zero in Worked example 177 part(a) means that the area above the axis is exactly cancelled by thearea below the axis
Tis example warns us that when asked to 1047297nd an area we mustalways sketch the graph and identify exactly where each partof the area is If we are evaluating the area on the calculator wecan use the modulus function to ensure that the entire graph isabove the x -axis Using the function from Worked example 177
1
4
int 1y
x
y = |x2 minus 4x + 3|
1 3 4
continued
The area is made up of twoparts so evaluate each of
them separately
x
1
33
xminusint
( ) minus = minus
there4 rea below the axis is
x x3
4
2xminus int
minus ( ) =
4
there4 rea above the axis is
Total area = + =3 3
Transformationsof graphs using the
modulus function
were covered inchapter 7
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586 Topic 6 Calculus
KEY POINT 179
Te area bounded by the curve y f )x the x-axis and the
lines x = a and x = b is given byb
x int
When working without a calculator if the curve crosses
the x -axis between a and b we need to split the area intoseveral parts and 1047297nd each one separately
Te interpretation of integrals as areas causes one inconsistency
with our previous work Consider the integral1
2
1
x dminus
int
Graphically we can see that this area should exist
x
y
minus1minus2
However if we do the integration we 1047297nd that
12
1
2
1
]minus
minus
minus
minus
minus n minus
minusn
1
= ln
1
minus n
Tis is the correct answer (which we could have found usingthe symmetry of the curve) but it goes through a stage wherewe had to take logarithms of negative numbers and this issomething we are not allowed to do We avoid this by rede1047297ning
the integral ofx
as
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17 Basic integration and its applications 587
KEY POINT 174 AGAIN
minusint =
With this de1047297nition we can integrate y =1
over negative
numbers and the integral above becomes
12
1
2
1
x
e oreminus
minusint
n
Notice that the answer is negative since the required area is below
the x -axis We can still not integrate1
with a negative lower and
positive upper limit since the graph has an asymptote at x = 0
You may rightly be alittle uncomfortablewith inserting amodulus function lsquojust
because it worksrsquo Inmathematics do the ends
justify the means
Exercise 17H
1 Find the shaded areas
(a) (i)y =x2
y
x1 2
(ii)y = 1
x2
y
x2 4
(b) (i) y =x2
minus4x+ 3
y
x1 2
(ii)y =x2
minus4
y
x1
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588 Topic 6 Calculus
(c) (i)y = x3 minus x
y
xminus1 2
(ii)y = x2minus3x
y
x
5
2 Te area enclosed by the x -axis the curve andthe line x = k is 18 Find the value of k [6 marks]
3 (a) Find3
int (b) Find the area between the curve minus2 and the
x -axis between x = 0 and x = 3 [5 marks]
4 Between x = 0 and x = 3 the area of the graph y x minus2 below the x -axis equals the area above thex -axis Find the value of k [6 marks]
5 Find the area enclosed by the curve 1x minus minus 0
and the x -axis [7 marks]
lsquo Find t he area
enc losedrsquo means firs t
find a c losed region
bounded b y t he
cur ves men tioned
t hen find i ts area
A s ke tc h is a ver y
use fu l too l
e x a m h i n t
17I The area between a curve andthe y -axis
Consider the diagram alongside How can we 1047297nd the shadedarea A
One possible strategy is to construct a box around the graph todivide up the regions of interest You can integrate to 1047297nd thearea labelled A1 and then by adding and subtracting the areas ofthe blue and red rectangles shown calculate A
y
x
y = f (x)
c
d
A
y
x
y = f (x)
c
d A1
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17 Basic integration and its applications 589
Happily there is a quicker way we can treat x as a functionof y effectively re1047298ecting the whole diagram in the line y = x and then use the same method as in the previous section
KEY POINT 1710
Te area bounded by the curve y the y -axis and the
lines y = c and y = d is given by y d
) dint where is
the expression for x in terms of y
You may have realised that this is related to inverse functions from Section 5E
x
y
x = f (y)
cd
A
Worked example 178
Te curve shown has equation y 1minus Find the shaded area
y
x
y = 2radic x minus 1
10
Express x in terms of y x = 2
rArr = + y
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590 Topic 6 Calculus
Exercise 17I
continued
Find the limits on the y -axisIt may help to label them on the
graph
When x = minusWhen x = 1 = =minus 6
y
y = 2radic minus
1
10
Write down the integral andevaluate using calculator
Area fro GDC= +2
1 2=d
1 Find the shaded areas
(a) (i) y
x
y = x2
1
6
(ii) y
x
y = x3
1
3
(b) (i) y
xy = 1
x2
2
1
(ii) y
x
y =radic x
1
5
(c) (i)y
x
y = ln x
1e
e
(ii) y
x
y = 3x
1 6
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17 Basic integration and its applications 591
2 Te diagram shows the curve If the shadedarea is 504 1047297nd the value of a [6 marks]
y
x
y = radic
x
a
2a
0
3 Find the exact value of the area enclosed by the graphof + the line y = 2 and the y -axis [6 marks]
4 Te diagram shows the graph of
Te shaded area is 39 units Find the value of a [7 marks]y
x
y =radic x
4 a
5 Te diagram shows the graph of 2 where a isin infin Te area of the pink region is equal to the area of the blue
region Give two equations for a in terms of b and hencegive a in exact form and determine the size of theblue area [8 marks]y
x
y = x2
b
1 a
17J The area between two curves
So far we have only looked at areas bounded by a curve and oneof the coordinate axes but we can also 1047297nd areas bounded bytwo curvesTe area A in the diagram can be found by taking the areabounded by f )x and the x -axis and subtracting the areabounded by and the x -axis that is
A
b
minus
y
x
y = f (x)
y = g(x)
a b
A
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592 Topic 6 Calculus
We can do the subtraction before integrating so that we onlyhave to integrate one expression instead of two Tis gives analternative formula for the area
KEY POINT 1711
Te area A between two curves f (x ) and g (x ) is
Ab
(
where a and b are the x-coordinates of the intersectionpoints of the two curves
Worked example 179
Find the area A enclosed between + and minus2 +
First find the x -coordinates ofintersection
For intersection
x x x
rArr =
rArr ) =rArr =
minus
minus
+x
4minusx 0
Make a rough sketch to see therelative positions of the two curves
y
x1 4
A
2x + 1y =
2 minus 3x + 5y = x
Subtract the lower curve from thehigher before integrating
= x
minusx x
4
minus
minus minusx
2
1
4
2
minus =
8
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17 Basic integration and its applications 593
Subtracting the two equations before integrating is particularlyuseful when one of the curves is partly below the x-axis If x is always above then the expression we are integrating f g )x minus )x is always positive so we do not have to worryabout the signs of f )x and g )x themselves
Worked example 1710
Find the area bounded by the curves y x minus and y x 2
Sketch the graph to see therelative position of two
curves
Using GDC
= ex minus 5
y = 3 2
y
minus2 2
A
Find the intersection points ndash
use calculatorintersections x = minus2 658
Write down the integralrepresenting the area
rea =minus
minus818
1 658
xminus
= minus
1 6 8
x
Evaluate the integral usingcalculator = 21 6 (3SF)
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594 Topic 6 Calculus
Exercise 17J
1 Find the shaded areas
(a) (i)
y = 2x+ 1
y = (xminus1)2
y
x
(ii) y =x+ 1
y = 4xminusx2minus1
y
x
(b) (i)
y =minusx2minus4x+ 12
y
x
y =x2 + 2x+ 12
(ii)
y =x2minus2x+ 9
y
x
y = 4xminusx2 + 5
(c) (i)y =x2 minusx
y = 2xminusx2
y
x
(ii)y =x2minus7x+ 7
y = 3minusxminusx2
y
x
2 Find the area enclosed between the graphs of y x +x minus and + [6 marks]
3 Find the area enclosed by the curve 2
the y-axisand the line [6 marks]
4 Find the area between the curves1
and in theregion lt lt π [6 marks]
5 Show that the area of the shaded region alongside is2
[6 marks]
y = x2
y
xminus1 2
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17 Basic integration and its applications 595
6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]
7 Find the total area enclosed between the graphs of
)minus 2 and minus2 7 1+ [6 marks]
8 Te area enclosed between the curve y x 2 and the line
y mx is 10 Find the value of m if m 0 [7 marks]
9 Show that the shaded area in the diagram below is [8 marks]
y = 2 minus xy2 = x
y
x
y = cos x y = sin x
y
x
Summary
bull Integration is the reverse process of differentiation
bull Any integral without limits (inde1047297nite) will generate a constant of integration
bull For all rational n ne minus1 int +x c
1
1
bull If minus we get the natural logarithm function int minus =
bull Te integral of the exponential function is int e c
bull Te integrals of the trigonometric functions are
int si minus=
int tan =
bull Te de1047297nite integral has limits f( ) x b
dint is found by evaluating the integrated expression
at b and then subtracting the integrated expression evaluated at a
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596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
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17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
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7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
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570 Topic 6 Calculus
Let us look at two particular cases to get a feel for this process
Each time we will be given and need to answer the question
lsquoWhat was differentiated to give thisrsquo
If then the original function y must have contained
x 2 as we know that differentiation decreases the power by 1Differentiating x 2 gives exactly 2x so we have found that ifd y
x x = then 2
If12 then the original function y must have contained x
3
2
Differentiating32 will give
12 and we do not
want the3
However if we multiply the3
2 by then when we
differentiate the coeffi cient cancels to 1 so if x 1
2
then2 3
2
Writing out lsquoifd
d
y
x
1
2 then y x 32 is descriptive but rather
laborious and so the notation used for integration is
1 32
Here the dx simply states that the integration is taking place
with respect to the variable x in exactly the same way that in y
it states that the differentiation is taking place with respect to x
We could equally well write for example int 1 3
2
The integration symbolcomes from the oldEnglish way ofwriting the letter lsquoSrsquoOriginally it stood for theword lsquoSumrsquo (or rather m)
As you will see in latersections the integral doesindeed represent a sum ofinfinitesimally small quantities
Exercise 17A1 Find a possible expression for y in terms of x
(a) (i) 2 (ii) 4
(b) (i) = minus1
2 (ii)
x x = minus
5
You may have heardof the term lsquodifferentialequationrsquo These arethe simplest types ofdifferential equation
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17 Basic integration and its applications 571
(c) (i)d
d
y
x x
1 (ii)
d
d
y
x x
12
(d) (i)x
10 4 (ii) 1 2
17B Constant of integration
We have seen how to integrate some functions of the form x n byreversing the process of differentiation but the process as carriedout above was not complete
Let us consider again the 1047297rst example where we stated that
int 2 2
Were there any other possible answers here
We could have given int 2 or
int = minus3
2x x x
Both of these are just as valid as our original answer we know
that when we differentiate the constant ( +1 or minus ) we just
get 0 We could therefore have given any constant withoutfurther information we cannot know what this constant on the
original function was before it was differentiatedHence our complete answers to the integrals considered inSection 17A are
int =2
x c1 32
3
where the c is an unknown constant of integration
We will see later that given further information we can 1047297nd
this constant
Exercise 17B
1 Give three possible functions which when differentiatedwith respect to x give the following
(a) 3
(b) 0
We will see how to 1047297nd the constant
of integration in
Section 17F
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572 Topic 6 Calculus
2 Find the integrals
(a) (i) 7 (ii) int 1
x x
(b) (i)1
2 (ii)
3
17C Rules of integration
o 1047297nd integrals so far we have used the idea of reversingdifferentiation for each speci1047297c function Let us now thinkabout applying the reverse process to the general rule ofdifferentiation
We know that for n n nminus1 or in words
o differentiate x n multiply by the old power then decrease thepower by 1
We can express the reverse of this process as follows
o integrate x n increase the power by 1 then divide by thenew power
Using integral notation
KEY POINT 171
Te general rule for integrating x n for any rational powern ne minus1 is
x c+int 1
1
Note the condition n ne minus1 which ensures that we are notdividing by zero
It is worth remembering the formula below for integrating aconstant = +k x c which is a special case of the rule inKey point 171
1
In Key point 163 we saw that if we differentiate kf )x we getkf prime( )x
we can reverse this logic to show that
KEY POINT 172
o integrate multiples of functions
We will see how
to integrate x ndash1 inSection 17D
T he + c is a par t o f
t he ans wer and you
mus t wri te i t e ver y
time
e x a m h i n t
T his ru le on l y wor ks
i f k is a cons tan t
e x a m h i n t
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17 Basic integration and its applications 573
Since we can differentiate term by term (also in Key point 164)then we can also split up integrals of sums
KEY POINT 173
For the sum of integrals
int int +
By combining Key points 172 and 173 with k = minus1 we canalso show that the integral of a difference is the difference of theintegrals of the separate parts
Tese ideas are demonstrated in the following examples
Be warned You
canno t in tegra te
produc ts or quo tien ts
b y in tegra ting eac h
par t separa te l y
e x a m h i n t
Worked example 171
Find (a) int x x (b) int minus
( )+4
minus x d
Add one to the power anddivide by this new power
(a) 6xint minus
minus c x = +xminus
Tidy up
= minus
22
x
= minus +minus2
x
Go through term by termadding one to the powerof x and dividing by this
new powerRemember the rule forintegrating a constant
(b) 3 3 8
4
1x x+x minus +x
minus +minus minus+ +
int c
Tidy up
=minus
+minus
5 1xminus c
= + +
minus 1
3x c
Just as for differentiation it may be necessary to change termsinto the form n before integrating
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574 Topic 6 Calculus
Exercise 17C
1 Find the following integrals
(a) (i) int 9x x (ii) int 12x x
(b) (i) int (ii) int (c) (i) 9 (ii) int
1dx
(d) (i) (ii)
(e) (i) int x x (ii) int 33 x x
(f) (i)2
(ii)3
2 Find the following integrals
(a) (i) int 3 (ii) int 7 z
(b) (i) int q (ii) int r r
In t he in tegra l do no t
forge t t he d x or t he
equi va len t We wi l l
ma ke more use o f i t
la ter T he func tion
you are in tegra ting
is ac tua l l y being
mu l tip lied b y d x
so you cou ld
wri te ques tion
1( f )(ii ) as int2 x
e x a m h i n t
Worked example 172
Find (a) 3
(b) int
( )minusx
2
d
Write the cube root as a powerand use rules of exponents
(a)
int xx
= int x x
Dividing by10
3
7
31+ is the
same as multiplying by3
10
Expand the brackets first thenuse rules of exponents
(b) int +
x
x minus
Dividing by a fraction isthe same as multiplying by
its reciprocal
times10
10
3x c = +10
= minusminus xx 2
= + +5
95 3
minus times c
= ++5 3 1
minus c
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17 Basic integration and its applications 575
(c) (i) 13
(ii) int 57
(d) (i) int 2h (ii) int
d p
p4
3 Find the following integrals (a) (i) int +x minus x 2 d (ii) int x minus x + d
(b) (i) int +1 1
4d (ii) int times minus times
1 15
v d
(c) (i) int x x x d (ii) int 3
(d) (i) int ( 3 (ii) int )2
4 Find int 1
x [4 marks]
17D Integrating x ndash1 and ex
When integrating int +n
1
1 we were careful to exclude
the case n = minus1
In Key point 168 we saw that ( )n Reversing this gives
KEY POINT 174
int minusx c=x
In Key point 167 we saw that x e( We can use this to
integrate the exponential function
KEY POINT 175
e
We will modify
this rule in Section17H
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576 Topic 6 Calculus
Exercise 17D 1 Find the following integrals
(a) (i) (ii)
(b) (i) int 1
x (ii) int
1
x
(c) (i) int minus x 2 1
d (ii) int + x 3
d
(d) (i) int +
2x (ii) int
x x minusx 2
2 Find the following integrals
(a) (i) int ex (ii) int ex x
(b) (i)x
(ii)7 x
(c) (i) int + x
(ii) int +
17E Integrating trigonometric functions
We can expand the set of functions that we can integrate bycontinuing to refer back to work covered in chapter 16
We saw in Key point 166 that (s x which means
that
Similarly as ( then int si minus
KEY POINT 176
Te integrals of trigonometric functions
int s minus s
=
We do not have a function whose derivative is ta and so
have no way (yet) of 1047297nding ta We will meet a method that
enables us to establish this in chapter 19 but for completeness
the result is given here
KEY POINT 176a
tan
See Exercise 19B
for establishing this
result
T he in tegra l o f
tan x is no t gi ven
in t he Formu la
boo k le t and is wor t h
remem bering
e x a m h i n t
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17 Basic integration and its applications 577
Exercise 17E
1 Find the following integrals
(a) (i) int s n x cminus os x d (ii) x s n x d
(b) (i) int t n (ii)
sin
2 3
(c) (i)7
(ii)
(d) (i) int minus x s n x d (ii) cos minus x sminus ncos x
2 Find int s n
cos
x cos
x x d [5 marks]
3 Find int cos x sminus n [5 marks]
17F Finding the equation of a curve
We have seen how we can integrate the function y
to 1047297nd the
equation of the original curve except for the unknown constant
of integration Tis is because the gradientd y
x determines the
shape of the curve but not exactly where it is However if weare also given the coordinates of a point on the curve we can
then determine the constant and hence specify the originalfunction precisely
If we again consider x 2 which we met at the start of this
chapter we know that the original function must have equation y = x 2 + c for some constant value c
If we are also told that the curve passes through the point(1 minus1) we can 1047297nd c and specify which of the family of curvesour function must be
(1minus1)
c = minus6
c = minus4
c = minus2
c = 4
c = 2
c = 0
y
x0
Look back to Worked
example 162 where given the gradient
we could draw many
different curves bychanging the starting
point
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578 Topic 6 Calculus
Te above example illustrates the general procedure for 1047297ndingthe equation of a curve from its gradient function
KEY POINT 177
o 1047297nd the equation for y given the gradientx
and onepoint ( p q) on the curve
1 Integrate remembering +c
2 Find the constant of integration by substitutingx = p y = q
Exercise 17F
1 Find the equation of the original curve if
(a) (i) and the curve passes through (ndash2 7)
(ii) x = 2 and the curve passes through (0 5)
(b) (i)d
d
y
x x
1 and the curve passes through (4 8)
(ii)1
2 and the curve passes through (1 3)
(c) (i) y
= + and the curve passes through (1 1)
(ii)d y
x = e and the curve passes through (ln5 0)
Worked example 173
Te gradient of a curve is given by = + and the curve passes through the point
(1 ndash4) Find the equation of the curve
To find y from ddy x
we need tointegrate
Donrsquot forget + c
= + 5xminus d
= +minus +
The coordinates of the givenpoint must satisfy this
equation so we can find c
When x = = minus so
minus ) + ) += 53 2
c
rArr minus minus + == 4 + 6c rArr
there4 = minusx minus x5+ 6
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17 Basic integration and its applications 579
(d) (i)+
and the curve passes through (e e)
(ii)1
and the curve passes through (e2 5)
(e) (i)d y
x x = x and the curve passesthrough (π 1)
(ii) x 3ta and the curve passes through (0 4)
2 Te derivative of the curve y f )x is1
(a) Find an expression for all possible functions f(x)
(b) If the curve passes through the point (2 7) 1047297nd theequation of the curve [5 marks]
3 Te gradient of a curve is found to be minus2
(a) Find the x -coordinate of the maximum point justifyingthat it is a maximum
(b) Given that the curve passes through the point (0 2)show that the y -coordinate of the maximum point
is minus7 [5 marks]
4 Te gradient of the normal to a curve at any point isequal to the x -coordinate at that point If the curve passesthrough the point (e2 3) 1047297nd the equation of the curvein the form ( ) where is a rationalfunction x gt 0 [6 marks]
17G Definite integration
Until now we have been carrying out a process known as
inde1047297nite integration inde1047297nite in the sense that we have anunknown constant each time for example
1
However there is also a process called de1047297nite integration which yields a numerical answer without the involvementof the constant of integration for example
bb
3 3 3
3 3 minus
int
Here a and b are known as the limits of integration a is the
lower limit and b the upper limit
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580 Topic 6 Calculus
Te square bracket notation means that the integration hastaken place but the limits have not yet been applied o do thiswe simply evaluate the integrated expression at the upper limitand subtract the integrated expression evaluated at the lowerlimit
You may be wondering where the constant of integration hasgone We could write it in as before but we quickly realise thatthis is unnecessary as it will just cancel out at the upper andlower limit each time
b c
bb
3
3
1
1
1
Te value of x is a dummy variable it does not come intothe answer But both a and b can vary and affect the resultChanging x to a different variable does not change the answerFor example
u a x bb
31 1
int
Worked example 174
Find the exact value of1
1+
Integrate and write in squarebrackets x
+ ]xint
Evaluate the integratedexpression at the upper
and lower limits andsubtract the lower from
the upper
=(In (e) + 4 (e)) minus (In (1) + 4 (1))
= + minusminus =
Ma ke sure you
kno w ho w to e va lua te
de fini te in tegra ls on your
ca lcu la tor as e xp lained
on Ca lcu la tor s ki l ls s hee t
1 0 on t he C D - R OM
I t can sa ve you time
and you can e va lua te
in tegra ls you donrsquo t kno w
ho w to do a lge braica l l y
E ven w hen you are
as ked to find t he e xac t
va lue o f t he in tegra l you
can c hec k your ans wer
on t he ca lcu la tor
e x a m h i n t
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17 Basic integration and its applications 581
Exercise 17G
1 Evaluate the following de1047297nite integrals giving exact answers
(a) (i) x x 2
(ii)1
4
1
(b) (i)2π
int (ii) sπ
π2
(c) (i) ex 1
int (ii) 31
1
eminusint
2 Evaluate correct to three signi1047297cant 1047297gures
(a) (i)3
1 4
int (ii)1
int
(b) (i) e1
int (ii) ne
1int 13 Find the exact value of the integral e x int
π [5 marks]
4 Show that the value of the integral12
x k
k
is independentof k [4 marks]
5 If x ) 73
evaluate 13
[4 marks]
6 Solve the equation 1 [5 marks]
y
x
y = f (x)
a b
A
17H Geometrical significance of definiteintegration
Now we have a method that gives a numerical value for anintegral the natural question to ask is what does this numbermean
On Fill-in proof sheet 20 on the CD-ROM Te fundamentaltheorem of calculus we show that as long as f )x is positivethe de1047297nite integral of f )x between the limits a and b is thearea enclosed between the curve the x -axis and the lines x = a and x = b
KEY POINT 178
rea )x b
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582 Topic 6 Calculus
If you are sketching the graph on the calculator you can get it toshade and evaluate the required area see Calculator skillssheet 10 on the CD-ROM You need to show the sketch as a partof your working if it is not already shown in the question
Worked example 175
Find the exact area enclosed between the x -axis the curve y = sin x and the lines x = 0 and x = π
Sketch the graph and identifythe area required = in
3
Integrate and write in squarebrackets
A = ]= minus
0
3ππ
Evaluate the integratedexpression at the upper andlower limit and subtract the
lower from the upper
= minus )cos cminus minus os
= minus minus ( )minus =
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17 Basic integration and its applications 583
The Ancient Greekshad developed ideasof limiting processessimilar to those used
in calculus but it took nearly2000 years for these ideas
to be formalised This wasdone almost simultaneouslyby Isaac Newton andGottried Leibniz in the 17thCentury Is this a coincidenceor is it often the case that along period of slow progressis often necessary to get tothe stage of majorbreakthroughsSupplementary sheet 10
looks at some other peoplewho can claim to haveinvented calculus
In the 17th Century integration was defined as thearea under a curve The area was broken downinto small rectangles each with a height f (x ) and a
width of a small bit of x called ∆x The total areawas approximately the sum of all of these rectangles
x a
x
f x x sum )∆
Isaac Newton one of the pioneers of calculus was also abig fan of writing in English rather than Greek So sigmabecame the English letter lsquoSrsquo and delta became the Englishletter d so that when the limit is taken as the width of therectangles become vanishingly small then the expressionbecomes
x a
)x dint This illustrates another very important interpretation of
integration ndash the infinite sum of infinitesimally small parts
Worked example 176
Find the area A in this graphy
x
y = x(xminus1)(2minusx)
1 2A
0
Write down the integral to beevaluated then use calculator
x x xminusint 1
x = minus y D
The area must be positive there4 =
When the curve is entirely below the x -axis the integral will givea negative value Te modulus of this value is the area
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584 Topic 6 Calculus
Unfortunately the relationship between integrals and areas isnot so simple when there are parts of the curve above and belowthe axis Tose bits above the axis contribute positively to thearea but bits below the axis contribute negatively to the areaWe must separate out the sections above the axis and belowthe axis
Worked example 177
(a) Find x x 1
4
x int 1 d
(b) Find the area enclosed between the x -axis the curve y minus x + and the lines= 1 and 4
Apply standard integration (a) x x
4
xminus1
= ( ( ) ( ( )minus + minus +
= minus =3
4
The value found above canrsquot bethe correct area for (b)
Sketch the curve to see exactlywhich area we are being asked
to find
(b) y
2 minus + 3
1 3
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17 Basic integration and its applications 585
Te fact that the integral was zero in Worked example 177 part(a) means that the area above the axis is exactly cancelled by thearea below the axis
Tis example warns us that when asked to 1047297nd an area we mustalways sketch the graph and identify exactly where each partof the area is If we are evaluating the area on the calculator wecan use the modulus function to ensure that the entire graph isabove the x -axis Using the function from Worked example 177
1
4
int 1y
x
y = |x2 minus 4x + 3|
1 3 4
continued
The area is made up of twoparts so evaluate each of
them separately
x
1
33
xminusint
( ) minus = minus
there4 rea below the axis is
x x3
4
2xminus int
minus ( ) =
4
there4 rea above the axis is
Total area = + =3 3
Transformationsof graphs using the
modulus function
were covered inchapter 7
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586 Topic 6 Calculus
KEY POINT 179
Te area bounded by the curve y f )x the x-axis and the
lines x = a and x = b is given byb
x int
When working without a calculator if the curve crosses
the x -axis between a and b we need to split the area intoseveral parts and 1047297nd each one separately
Te interpretation of integrals as areas causes one inconsistency
with our previous work Consider the integral1
2
1
x dminus
int
Graphically we can see that this area should exist
x
y
minus1minus2
However if we do the integration we 1047297nd that
12
1
2
1
]minus
minus
minus
minus
minus n minus
minusn
1
= ln
1
minus n
Tis is the correct answer (which we could have found usingthe symmetry of the curve) but it goes through a stage wherewe had to take logarithms of negative numbers and this issomething we are not allowed to do We avoid this by rede1047297ning
the integral ofx
as
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17 Basic integration and its applications 587
KEY POINT 174 AGAIN
minusint =
With this de1047297nition we can integrate y =1
over negative
numbers and the integral above becomes
12
1
2
1
x
e oreminus
minusint
n
Notice that the answer is negative since the required area is below
the x -axis We can still not integrate1
with a negative lower and
positive upper limit since the graph has an asymptote at x = 0
You may rightly be alittle uncomfortablewith inserting amodulus function lsquojust
because it worksrsquo Inmathematics do the ends
justify the means
Exercise 17H
1 Find the shaded areas
(a) (i)y =x2
y
x1 2
(ii)y = 1
x2
y
x2 4
(b) (i) y =x2
minus4x+ 3
y
x1 2
(ii)y =x2
minus4
y
x1
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588 Topic 6 Calculus
(c) (i)y = x3 minus x
y
xminus1 2
(ii)y = x2minus3x
y
x
5
2 Te area enclosed by the x -axis the curve andthe line x = k is 18 Find the value of k [6 marks]
3 (a) Find3
int (b) Find the area between the curve minus2 and the
x -axis between x = 0 and x = 3 [5 marks]
4 Between x = 0 and x = 3 the area of the graph y x minus2 below the x -axis equals the area above thex -axis Find the value of k [6 marks]
5 Find the area enclosed by the curve 1x minus minus 0
and the x -axis [7 marks]
lsquo Find t he area
enc losedrsquo means firs t
find a c losed region
bounded b y t he
cur ves men tioned
t hen find i ts area
A s ke tc h is a ver y
use fu l too l
e x a m h i n t
17I The area between a curve andthe y -axis
Consider the diagram alongside How can we 1047297nd the shadedarea A
One possible strategy is to construct a box around the graph todivide up the regions of interest You can integrate to 1047297nd thearea labelled A1 and then by adding and subtracting the areas ofthe blue and red rectangles shown calculate A
y
x
y = f (x)
c
d
A
y
x
y = f (x)
c
d A1
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17 Basic integration and its applications 589
Happily there is a quicker way we can treat x as a functionof y effectively re1047298ecting the whole diagram in the line y = x and then use the same method as in the previous section
KEY POINT 1710
Te area bounded by the curve y the y -axis and the
lines y = c and y = d is given by y d
) dint where is
the expression for x in terms of y
You may have realised that this is related to inverse functions from Section 5E
x
y
x = f (y)
cd
A
Worked example 178
Te curve shown has equation y 1minus Find the shaded area
y
x
y = 2radic x minus 1
10
Express x in terms of y x = 2
rArr = + y
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590 Topic 6 Calculus
Exercise 17I
continued
Find the limits on the y -axisIt may help to label them on the
graph
When x = minusWhen x = 1 = =minus 6
y
y = 2radic minus
1
10
Write down the integral andevaluate using calculator
Area fro GDC= +2
1 2=d
1 Find the shaded areas
(a) (i) y
x
y = x2
1
6
(ii) y
x
y = x3
1
3
(b) (i) y
xy = 1
x2
2
1
(ii) y
x
y =radic x
1
5
(c) (i)y
x
y = ln x
1e
e
(ii) y
x
y = 3x
1 6
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17 Basic integration and its applications 591
2 Te diagram shows the curve If the shadedarea is 504 1047297nd the value of a [6 marks]
y
x
y = radic
x
a
2a
0
3 Find the exact value of the area enclosed by the graphof + the line y = 2 and the y -axis [6 marks]
4 Te diagram shows the graph of
Te shaded area is 39 units Find the value of a [7 marks]y
x
y =radic x
4 a
5 Te diagram shows the graph of 2 where a isin infin Te area of the pink region is equal to the area of the blue
region Give two equations for a in terms of b and hencegive a in exact form and determine the size of theblue area [8 marks]y
x
y = x2
b
1 a
17J The area between two curves
So far we have only looked at areas bounded by a curve and oneof the coordinate axes but we can also 1047297nd areas bounded bytwo curvesTe area A in the diagram can be found by taking the areabounded by f )x and the x -axis and subtracting the areabounded by and the x -axis that is
A
b
minus
y
x
y = f (x)
y = g(x)
a b
A
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592 Topic 6 Calculus
We can do the subtraction before integrating so that we onlyhave to integrate one expression instead of two Tis gives analternative formula for the area
KEY POINT 1711
Te area A between two curves f (x ) and g (x ) is
Ab
(
where a and b are the x-coordinates of the intersectionpoints of the two curves
Worked example 179
Find the area A enclosed between + and minus2 +
First find the x -coordinates ofintersection
For intersection
x x x
rArr =
rArr ) =rArr =
minus
minus
+x
4minusx 0
Make a rough sketch to see therelative positions of the two curves
y
x1 4
A
2x + 1y =
2 minus 3x + 5y = x
Subtract the lower curve from thehigher before integrating
= x
minusx x
4
minus
minus minusx
2
1
4
2
minus =
8
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17 Basic integration and its applications 593
Subtracting the two equations before integrating is particularlyuseful when one of the curves is partly below the x-axis If x is always above then the expression we are integrating f g )x minus )x is always positive so we do not have to worryabout the signs of f )x and g )x themselves
Worked example 1710
Find the area bounded by the curves y x minus and y x 2
Sketch the graph to see therelative position of two
curves
Using GDC
= ex minus 5
y = 3 2
y
minus2 2
A
Find the intersection points ndash
use calculatorintersections x = minus2 658
Write down the integralrepresenting the area
rea =minus
minus818
1 658
xminus
= minus
1 6 8
x
Evaluate the integral usingcalculator = 21 6 (3SF)
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594 Topic 6 Calculus
Exercise 17J
1 Find the shaded areas
(a) (i)
y = 2x+ 1
y = (xminus1)2
y
x
(ii) y =x+ 1
y = 4xminusx2minus1
y
x
(b) (i)
y =minusx2minus4x+ 12
y
x
y =x2 + 2x+ 12
(ii)
y =x2minus2x+ 9
y
x
y = 4xminusx2 + 5
(c) (i)y =x2 minusx
y = 2xminusx2
y
x
(ii)y =x2minus7x+ 7
y = 3minusxminusx2
y
x
2 Find the area enclosed between the graphs of y x +x minus and + [6 marks]
3 Find the area enclosed by the curve 2
the y-axisand the line [6 marks]
4 Find the area between the curves1
and in theregion lt lt π [6 marks]
5 Show that the area of the shaded region alongside is2
[6 marks]
y = x2
y
xminus1 2
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17 Basic integration and its applications 595
6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]
7 Find the total area enclosed between the graphs of
)minus 2 and minus2 7 1+ [6 marks]
8 Te area enclosed between the curve y x 2 and the line
y mx is 10 Find the value of m if m 0 [7 marks]
9 Show that the shaded area in the diagram below is [8 marks]
y = 2 minus xy2 = x
y
x
y = cos x y = sin x
y
x
Summary
bull Integration is the reverse process of differentiation
bull Any integral without limits (inde1047297nite) will generate a constant of integration
bull For all rational n ne minus1 int +x c
1
1
bull If minus we get the natural logarithm function int minus =
bull Te integral of the exponential function is int e c
bull Te integrals of the trigonometric functions are
int si minus=
int tan =
bull Te de1047297nite integral has limits f( ) x b
dint is found by evaluating the integrated expression
at b and then subtracting the integrated expression evaluated at a
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596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
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17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
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7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
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17 Basic integration and its applications 571
(c) (i)d
d
y
x x
1 (ii)
d
d
y
x x
12
(d) (i)x
10 4 (ii) 1 2
17B Constant of integration
We have seen how to integrate some functions of the form x n byreversing the process of differentiation but the process as carriedout above was not complete
Let us consider again the 1047297rst example where we stated that
int 2 2
Were there any other possible answers here
We could have given int 2 or
int = minus3
2x x x
Both of these are just as valid as our original answer we know
that when we differentiate the constant ( +1 or minus ) we just
get 0 We could therefore have given any constant withoutfurther information we cannot know what this constant on the
original function was before it was differentiatedHence our complete answers to the integrals considered inSection 17A are
int =2
x c1 32
3
where the c is an unknown constant of integration
We will see later that given further information we can 1047297nd
this constant
Exercise 17B
1 Give three possible functions which when differentiatedwith respect to x give the following
(a) 3
(b) 0
We will see how to 1047297nd the constant
of integration in
Section 17F
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572 Topic 6 Calculus
2 Find the integrals
(a) (i) 7 (ii) int 1
x x
(b) (i)1
2 (ii)
3
17C Rules of integration
o 1047297nd integrals so far we have used the idea of reversingdifferentiation for each speci1047297c function Let us now thinkabout applying the reverse process to the general rule ofdifferentiation
We know that for n n nminus1 or in words
o differentiate x n multiply by the old power then decrease thepower by 1
We can express the reverse of this process as follows
o integrate x n increase the power by 1 then divide by thenew power
Using integral notation
KEY POINT 171
Te general rule for integrating x n for any rational powern ne minus1 is
x c+int 1
1
Note the condition n ne minus1 which ensures that we are notdividing by zero
It is worth remembering the formula below for integrating aconstant = +k x c which is a special case of the rule inKey point 171
1
In Key point 163 we saw that if we differentiate kf )x we getkf prime( )x
we can reverse this logic to show that
KEY POINT 172
o integrate multiples of functions
We will see how
to integrate x ndash1 inSection 17D
T he + c is a par t o f
t he ans wer and you
mus t wri te i t e ver y
time
e x a m h i n t
T his ru le on l y wor ks
i f k is a cons tan t
e x a m h i n t
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17 Basic integration and its applications 573
Since we can differentiate term by term (also in Key point 164)then we can also split up integrals of sums
KEY POINT 173
For the sum of integrals
int int +
By combining Key points 172 and 173 with k = minus1 we canalso show that the integral of a difference is the difference of theintegrals of the separate parts
Tese ideas are demonstrated in the following examples
Be warned You
canno t in tegra te
produc ts or quo tien ts
b y in tegra ting eac h
par t separa te l y
e x a m h i n t
Worked example 171
Find (a) int x x (b) int minus
( )+4
minus x d
Add one to the power anddivide by this new power
(a) 6xint minus
minus c x = +xminus
Tidy up
= minus
22
x
= minus +minus2
x
Go through term by termadding one to the powerof x and dividing by this
new powerRemember the rule forintegrating a constant
(b) 3 3 8
4
1x x+x minus +x
minus +minus minus+ +
int c
Tidy up
=minus
+minus
5 1xminus c
= + +
minus 1
3x c
Just as for differentiation it may be necessary to change termsinto the form n before integrating
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574 Topic 6 Calculus
Exercise 17C
1 Find the following integrals
(a) (i) int 9x x (ii) int 12x x
(b) (i) int (ii) int (c) (i) 9 (ii) int
1dx
(d) (i) (ii)
(e) (i) int x x (ii) int 33 x x
(f) (i)2
(ii)3
2 Find the following integrals
(a) (i) int 3 (ii) int 7 z
(b) (i) int q (ii) int r r
In t he in tegra l do no t
forge t t he d x or t he
equi va len t We wi l l
ma ke more use o f i t
la ter T he func tion
you are in tegra ting
is ac tua l l y being
mu l tip lied b y d x
so you cou ld
wri te ques tion
1( f )(ii ) as int2 x
e x a m h i n t
Worked example 172
Find (a) 3
(b) int
( )minusx
2
d
Write the cube root as a powerand use rules of exponents
(a)
int xx
= int x x
Dividing by10
3
7
31+ is the
same as multiplying by3
10
Expand the brackets first thenuse rules of exponents
(b) int +
x
x minus
Dividing by a fraction isthe same as multiplying by
its reciprocal
times10
10
3x c = +10
= minusminus xx 2
= + +5
95 3
minus times c
= ++5 3 1
minus c
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17 Basic integration and its applications 575
(c) (i) 13
(ii) int 57
(d) (i) int 2h (ii) int
d p
p4
3 Find the following integrals (a) (i) int +x minus x 2 d (ii) int x minus x + d
(b) (i) int +1 1
4d (ii) int times minus times
1 15
v d
(c) (i) int x x x d (ii) int 3
(d) (i) int ( 3 (ii) int )2
4 Find int 1
x [4 marks]
17D Integrating x ndash1 and ex
When integrating int +n
1
1 we were careful to exclude
the case n = minus1
In Key point 168 we saw that ( )n Reversing this gives
KEY POINT 174
int minusx c=x
In Key point 167 we saw that x e( We can use this to
integrate the exponential function
KEY POINT 175
e
We will modify
this rule in Section17H
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576 Topic 6 Calculus
Exercise 17D 1 Find the following integrals
(a) (i) (ii)
(b) (i) int 1
x (ii) int
1
x
(c) (i) int minus x 2 1
d (ii) int + x 3
d
(d) (i) int +
2x (ii) int
x x minusx 2
2 Find the following integrals
(a) (i) int ex (ii) int ex x
(b) (i)x
(ii)7 x
(c) (i) int + x
(ii) int +
17E Integrating trigonometric functions
We can expand the set of functions that we can integrate bycontinuing to refer back to work covered in chapter 16
We saw in Key point 166 that (s x which means
that
Similarly as ( then int si minus
KEY POINT 176
Te integrals of trigonometric functions
int s minus s
=
We do not have a function whose derivative is ta and so
have no way (yet) of 1047297nding ta We will meet a method that
enables us to establish this in chapter 19 but for completeness
the result is given here
KEY POINT 176a
tan
See Exercise 19B
for establishing this
result
T he in tegra l o f
tan x is no t gi ven
in t he Formu la
boo k le t and is wor t h
remem bering
e x a m h i n t
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17 Basic integration and its applications 577
Exercise 17E
1 Find the following integrals
(a) (i) int s n x cminus os x d (ii) x s n x d
(b) (i) int t n (ii)
sin
2 3
(c) (i)7
(ii)
(d) (i) int minus x s n x d (ii) cos minus x sminus ncos x
2 Find int s n
cos
x cos
x x d [5 marks]
3 Find int cos x sminus n [5 marks]
17F Finding the equation of a curve
We have seen how we can integrate the function y
to 1047297nd the
equation of the original curve except for the unknown constant
of integration Tis is because the gradientd y
x determines the
shape of the curve but not exactly where it is However if weare also given the coordinates of a point on the curve we can
then determine the constant and hence specify the originalfunction precisely
If we again consider x 2 which we met at the start of this
chapter we know that the original function must have equation y = x 2 + c for some constant value c
If we are also told that the curve passes through the point(1 minus1) we can 1047297nd c and specify which of the family of curvesour function must be
(1minus1)
c = minus6
c = minus4
c = minus2
c = 4
c = 2
c = 0
y
x0
Look back to Worked
example 162 where given the gradient
we could draw many
different curves bychanging the starting
point
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578 Topic 6 Calculus
Te above example illustrates the general procedure for 1047297ndingthe equation of a curve from its gradient function
KEY POINT 177
o 1047297nd the equation for y given the gradientx
and onepoint ( p q) on the curve
1 Integrate remembering +c
2 Find the constant of integration by substitutingx = p y = q
Exercise 17F
1 Find the equation of the original curve if
(a) (i) and the curve passes through (ndash2 7)
(ii) x = 2 and the curve passes through (0 5)
(b) (i)d
d
y
x x
1 and the curve passes through (4 8)
(ii)1
2 and the curve passes through (1 3)
(c) (i) y
= + and the curve passes through (1 1)
(ii)d y
x = e and the curve passes through (ln5 0)
Worked example 173
Te gradient of a curve is given by = + and the curve passes through the point
(1 ndash4) Find the equation of the curve
To find y from ddy x
we need tointegrate
Donrsquot forget + c
= + 5xminus d
= +minus +
The coordinates of the givenpoint must satisfy this
equation so we can find c
When x = = minus so
minus ) + ) += 53 2
c
rArr minus minus + == 4 + 6c rArr
there4 = minusx minus x5+ 6
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17 Basic integration and its applications 579
(d) (i)+
and the curve passes through (e e)
(ii)1
and the curve passes through (e2 5)
(e) (i)d y
x x = x and the curve passesthrough (π 1)
(ii) x 3ta and the curve passes through (0 4)
2 Te derivative of the curve y f )x is1
(a) Find an expression for all possible functions f(x)
(b) If the curve passes through the point (2 7) 1047297nd theequation of the curve [5 marks]
3 Te gradient of a curve is found to be minus2
(a) Find the x -coordinate of the maximum point justifyingthat it is a maximum
(b) Given that the curve passes through the point (0 2)show that the y -coordinate of the maximum point
is minus7 [5 marks]
4 Te gradient of the normal to a curve at any point isequal to the x -coordinate at that point If the curve passesthrough the point (e2 3) 1047297nd the equation of the curvein the form ( ) where is a rationalfunction x gt 0 [6 marks]
17G Definite integration
Until now we have been carrying out a process known as
inde1047297nite integration inde1047297nite in the sense that we have anunknown constant each time for example
1
However there is also a process called de1047297nite integration which yields a numerical answer without the involvementof the constant of integration for example
bb
3 3 3
3 3 minus
int
Here a and b are known as the limits of integration a is the
lower limit and b the upper limit
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580 Topic 6 Calculus
Te square bracket notation means that the integration hastaken place but the limits have not yet been applied o do thiswe simply evaluate the integrated expression at the upper limitand subtract the integrated expression evaluated at the lowerlimit
You may be wondering where the constant of integration hasgone We could write it in as before but we quickly realise thatthis is unnecessary as it will just cancel out at the upper andlower limit each time
b c
bb
3
3
1
1
1
Te value of x is a dummy variable it does not come intothe answer But both a and b can vary and affect the resultChanging x to a different variable does not change the answerFor example
u a x bb
31 1
int
Worked example 174
Find the exact value of1
1+
Integrate and write in squarebrackets x
+ ]xint
Evaluate the integratedexpression at the upper
and lower limits andsubtract the lower from
the upper
=(In (e) + 4 (e)) minus (In (1) + 4 (1))
= + minusminus =
Ma ke sure you
kno w ho w to e va lua te
de fini te in tegra ls on your
ca lcu la tor as e xp lained
on Ca lcu la tor s ki l ls s hee t
1 0 on t he C D - R OM
I t can sa ve you time
and you can e va lua te
in tegra ls you donrsquo t kno w
ho w to do a lge braica l l y
E ven w hen you are
as ked to find t he e xac t
va lue o f t he in tegra l you
can c hec k your ans wer
on t he ca lcu la tor
e x a m h i n t
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17 Basic integration and its applications 581
Exercise 17G
1 Evaluate the following de1047297nite integrals giving exact answers
(a) (i) x x 2
(ii)1
4
1
(b) (i)2π
int (ii) sπ
π2
(c) (i) ex 1
int (ii) 31
1
eminusint
2 Evaluate correct to three signi1047297cant 1047297gures
(a) (i)3
1 4
int (ii)1
int
(b) (i) e1
int (ii) ne
1int 13 Find the exact value of the integral e x int
π [5 marks]
4 Show that the value of the integral12
x k
k
is independentof k [4 marks]
5 If x ) 73
evaluate 13
[4 marks]
6 Solve the equation 1 [5 marks]
y
x
y = f (x)
a b
A
17H Geometrical significance of definiteintegration
Now we have a method that gives a numerical value for anintegral the natural question to ask is what does this numbermean
On Fill-in proof sheet 20 on the CD-ROM Te fundamentaltheorem of calculus we show that as long as f )x is positivethe de1047297nite integral of f )x between the limits a and b is thearea enclosed between the curve the x -axis and the lines x = a and x = b
KEY POINT 178
rea )x b
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582 Topic 6 Calculus
If you are sketching the graph on the calculator you can get it toshade and evaluate the required area see Calculator skillssheet 10 on the CD-ROM You need to show the sketch as a partof your working if it is not already shown in the question
Worked example 175
Find the exact area enclosed between the x -axis the curve y = sin x and the lines x = 0 and x = π
Sketch the graph and identifythe area required = in
3
Integrate and write in squarebrackets
A = ]= minus
0
3ππ
Evaluate the integratedexpression at the upper andlower limit and subtract the
lower from the upper
= minus )cos cminus minus os
= minus minus ( )minus =
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17 Basic integration and its applications 583
The Ancient Greekshad developed ideasof limiting processessimilar to those used
in calculus but it took nearly2000 years for these ideas
to be formalised This wasdone almost simultaneouslyby Isaac Newton andGottried Leibniz in the 17thCentury Is this a coincidenceor is it often the case that along period of slow progressis often necessary to get tothe stage of majorbreakthroughsSupplementary sheet 10
looks at some other peoplewho can claim to haveinvented calculus
In the 17th Century integration was defined as thearea under a curve The area was broken downinto small rectangles each with a height f (x ) and a
width of a small bit of x called ∆x The total areawas approximately the sum of all of these rectangles
x a
x
f x x sum )∆
Isaac Newton one of the pioneers of calculus was also abig fan of writing in English rather than Greek So sigmabecame the English letter lsquoSrsquo and delta became the Englishletter d so that when the limit is taken as the width of therectangles become vanishingly small then the expressionbecomes
x a
)x dint This illustrates another very important interpretation of
integration ndash the infinite sum of infinitesimally small parts
Worked example 176
Find the area A in this graphy
x
y = x(xminus1)(2minusx)
1 2A
0
Write down the integral to beevaluated then use calculator
x x xminusint 1
x = minus y D
The area must be positive there4 =
When the curve is entirely below the x -axis the integral will givea negative value Te modulus of this value is the area
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584 Topic 6 Calculus
Unfortunately the relationship between integrals and areas isnot so simple when there are parts of the curve above and belowthe axis Tose bits above the axis contribute positively to thearea but bits below the axis contribute negatively to the areaWe must separate out the sections above the axis and belowthe axis
Worked example 177
(a) Find x x 1
4
x int 1 d
(b) Find the area enclosed between the x -axis the curve y minus x + and the lines= 1 and 4
Apply standard integration (a) x x
4
xminus1
= ( ( ) ( ( )minus + minus +
= minus =3
4
The value found above canrsquot bethe correct area for (b)
Sketch the curve to see exactlywhich area we are being asked
to find
(b) y
2 minus + 3
1 3
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17 Basic integration and its applications 585
Te fact that the integral was zero in Worked example 177 part(a) means that the area above the axis is exactly cancelled by thearea below the axis
Tis example warns us that when asked to 1047297nd an area we mustalways sketch the graph and identify exactly where each partof the area is If we are evaluating the area on the calculator wecan use the modulus function to ensure that the entire graph isabove the x -axis Using the function from Worked example 177
1
4
int 1y
x
y = |x2 minus 4x + 3|
1 3 4
continued
The area is made up of twoparts so evaluate each of
them separately
x
1
33
xminusint
( ) minus = minus
there4 rea below the axis is
x x3
4
2xminus int
minus ( ) =
4
there4 rea above the axis is
Total area = + =3 3
Transformationsof graphs using the
modulus function
were covered inchapter 7
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586 Topic 6 Calculus
KEY POINT 179
Te area bounded by the curve y f )x the x-axis and the
lines x = a and x = b is given byb
x int
When working without a calculator if the curve crosses
the x -axis between a and b we need to split the area intoseveral parts and 1047297nd each one separately
Te interpretation of integrals as areas causes one inconsistency
with our previous work Consider the integral1
2
1
x dminus
int
Graphically we can see that this area should exist
x
y
minus1minus2
However if we do the integration we 1047297nd that
12
1
2
1
]minus
minus
minus
minus
minus n minus
minusn
1
= ln
1
minus n
Tis is the correct answer (which we could have found usingthe symmetry of the curve) but it goes through a stage wherewe had to take logarithms of negative numbers and this issomething we are not allowed to do We avoid this by rede1047297ning
the integral ofx
as
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17 Basic integration and its applications 587
KEY POINT 174 AGAIN
minusint =
With this de1047297nition we can integrate y =1
over negative
numbers and the integral above becomes
12
1
2
1
x
e oreminus
minusint
n
Notice that the answer is negative since the required area is below
the x -axis We can still not integrate1
with a negative lower and
positive upper limit since the graph has an asymptote at x = 0
You may rightly be alittle uncomfortablewith inserting amodulus function lsquojust
because it worksrsquo Inmathematics do the ends
justify the means
Exercise 17H
1 Find the shaded areas
(a) (i)y =x2
y
x1 2
(ii)y = 1
x2
y
x2 4
(b) (i) y =x2
minus4x+ 3
y
x1 2
(ii)y =x2
minus4
y
x1
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588 Topic 6 Calculus
(c) (i)y = x3 minus x
y
xminus1 2
(ii)y = x2minus3x
y
x
5
2 Te area enclosed by the x -axis the curve andthe line x = k is 18 Find the value of k [6 marks]
3 (a) Find3
int (b) Find the area between the curve minus2 and the
x -axis between x = 0 and x = 3 [5 marks]
4 Between x = 0 and x = 3 the area of the graph y x minus2 below the x -axis equals the area above thex -axis Find the value of k [6 marks]
5 Find the area enclosed by the curve 1x minus minus 0
and the x -axis [7 marks]
lsquo Find t he area
enc losedrsquo means firs t
find a c losed region
bounded b y t he
cur ves men tioned
t hen find i ts area
A s ke tc h is a ver y
use fu l too l
e x a m h i n t
17I The area between a curve andthe y -axis
Consider the diagram alongside How can we 1047297nd the shadedarea A
One possible strategy is to construct a box around the graph todivide up the regions of interest You can integrate to 1047297nd thearea labelled A1 and then by adding and subtracting the areas ofthe blue and red rectangles shown calculate A
y
x
y = f (x)
c
d
A
y
x
y = f (x)
c
d A1
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17 Basic integration and its applications 589
Happily there is a quicker way we can treat x as a functionof y effectively re1047298ecting the whole diagram in the line y = x and then use the same method as in the previous section
KEY POINT 1710
Te area bounded by the curve y the y -axis and the
lines y = c and y = d is given by y d
) dint where is
the expression for x in terms of y
You may have realised that this is related to inverse functions from Section 5E
x
y
x = f (y)
cd
A
Worked example 178
Te curve shown has equation y 1minus Find the shaded area
y
x
y = 2radic x minus 1
10
Express x in terms of y x = 2
rArr = + y
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590 Topic 6 Calculus
Exercise 17I
continued
Find the limits on the y -axisIt may help to label them on the
graph
When x = minusWhen x = 1 = =minus 6
y
y = 2radic minus
1
10
Write down the integral andevaluate using calculator
Area fro GDC= +2
1 2=d
1 Find the shaded areas
(a) (i) y
x
y = x2
1
6
(ii) y
x
y = x3
1
3
(b) (i) y
xy = 1
x2
2
1
(ii) y
x
y =radic x
1
5
(c) (i)y
x
y = ln x
1e
e
(ii) y
x
y = 3x
1 6
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17 Basic integration and its applications 591
2 Te diagram shows the curve If the shadedarea is 504 1047297nd the value of a [6 marks]
y
x
y = radic
x
a
2a
0
3 Find the exact value of the area enclosed by the graphof + the line y = 2 and the y -axis [6 marks]
4 Te diagram shows the graph of
Te shaded area is 39 units Find the value of a [7 marks]y
x
y =radic x
4 a
5 Te diagram shows the graph of 2 where a isin infin Te area of the pink region is equal to the area of the blue
region Give two equations for a in terms of b and hencegive a in exact form and determine the size of theblue area [8 marks]y
x
y = x2
b
1 a
17J The area between two curves
So far we have only looked at areas bounded by a curve and oneof the coordinate axes but we can also 1047297nd areas bounded bytwo curvesTe area A in the diagram can be found by taking the areabounded by f )x and the x -axis and subtracting the areabounded by and the x -axis that is
A
b
minus
y
x
y = f (x)
y = g(x)
a b
A
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592 Topic 6 Calculus
We can do the subtraction before integrating so that we onlyhave to integrate one expression instead of two Tis gives analternative formula for the area
KEY POINT 1711
Te area A between two curves f (x ) and g (x ) is
Ab
(
where a and b are the x-coordinates of the intersectionpoints of the two curves
Worked example 179
Find the area A enclosed between + and minus2 +
First find the x -coordinates ofintersection
For intersection
x x x
rArr =
rArr ) =rArr =
minus
minus
+x
4minusx 0
Make a rough sketch to see therelative positions of the two curves
y
x1 4
A
2x + 1y =
2 minus 3x + 5y = x
Subtract the lower curve from thehigher before integrating
= x
minusx x
4
minus
minus minusx
2
1
4
2
minus =
8
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17 Basic integration and its applications 593
Subtracting the two equations before integrating is particularlyuseful when one of the curves is partly below the x-axis If x is always above then the expression we are integrating f g )x minus )x is always positive so we do not have to worryabout the signs of f )x and g )x themselves
Worked example 1710
Find the area bounded by the curves y x minus and y x 2
Sketch the graph to see therelative position of two
curves
Using GDC
= ex minus 5
y = 3 2
y
minus2 2
A
Find the intersection points ndash
use calculatorintersections x = minus2 658
Write down the integralrepresenting the area
rea =minus
minus818
1 658
xminus
= minus
1 6 8
x
Evaluate the integral usingcalculator = 21 6 (3SF)
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594 Topic 6 Calculus
Exercise 17J
1 Find the shaded areas
(a) (i)
y = 2x+ 1
y = (xminus1)2
y
x
(ii) y =x+ 1
y = 4xminusx2minus1
y
x
(b) (i)
y =minusx2minus4x+ 12
y
x
y =x2 + 2x+ 12
(ii)
y =x2minus2x+ 9
y
x
y = 4xminusx2 + 5
(c) (i)y =x2 minusx
y = 2xminusx2
y
x
(ii)y =x2minus7x+ 7
y = 3minusxminusx2
y
x
2 Find the area enclosed between the graphs of y x +x minus and + [6 marks]
3 Find the area enclosed by the curve 2
the y-axisand the line [6 marks]
4 Find the area between the curves1
and in theregion lt lt π [6 marks]
5 Show that the area of the shaded region alongside is2
[6 marks]
y = x2
y
xminus1 2
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17 Basic integration and its applications 595
6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]
7 Find the total area enclosed between the graphs of
)minus 2 and minus2 7 1+ [6 marks]
8 Te area enclosed between the curve y x 2 and the line
y mx is 10 Find the value of m if m 0 [7 marks]
9 Show that the shaded area in the diagram below is [8 marks]
y = 2 minus xy2 = x
y
x
y = cos x y = sin x
y
x
Summary
bull Integration is the reverse process of differentiation
bull Any integral without limits (inde1047297nite) will generate a constant of integration
bull For all rational n ne minus1 int +x c
1
1
bull If minus we get the natural logarithm function int minus =
bull Te integral of the exponential function is int e c
bull Te integrals of the trigonometric functions are
int si minus=
int tan =
bull Te de1047297nite integral has limits f( ) x b
dint is found by evaluating the integrated expression
at b and then subtracting the integrated expression evaluated at a
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596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
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17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
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7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
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572 Topic 6 Calculus
2 Find the integrals
(a) (i) 7 (ii) int 1
x x
(b) (i)1
2 (ii)
3
17C Rules of integration
o 1047297nd integrals so far we have used the idea of reversingdifferentiation for each speci1047297c function Let us now thinkabout applying the reverse process to the general rule ofdifferentiation
We know that for n n nminus1 or in words
o differentiate x n multiply by the old power then decrease thepower by 1
We can express the reverse of this process as follows
o integrate x n increase the power by 1 then divide by thenew power
Using integral notation
KEY POINT 171
Te general rule for integrating x n for any rational powern ne minus1 is
x c+int 1
1
Note the condition n ne minus1 which ensures that we are notdividing by zero
It is worth remembering the formula below for integrating aconstant = +k x c which is a special case of the rule inKey point 171
1
In Key point 163 we saw that if we differentiate kf )x we getkf prime( )x
we can reverse this logic to show that
KEY POINT 172
o integrate multiples of functions
We will see how
to integrate x ndash1 inSection 17D
T he + c is a par t o f
t he ans wer and you
mus t wri te i t e ver y
time
e x a m h i n t
T his ru le on l y wor ks
i f k is a cons tan t
e x a m h i n t
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17 Basic integration and its applications 573
Since we can differentiate term by term (also in Key point 164)then we can also split up integrals of sums
KEY POINT 173
For the sum of integrals
int int +
By combining Key points 172 and 173 with k = minus1 we canalso show that the integral of a difference is the difference of theintegrals of the separate parts
Tese ideas are demonstrated in the following examples
Be warned You
canno t in tegra te
produc ts or quo tien ts
b y in tegra ting eac h
par t separa te l y
e x a m h i n t
Worked example 171
Find (a) int x x (b) int minus
( )+4
minus x d
Add one to the power anddivide by this new power
(a) 6xint minus
minus c x = +xminus
Tidy up
= minus
22
x
= minus +minus2
x
Go through term by termadding one to the powerof x and dividing by this
new powerRemember the rule forintegrating a constant
(b) 3 3 8
4
1x x+x minus +x
minus +minus minus+ +
int c
Tidy up
=minus
+minus
5 1xminus c
= + +
minus 1
3x c
Just as for differentiation it may be necessary to change termsinto the form n before integrating
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574 Topic 6 Calculus
Exercise 17C
1 Find the following integrals
(a) (i) int 9x x (ii) int 12x x
(b) (i) int (ii) int (c) (i) 9 (ii) int
1dx
(d) (i) (ii)
(e) (i) int x x (ii) int 33 x x
(f) (i)2
(ii)3
2 Find the following integrals
(a) (i) int 3 (ii) int 7 z
(b) (i) int q (ii) int r r
In t he in tegra l do no t
forge t t he d x or t he
equi va len t We wi l l
ma ke more use o f i t
la ter T he func tion
you are in tegra ting
is ac tua l l y being
mu l tip lied b y d x
so you cou ld
wri te ques tion
1( f )(ii ) as int2 x
e x a m h i n t
Worked example 172
Find (a) 3
(b) int
( )minusx
2
d
Write the cube root as a powerand use rules of exponents
(a)
int xx
= int x x
Dividing by10
3
7
31+ is the
same as multiplying by3
10
Expand the brackets first thenuse rules of exponents
(b) int +
x
x minus
Dividing by a fraction isthe same as multiplying by
its reciprocal
times10
10
3x c = +10
= minusminus xx 2
= + +5
95 3
minus times c
= ++5 3 1
minus c
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17 Basic integration and its applications 575
(c) (i) 13
(ii) int 57
(d) (i) int 2h (ii) int
d p
p4
3 Find the following integrals (a) (i) int +x minus x 2 d (ii) int x minus x + d
(b) (i) int +1 1
4d (ii) int times minus times
1 15
v d
(c) (i) int x x x d (ii) int 3
(d) (i) int ( 3 (ii) int )2
4 Find int 1
x [4 marks]
17D Integrating x ndash1 and ex
When integrating int +n
1
1 we were careful to exclude
the case n = minus1
In Key point 168 we saw that ( )n Reversing this gives
KEY POINT 174
int minusx c=x
In Key point 167 we saw that x e( We can use this to
integrate the exponential function
KEY POINT 175
e
We will modify
this rule in Section17H
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576 Topic 6 Calculus
Exercise 17D 1 Find the following integrals
(a) (i) (ii)
(b) (i) int 1
x (ii) int
1
x
(c) (i) int minus x 2 1
d (ii) int + x 3
d
(d) (i) int +
2x (ii) int
x x minusx 2
2 Find the following integrals
(a) (i) int ex (ii) int ex x
(b) (i)x
(ii)7 x
(c) (i) int + x
(ii) int +
17E Integrating trigonometric functions
We can expand the set of functions that we can integrate bycontinuing to refer back to work covered in chapter 16
We saw in Key point 166 that (s x which means
that
Similarly as ( then int si minus
KEY POINT 176
Te integrals of trigonometric functions
int s minus s
=
We do not have a function whose derivative is ta and so
have no way (yet) of 1047297nding ta We will meet a method that
enables us to establish this in chapter 19 but for completeness
the result is given here
KEY POINT 176a
tan
See Exercise 19B
for establishing this
result
T he in tegra l o f
tan x is no t gi ven
in t he Formu la
boo k le t and is wor t h
remem bering
e x a m h i n t
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17 Basic integration and its applications 577
Exercise 17E
1 Find the following integrals
(a) (i) int s n x cminus os x d (ii) x s n x d
(b) (i) int t n (ii)
sin
2 3
(c) (i)7
(ii)
(d) (i) int minus x s n x d (ii) cos minus x sminus ncos x
2 Find int s n
cos
x cos
x x d [5 marks]
3 Find int cos x sminus n [5 marks]
17F Finding the equation of a curve
We have seen how we can integrate the function y
to 1047297nd the
equation of the original curve except for the unknown constant
of integration Tis is because the gradientd y
x determines the
shape of the curve but not exactly where it is However if weare also given the coordinates of a point on the curve we can
then determine the constant and hence specify the originalfunction precisely
If we again consider x 2 which we met at the start of this
chapter we know that the original function must have equation y = x 2 + c for some constant value c
If we are also told that the curve passes through the point(1 minus1) we can 1047297nd c and specify which of the family of curvesour function must be
(1minus1)
c = minus6
c = minus4
c = minus2
c = 4
c = 2
c = 0
y
x0
Look back to Worked
example 162 where given the gradient
we could draw many
different curves bychanging the starting
point
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578 Topic 6 Calculus
Te above example illustrates the general procedure for 1047297ndingthe equation of a curve from its gradient function
KEY POINT 177
o 1047297nd the equation for y given the gradientx
and onepoint ( p q) on the curve
1 Integrate remembering +c
2 Find the constant of integration by substitutingx = p y = q
Exercise 17F
1 Find the equation of the original curve if
(a) (i) and the curve passes through (ndash2 7)
(ii) x = 2 and the curve passes through (0 5)
(b) (i)d
d
y
x x
1 and the curve passes through (4 8)
(ii)1
2 and the curve passes through (1 3)
(c) (i) y
= + and the curve passes through (1 1)
(ii)d y
x = e and the curve passes through (ln5 0)
Worked example 173
Te gradient of a curve is given by = + and the curve passes through the point
(1 ndash4) Find the equation of the curve
To find y from ddy x
we need tointegrate
Donrsquot forget + c
= + 5xminus d
= +minus +
The coordinates of the givenpoint must satisfy this
equation so we can find c
When x = = minus so
minus ) + ) += 53 2
c
rArr minus minus + == 4 + 6c rArr
there4 = minusx minus x5+ 6
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17 Basic integration and its applications 579
(d) (i)+
and the curve passes through (e e)
(ii)1
and the curve passes through (e2 5)
(e) (i)d y
x x = x and the curve passesthrough (π 1)
(ii) x 3ta and the curve passes through (0 4)
2 Te derivative of the curve y f )x is1
(a) Find an expression for all possible functions f(x)
(b) If the curve passes through the point (2 7) 1047297nd theequation of the curve [5 marks]
3 Te gradient of a curve is found to be minus2
(a) Find the x -coordinate of the maximum point justifyingthat it is a maximum
(b) Given that the curve passes through the point (0 2)show that the y -coordinate of the maximum point
is minus7 [5 marks]
4 Te gradient of the normal to a curve at any point isequal to the x -coordinate at that point If the curve passesthrough the point (e2 3) 1047297nd the equation of the curvein the form ( ) where is a rationalfunction x gt 0 [6 marks]
17G Definite integration
Until now we have been carrying out a process known as
inde1047297nite integration inde1047297nite in the sense that we have anunknown constant each time for example
1
However there is also a process called de1047297nite integration which yields a numerical answer without the involvementof the constant of integration for example
bb
3 3 3
3 3 minus
int
Here a and b are known as the limits of integration a is the
lower limit and b the upper limit
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580 Topic 6 Calculus
Te square bracket notation means that the integration hastaken place but the limits have not yet been applied o do thiswe simply evaluate the integrated expression at the upper limitand subtract the integrated expression evaluated at the lowerlimit
You may be wondering where the constant of integration hasgone We could write it in as before but we quickly realise thatthis is unnecessary as it will just cancel out at the upper andlower limit each time
b c
bb
3
3
1
1
1
Te value of x is a dummy variable it does not come intothe answer But both a and b can vary and affect the resultChanging x to a different variable does not change the answerFor example
u a x bb
31 1
int
Worked example 174
Find the exact value of1
1+
Integrate and write in squarebrackets x
+ ]xint
Evaluate the integratedexpression at the upper
and lower limits andsubtract the lower from
the upper
=(In (e) + 4 (e)) minus (In (1) + 4 (1))
= + minusminus =
Ma ke sure you
kno w ho w to e va lua te
de fini te in tegra ls on your
ca lcu la tor as e xp lained
on Ca lcu la tor s ki l ls s hee t
1 0 on t he C D - R OM
I t can sa ve you time
and you can e va lua te
in tegra ls you donrsquo t kno w
ho w to do a lge braica l l y
E ven w hen you are
as ked to find t he e xac t
va lue o f t he in tegra l you
can c hec k your ans wer
on t he ca lcu la tor
e x a m h i n t
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17 Basic integration and its applications 581
Exercise 17G
1 Evaluate the following de1047297nite integrals giving exact answers
(a) (i) x x 2
(ii)1
4
1
(b) (i)2π
int (ii) sπ
π2
(c) (i) ex 1
int (ii) 31
1
eminusint
2 Evaluate correct to three signi1047297cant 1047297gures
(a) (i)3
1 4
int (ii)1
int
(b) (i) e1
int (ii) ne
1int 13 Find the exact value of the integral e x int
π [5 marks]
4 Show that the value of the integral12
x k
k
is independentof k [4 marks]
5 If x ) 73
evaluate 13
[4 marks]
6 Solve the equation 1 [5 marks]
y
x
y = f (x)
a b
A
17H Geometrical significance of definiteintegration
Now we have a method that gives a numerical value for anintegral the natural question to ask is what does this numbermean
On Fill-in proof sheet 20 on the CD-ROM Te fundamentaltheorem of calculus we show that as long as f )x is positivethe de1047297nite integral of f )x between the limits a and b is thearea enclosed between the curve the x -axis and the lines x = a and x = b
KEY POINT 178
rea )x b
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582 Topic 6 Calculus
If you are sketching the graph on the calculator you can get it toshade and evaluate the required area see Calculator skillssheet 10 on the CD-ROM You need to show the sketch as a partof your working if it is not already shown in the question
Worked example 175
Find the exact area enclosed between the x -axis the curve y = sin x and the lines x = 0 and x = π
Sketch the graph and identifythe area required = in
3
Integrate and write in squarebrackets
A = ]= minus
0
3ππ
Evaluate the integratedexpression at the upper andlower limit and subtract the
lower from the upper
= minus )cos cminus minus os
= minus minus ( )minus =
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17 Basic integration and its applications 583
The Ancient Greekshad developed ideasof limiting processessimilar to those used
in calculus but it took nearly2000 years for these ideas
to be formalised This wasdone almost simultaneouslyby Isaac Newton andGottried Leibniz in the 17thCentury Is this a coincidenceor is it often the case that along period of slow progressis often necessary to get tothe stage of majorbreakthroughsSupplementary sheet 10
looks at some other peoplewho can claim to haveinvented calculus
In the 17th Century integration was defined as thearea under a curve The area was broken downinto small rectangles each with a height f (x ) and a
width of a small bit of x called ∆x The total areawas approximately the sum of all of these rectangles
x a
x
f x x sum )∆
Isaac Newton one of the pioneers of calculus was also abig fan of writing in English rather than Greek So sigmabecame the English letter lsquoSrsquo and delta became the Englishletter d so that when the limit is taken as the width of therectangles become vanishingly small then the expressionbecomes
x a
)x dint This illustrates another very important interpretation of
integration ndash the infinite sum of infinitesimally small parts
Worked example 176
Find the area A in this graphy
x
y = x(xminus1)(2minusx)
1 2A
0
Write down the integral to beevaluated then use calculator
x x xminusint 1
x = minus y D
The area must be positive there4 =
When the curve is entirely below the x -axis the integral will givea negative value Te modulus of this value is the area
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584 Topic 6 Calculus
Unfortunately the relationship between integrals and areas isnot so simple when there are parts of the curve above and belowthe axis Tose bits above the axis contribute positively to thearea but bits below the axis contribute negatively to the areaWe must separate out the sections above the axis and belowthe axis
Worked example 177
(a) Find x x 1
4
x int 1 d
(b) Find the area enclosed between the x -axis the curve y minus x + and the lines= 1 and 4
Apply standard integration (a) x x
4
xminus1
= ( ( ) ( ( )minus + minus +
= minus =3
4
The value found above canrsquot bethe correct area for (b)
Sketch the curve to see exactlywhich area we are being asked
to find
(b) y
2 minus + 3
1 3
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17 Basic integration and its applications 585
Te fact that the integral was zero in Worked example 177 part(a) means that the area above the axis is exactly cancelled by thearea below the axis
Tis example warns us that when asked to 1047297nd an area we mustalways sketch the graph and identify exactly where each partof the area is If we are evaluating the area on the calculator wecan use the modulus function to ensure that the entire graph isabove the x -axis Using the function from Worked example 177
1
4
int 1y
x
y = |x2 minus 4x + 3|
1 3 4
continued
The area is made up of twoparts so evaluate each of
them separately
x
1
33
xminusint
( ) minus = minus
there4 rea below the axis is
x x3
4
2xminus int
minus ( ) =
4
there4 rea above the axis is
Total area = + =3 3
Transformationsof graphs using the
modulus function
were covered inchapter 7
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586 Topic 6 Calculus
KEY POINT 179
Te area bounded by the curve y f )x the x-axis and the
lines x = a and x = b is given byb
x int
When working without a calculator if the curve crosses
the x -axis between a and b we need to split the area intoseveral parts and 1047297nd each one separately
Te interpretation of integrals as areas causes one inconsistency
with our previous work Consider the integral1
2
1
x dminus
int
Graphically we can see that this area should exist
x
y
minus1minus2
However if we do the integration we 1047297nd that
12
1
2
1
]minus
minus
minus
minus
minus n minus
minusn
1
= ln
1
minus n
Tis is the correct answer (which we could have found usingthe symmetry of the curve) but it goes through a stage wherewe had to take logarithms of negative numbers and this issomething we are not allowed to do We avoid this by rede1047297ning
the integral ofx
as
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17 Basic integration and its applications 587
KEY POINT 174 AGAIN
minusint =
With this de1047297nition we can integrate y =1
over negative
numbers and the integral above becomes
12
1
2
1
x
e oreminus
minusint
n
Notice that the answer is negative since the required area is below
the x -axis We can still not integrate1
with a negative lower and
positive upper limit since the graph has an asymptote at x = 0
You may rightly be alittle uncomfortablewith inserting amodulus function lsquojust
because it worksrsquo Inmathematics do the ends
justify the means
Exercise 17H
1 Find the shaded areas
(a) (i)y =x2
y
x1 2
(ii)y = 1
x2
y
x2 4
(b) (i) y =x2
minus4x+ 3
y
x1 2
(ii)y =x2
minus4
y
x1
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588 Topic 6 Calculus
(c) (i)y = x3 minus x
y
xminus1 2
(ii)y = x2minus3x
y
x
5
2 Te area enclosed by the x -axis the curve andthe line x = k is 18 Find the value of k [6 marks]
3 (a) Find3
int (b) Find the area between the curve minus2 and the
x -axis between x = 0 and x = 3 [5 marks]
4 Between x = 0 and x = 3 the area of the graph y x minus2 below the x -axis equals the area above thex -axis Find the value of k [6 marks]
5 Find the area enclosed by the curve 1x minus minus 0
and the x -axis [7 marks]
lsquo Find t he area
enc losedrsquo means firs t
find a c losed region
bounded b y t he
cur ves men tioned
t hen find i ts area
A s ke tc h is a ver y
use fu l too l
e x a m h i n t
17I The area between a curve andthe y -axis
Consider the diagram alongside How can we 1047297nd the shadedarea A
One possible strategy is to construct a box around the graph todivide up the regions of interest You can integrate to 1047297nd thearea labelled A1 and then by adding and subtracting the areas ofthe blue and red rectangles shown calculate A
y
x
y = f (x)
c
d
A
y
x
y = f (x)
c
d A1
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17 Basic integration and its applications 589
Happily there is a quicker way we can treat x as a functionof y effectively re1047298ecting the whole diagram in the line y = x and then use the same method as in the previous section
KEY POINT 1710
Te area bounded by the curve y the y -axis and the
lines y = c and y = d is given by y d
) dint where is
the expression for x in terms of y
You may have realised that this is related to inverse functions from Section 5E
x
y
x = f (y)
cd
A
Worked example 178
Te curve shown has equation y 1minus Find the shaded area
y
x
y = 2radic x minus 1
10
Express x in terms of y x = 2
rArr = + y
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590 Topic 6 Calculus
Exercise 17I
continued
Find the limits on the y -axisIt may help to label them on the
graph
When x = minusWhen x = 1 = =minus 6
y
y = 2radic minus
1
10
Write down the integral andevaluate using calculator
Area fro GDC= +2
1 2=d
1 Find the shaded areas
(a) (i) y
x
y = x2
1
6
(ii) y
x
y = x3
1
3
(b) (i) y
xy = 1
x2
2
1
(ii) y
x
y =radic x
1
5
(c) (i)y
x
y = ln x
1e
e
(ii) y
x
y = 3x
1 6
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17 Basic integration and its applications 591
2 Te diagram shows the curve If the shadedarea is 504 1047297nd the value of a [6 marks]
y
x
y = radic
x
a
2a
0
3 Find the exact value of the area enclosed by the graphof + the line y = 2 and the y -axis [6 marks]
4 Te diagram shows the graph of
Te shaded area is 39 units Find the value of a [7 marks]y
x
y =radic x
4 a
5 Te diagram shows the graph of 2 where a isin infin Te area of the pink region is equal to the area of the blue
region Give two equations for a in terms of b and hencegive a in exact form and determine the size of theblue area [8 marks]y
x
y = x2
b
1 a
17J The area between two curves
So far we have only looked at areas bounded by a curve and oneof the coordinate axes but we can also 1047297nd areas bounded bytwo curvesTe area A in the diagram can be found by taking the areabounded by f )x and the x -axis and subtracting the areabounded by and the x -axis that is
A
b
minus
y
x
y = f (x)
y = g(x)
a b
A
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592 Topic 6 Calculus
We can do the subtraction before integrating so that we onlyhave to integrate one expression instead of two Tis gives analternative formula for the area
KEY POINT 1711
Te area A between two curves f (x ) and g (x ) is
Ab
(
where a and b are the x-coordinates of the intersectionpoints of the two curves
Worked example 179
Find the area A enclosed between + and minus2 +
First find the x -coordinates ofintersection
For intersection
x x x
rArr =
rArr ) =rArr =
minus
minus
+x
4minusx 0
Make a rough sketch to see therelative positions of the two curves
y
x1 4
A
2x + 1y =
2 minus 3x + 5y = x
Subtract the lower curve from thehigher before integrating
= x
minusx x
4
minus
minus minusx
2
1
4
2
minus =
8
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17 Basic integration and its applications 593
Subtracting the two equations before integrating is particularlyuseful when one of the curves is partly below the x-axis If x is always above then the expression we are integrating f g )x minus )x is always positive so we do not have to worryabout the signs of f )x and g )x themselves
Worked example 1710
Find the area bounded by the curves y x minus and y x 2
Sketch the graph to see therelative position of two
curves
Using GDC
= ex minus 5
y = 3 2
y
minus2 2
A
Find the intersection points ndash
use calculatorintersections x = minus2 658
Write down the integralrepresenting the area
rea =minus
minus818
1 658
xminus
= minus
1 6 8
x
Evaluate the integral usingcalculator = 21 6 (3SF)
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594 Topic 6 Calculus
Exercise 17J
1 Find the shaded areas
(a) (i)
y = 2x+ 1
y = (xminus1)2
y
x
(ii) y =x+ 1
y = 4xminusx2minus1
y
x
(b) (i)
y =minusx2minus4x+ 12
y
x
y =x2 + 2x+ 12
(ii)
y =x2minus2x+ 9
y
x
y = 4xminusx2 + 5
(c) (i)y =x2 minusx
y = 2xminusx2
y
x
(ii)y =x2minus7x+ 7
y = 3minusxminusx2
y
x
2 Find the area enclosed between the graphs of y x +x minus and + [6 marks]
3 Find the area enclosed by the curve 2
the y-axisand the line [6 marks]
4 Find the area between the curves1
and in theregion lt lt π [6 marks]
5 Show that the area of the shaded region alongside is2
[6 marks]
y = x2
y
xminus1 2
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17 Basic integration and its applications 595
6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]
7 Find the total area enclosed between the graphs of
)minus 2 and minus2 7 1+ [6 marks]
8 Te area enclosed between the curve y x 2 and the line
y mx is 10 Find the value of m if m 0 [7 marks]
9 Show that the shaded area in the diagram below is [8 marks]
y = 2 minus xy2 = x
y
x
y = cos x y = sin x
y
x
Summary
bull Integration is the reverse process of differentiation
bull Any integral without limits (inde1047297nite) will generate a constant of integration
bull For all rational n ne minus1 int +x c
1
1
bull If minus we get the natural logarithm function int minus =
bull Te integral of the exponential function is int e c
bull Te integrals of the trigonometric functions are
int si minus=
int tan =
bull Te de1047297nite integral has limits f( ) x b
dint is found by evaluating the integrated expression
at b and then subtracting the integrated expression evaluated at a
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596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
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17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
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7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
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17 Basic integration and its applications 573
Since we can differentiate term by term (also in Key point 164)then we can also split up integrals of sums
KEY POINT 173
For the sum of integrals
int int +
By combining Key points 172 and 173 with k = minus1 we canalso show that the integral of a difference is the difference of theintegrals of the separate parts
Tese ideas are demonstrated in the following examples
Be warned You
canno t in tegra te
produc ts or quo tien ts
b y in tegra ting eac h
par t separa te l y
e x a m h i n t
Worked example 171
Find (a) int x x (b) int minus
( )+4
minus x d
Add one to the power anddivide by this new power
(a) 6xint minus
minus c x = +xminus
Tidy up
= minus
22
x
= minus +minus2
x
Go through term by termadding one to the powerof x and dividing by this
new powerRemember the rule forintegrating a constant
(b) 3 3 8
4
1x x+x minus +x
minus +minus minus+ +
int c
Tidy up
=minus
+minus
5 1xminus c
= + +
minus 1
3x c
Just as for differentiation it may be necessary to change termsinto the form n before integrating
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574 Topic 6 Calculus
Exercise 17C
1 Find the following integrals
(a) (i) int 9x x (ii) int 12x x
(b) (i) int (ii) int (c) (i) 9 (ii) int
1dx
(d) (i) (ii)
(e) (i) int x x (ii) int 33 x x
(f) (i)2
(ii)3
2 Find the following integrals
(a) (i) int 3 (ii) int 7 z
(b) (i) int q (ii) int r r
In t he in tegra l do no t
forge t t he d x or t he
equi va len t We wi l l
ma ke more use o f i t
la ter T he func tion
you are in tegra ting
is ac tua l l y being
mu l tip lied b y d x
so you cou ld
wri te ques tion
1( f )(ii ) as int2 x
e x a m h i n t
Worked example 172
Find (a) 3
(b) int
( )minusx
2
d
Write the cube root as a powerand use rules of exponents
(a)
int xx
= int x x
Dividing by10
3
7
31+ is the
same as multiplying by3
10
Expand the brackets first thenuse rules of exponents
(b) int +
x
x minus
Dividing by a fraction isthe same as multiplying by
its reciprocal
times10
10
3x c = +10
= minusminus xx 2
= + +5
95 3
minus times c
= ++5 3 1
minus c
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17 Basic integration and its applications 575
(c) (i) 13
(ii) int 57
(d) (i) int 2h (ii) int
d p
p4
3 Find the following integrals (a) (i) int +x minus x 2 d (ii) int x minus x + d
(b) (i) int +1 1
4d (ii) int times minus times
1 15
v d
(c) (i) int x x x d (ii) int 3
(d) (i) int ( 3 (ii) int )2
4 Find int 1
x [4 marks]
17D Integrating x ndash1 and ex
When integrating int +n
1
1 we were careful to exclude
the case n = minus1
In Key point 168 we saw that ( )n Reversing this gives
KEY POINT 174
int minusx c=x
In Key point 167 we saw that x e( We can use this to
integrate the exponential function
KEY POINT 175
e
We will modify
this rule in Section17H
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576 Topic 6 Calculus
Exercise 17D 1 Find the following integrals
(a) (i) (ii)
(b) (i) int 1
x (ii) int
1
x
(c) (i) int minus x 2 1
d (ii) int + x 3
d
(d) (i) int +
2x (ii) int
x x minusx 2
2 Find the following integrals
(a) (i) int ex (ii) int ex x
(b) (i)x
(ii)7 x
(c) (i) int + x
(ii) int +
17E Integrating trigonometric functions
We can expand the set of functions that we can integrate bycontinuing to refer back to work covered in chapter 16
We saw in Key point 166 that (s x which means
that
Similarly as ( then int si minus
KEY POINT 176
Te integrals of trigonometric functions
int s minus s
=
We do not have a function whose derivative is ta and so
have no way (yet) of 1047297nding ta We will meet a method that
enables us to establish this in chapter 19 but for completeness
the result is given here
KEY POINT 176a
tan
See Exercise 19B
for establishing this
result
T he in tegra l o f
tan x is no t gi ven
in t he Formu la
boo k le t and is wor t h
remem bering
e x a m h i n t
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17 Basic integration and its applications 577
Exercise 17E
1 Find the following integrals
(a) (i) int s n x cminus os x d (ii) x s n x d
(b) (i) int t n (ii)
sin
2 3
(c) (i)7
(ii)
(d) (i) int minus x s n x d (ii) cos minus x sminus ncos x
2 Find int s n
cos
x cos
x x d [5 marks]
3 Find int cos x sminus n [5 marks]
17F Finding the equation of a curve
We have seen how we can integrate the function y
to 1047297nd the
equation of the original curve except for the unknown constant
of integration Tis is because the gradientd y
x determines the
shape of the curve but not exactly where it is However if weare also given the coordinates of a point on the curve we can
then determine the constant and hence specify the originalfunction precisely
If we again consider x 2 which we met at the start of this
chapter we know that the original function must have equation y = x 2 + c for some constant value c
If we are also told that the curve passes through the point(1 minus1) we can 1047297nd c and specify which of the family of curvesour function must be
(1minus1)
c = minus6
c = minus4
c = minus2
c = 4
c = 2
c = 0
y
x0
Look back to Worked
example 162 where given the gradient
we could draw many
different curves bychanging the starting
point
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578 Topic 6 Calculus
Te above example illustrates the general procedure for 1047297ndingthe equation of a curve from its gradient function
KEY POINT 177
o 1047297nd the equation for y given the gradientx
and onepoint ( p q) on the curve
1 Integrate remembering +c
2 Find the constant of integration by substitutingx = p y = q
Exercise 17F
1 Find the equation of the original curve if
(a) (i) and the curve passes through (ndash2 7)
(ii) x = 2 and the curve passes through (0 5)
(b) (i)d
d
y
x x
1 and the curve passes through (4 8)
(ii)1
2 and the curve passes through (1 3)
(c) (i) y
= + and the curve passes through (1 1)
(ii)d y
x = e and the curve passes through (ln5 0)
Worked example 173
Te gradient of a curve is given by = + and the curve passes through the point
(1 ndash4) Find the equation of the curve
To find y from ddy x
we need tointegrate
Donrsquot forget + c
= + 5xminus d
= +minus +
The coordinates of the givenpoint must satisfy this
equation so we can find c
When x = = minus so
minus ) + ) += 53 2
c
rArr minus minus + == 4 + 6c rArr
there4 = minusx minus x5+ 6
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17 Basic integration and its applications 579
(d) (i)+
and the curve passes through (e e)
(ii)1
and the curve passes through (e2 5)
(e) (i)d y
x x = x and the curve passesthrough (π 1)
(ii) x 3ta and the curve passes through (0 4)
2 Te derivative of the curve y f )x is1
(a) Find an expression for all possible functions f(x)
(b) If the curve passes through the point (2 7) 1047297nd theequation of the curve [5 marks]
3 Te gradient of a curve is found to be minus2
(a) Find the x -coordinate of the maximum point justifyingthat it is a maximum
(b) Given that the curve passes through the point (0 2)show that the y -coordinate of the maximum point
is minus7 [5 marks]
4 Te gradient of the normal to a curve at any point isequal to the x -coordinate at that point If the curve passesthrough the point (e2 3) 1047297nd the equation of the curvein the form ( ) where is a rationalfunction x gt 0 [6 marks]
17G Definite integration
Until now we have been carrying out a process known as
inde1047297nite integration inde1047297nite in the sense that we have anunknown constant each time for example
1
However there is also a process called de1047297nite integration which yields a numerical answer without the involvementof the constant of integration for example
bb
3 3 3
3 3 minus
int
Here a and b are known as the limits of integration a is the
lower limit and b the upper limit
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580 Topic 6 Calculus
Te square bracket notation means that the integration hastaken place but the limits have not yet been applied o do thiswe simply evaluate the integrated expression at the upper limitand subtract the integrated expression evaluated at the lowerlimit
You may be wondering where the constant of integration hasgone We could write it in as before but we quickly realise thatthis is unnecessary as it will just cancel out at the upper andlower limit each time
b c
bb
3
3
1
1
1
Te value of x is a dummy variable it does not come intothe answer But both a and b can vary and affect the resultChanging x to a different variable does not change the answerFor example
u a x bb
31 1
int
Worked example 174
Find the exact value of1
1+
Integrate and write in squarebrackets x
+ ]xint
Evaluate the integratedexpression at the upper
and lower limits andsubtract the lower from
the upper
=(In (e) + 4 (e)) minus (In (1) + 4 (1))
= + minusminus =
Ma ke sure you
kno w ho w to e va lua te
de fini te in tegra ls on your
ca lcu la tor as e xp lained
on Ca lcu la tor s ki l ls s hee t
1 0 on t he C D - R OM
I t can sa ve you time
and you can e va lua te
in tegra ls you donrsquo t kno w
ho w to do a lge braica l l y
E ven w hen you are
as ked to find t he e xac t
va lue o f t he in tegra l you
can c hec k your ans wer
on t he ca lcu la tor
e x a m h i n t
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17 Basic integration and its applications 581
Exercise 17G
1 Evaluate the following de1047297nite integrals giving exact answers
(a) (i) x x 2
(ii)1
4
1
(b) (i)2π
int (ii) sπ
π2
(c) (i) ex 1
int (ii) 31
1
eminusint
2 Evaluate correct to three signi1047297cant 1047297gures
(a) (i)3
1 4
int (ii)1
int
(b) (i) e1
int (ii) ne
1int 13 Find the exact value of the integral e x int
π [5 marks]
4 Show that the value of the integral12
x k
k
is independentof k [4 marks]
5 If x ) 73
evaluate 13
[4 marks]
6 Solve the equation 1 [5 marks]
y
x
y = f (x)
a b
A
17H Geometrical significance of definiteintegration
Now we have a method that gives a numerical value for anintegral the natural question to ask is what does this numbermean
On Fill-in proof sheet 20 on the CD-ROM Te fundamentaltheorem of calculus we show that as long as f )x is positivethe de1047297nite integral of f )x between the limits a and b is thearea enclosed between the curve the x -axis and the lines x = a and x = b
KEY POINT 178
rea )x b
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582 Topic 6 Calculus
If you are sketching the graph on the calculator you can get it toshade and evaluate the required area see Calculator skillssheet 10 on the CD-ROM You need to show the sketch as a partof your working if it is not already shown in the question
Worked example 175
Find the exact area enclosed between the x -axis the curve y = sin x and the lines x = 0 and x = π
Sketch the graph and identifythe area required = in
3
Integrate and write in squarebrackets
A = ]= minus
0
3ππ
Evaluate the integratedexpression at the upper andlower limit and subtract the
lower from the upper
= minus )cos cminus minus os
= minus minus ( )minus =
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17 Basic integration and its applications 583
The Ancient Greekshad developed ideasof limiting processessimilar to those used
in calculus but it took nearly2000 years for these ideas
to be formalised This wasdone almost simultaneouslyby Isaac Newton andGottried Leibniz in the 17thCentury Is this a coincidenceor is it often the case that along period of slow progressis often necessary to get tothe stage of majorbreakthroughsSupplementary sheet 10
looks at some other peoplewho can claim to haveinvented calculus
In the 17th Century integration was defined as thearea under a curve The area was broken downinto small rectangles each with a height f (x ) and a
width of a small bit of x called ∆x The total areawas approximately the sum of all of these rectangles
x a
x
f x x sum )∆
Isaac Newton one of the pioneers of calculus was also abig fan of writing in English rather than Greek So sigmabecame the English letter lsquoSrsquo and delta became the Englishletter d so that when the limit is taken as the width of therectangles become vanishingly small then the expressionbecomes
x a
)x dint This illustrates another very important interpretation of
integration ndash the infinite sum of infinitesimally small parts
Worked example 176
Find the area A in this graphy
x
y = x(xminus1)(2minusx)
1 2A
0
Write down the integral to beevaluated then use calculator
x x xminusint 1
x = minus y D
The area must be positive there4 =
When the curve is entirely below the x -axis the integral will givea negative value Te modulus of this value is the area
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584 Topic 6 Calculus
Unfortunately the relationship between integrals and areas isnot so simple when there are parts of the curve above and belowthe axis Tose bits above the axis contribute positively to thearea but bits below the axis contribute negatively to the areaWe must separate out the sections above the axis and belowthe axis
Worked example 177
(a) Find x x 1
4
x int 1 d
(b) Find the area enclosed between the x -axis the curve y minus x + and the lines= 1 and 4
Apply standard integration (a) x x
4
xminus1
= ( ( ) ( ( )minus + minus +
= minus =3
4
The value found above canrsquot bethe correct area for (b)
Sketch the curve to see exactlywhich area we are being asked
to find
(b) y
2 minus + 3
1 3
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17 Basic integration and its applications 585
Te fact that the integral was zero in Worked example 177 part(a) means that the area above the axis is exactly cancelled by thearea below the axis
Tis example warns us that when asked to 1047297nd an area we mustalways sketch the graph and identify exactly where each partof the area is If we are evaluating the area on the calculator wecan use the modulus function to ensure that the entire graph isabove the x -axis Using the function from Worked example 177
1
4
int 1y
x
y = |x2 minus 4x + 3|
1 3 4
continued
The area is made up of twoparts so evaluate each of
them separately
x
1
33
xminusint
( ) minus = minus
there4 rea below the axis is
x x3
4
2xminus int
minus ( ) =
4
there4 rea above the axis is
Total area = + =3 3
Transformationsof graphs using the
modulus function
were covered inchapter 7
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586 Topic 6 Calculus
KEY POINT 179
Te area bounded by the curve y f )x the x-axis and the
lines x = a and x = b is given byb
x int
When working without a calculator if the curve crosses
the x -axis between a and b we need to split the area intoseveral parts and 1047297nd each one separately
Te interpretation of integrals as areas causes one inconsistency
with our previous work Consider the integral1
2
1
x dminus
int
Graphically we can see that this area should exist
x
y
minus1minus2
However if we do the integration we 1047297nd that
12
1
2
1
]minus
minus
minus
minus
minus n minus
minusn
1
= ln
1
minus n
Tis is the correct answer (which we could have found usingthe symmetry of the curve) but it goes through a stage wherewe had to take logarithms of negative numbers and this issomething we are not allowed to do We avoid this by rede1047297ning
the integral ofx
as
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17 Basic integration and its applications 587
KEY POINT 174 AGAIN
minusint =
With this de1047297nition we can integrate y =1
over negative
numbers and the integral above becomes
12
1
2
1
x
e oreminus
minusint
n
Notice that the answer is negative since the required area is below
the x -axis We can still not integrate1
with a negative lower and
positive upper limit since the graph has an asymptote at x = 0
You may rightly be alittle uncomfortablewith inserting amodulus function lsquojust
because it worksrsquo Inmathematics do the ends
justify the means
Exercise 17H
1 Find the shaded areas
(a) (i)y =x2
y
x1 2
(ii)y = 1
x2
y
x2 4
(b) (i) y =x2
minus4x+ 3
y
x1 2
(ii)y =x2
minus4
y
x1
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588 Topic 6 Calculus
(c) (i)y = x3 minus x
y
xminus1 2
(ii)y = x2minus3x
y
x
5
2 Te area enclosed by the x -axis the curve andthe line x = k is 18 Find the value of k [6 marks]
3 (a) Find3
int (b) Find the area between the curve minus2 and the
x -axis between x = 0 and x = 3 [5 marks]
4 Between x = 0 and x = 3 the area of the graph y x minus2 below the x -axis equals the area above thex -axis Find the value of k [6 marks]
5 Find the area enclosed by the curve 1x minus minus 0
and the x -axis [7 marks]
lsquo Find t he area
enc losedrsquo means firs t
find a c losed region
bounded b y t he
cur ves men tioned
t hen find i ts area
A s ke tc h is a ver y
use fu l too l
e x a m h i n t
17I The area between a curve andthe y -axis
Consider the diagram alongside How can we 1047297nd the shadedarea A
One possible strategy is to construct a box around the graph todivide up the regions of interest You can integrate to 1047297nd thearea labelled A1 and then by adding and subtracting the areas ofthe blue and red rectangles shown calculate A
y
x
y = f (x)
c
d
A
y
x
y = f (x)
c
d A1
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17 Basic integration and its applications 589
Happily there is a quicker way we can treat x as a functionof y effectively re1047298ecting the whole diagram in the line y = x and then use the same method as in the previous section
KEY POINT 1710
Te area bounded by the curve y the y -axis and the
lines y = c and y = d is given by y d
) dint where is
the expression for x in terms of y
You may have realised that this is related to inverse functions from Section 5E
x
y
x = f (y)
cd
A
Worked example 178
Te curve shown has equation y 1minus Find the shaded area
y
x
y = 2radic x minus 1
10
Express x in terms of y x = 2
rArr = + y
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590 Topic 6 Calculus
Exercise 17I
continued
Find the limits on the y -axisIt may help to label them on the
graph
When x = minusWhen x = 1 = =minus 6
y
y = 2radic minus
1
10
Write down the integral andevaluate using calculator
Area fro GDC= +2
1 2=d
1 Find the shaded areas
(a) (i) y
x
y = x2
1
6
(ii) y
x
y = x3
1
3
(b) (i) y
xy = 1
x2
2
1
(ii) y
x
y =radic x
1
5
(c) (i)y
x
y = ln x
1e
e
(ii) y
x
y = 3x
1 6
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17 Basic integration and its applications 591
2 Te diagram shows the curve If the shadedarea is 504 1047297nd the value of a [6 marks]
y
x
y = radic
x
a
2a
0
3 Find the exact value of the area enclosed by the graphof + the line y = 2 and the y -axis [6 marks]
4 Te diagram shows the graph of
Te shaded area is 39 units Find the value of a [7 marks]y
x
y =radic x
4 a
5 Te diagram shows the graph of 2 where a isin infin Te area of the pink region is equal to the area of the blue
region Give two equations for a in terms of b and hencegive a in exact form and determine the size of theblue area [8 marks]y
x
y = x2
b
1 a
17J The area between two curves
So far we have only looked at areas bounded by a curve and oneof the coordinate axes but we can also 1047297nd areas bounded bytwo curvesTe area A in the diagram can be found by taking the areabounded by f )x and the x -axis and subtracting the areabounded by and the x -axis that is
A
b
minus
y
x
y = f (x)
y = g(x)
a b
A
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592 Topic 6 Calculus
We can do the subtraction before integrating so that we onlyhave to integrate one expression instead of two Tis gives analternative formula for the area
KEY POINT 1711
Te area A between two curves f (x ) and g (x ) is
Ab
(
where a and b are the x-coordinates of the intersectionpoints of the two curves
Worked example 179
Find the area A enclosed between + and minus2 +
First find the x -coordinates ofintersection
For intersection
x x x
rArr =
rArr ) =rArr =
minus
minus
+x
4minusx 0
Make a rough sketch to see therelative positions of the two curves
y
x1 4
A
2x + 1y =
2 minus 3x + 5y = x
Subtract the lower curve from thehigher before integrating
= x
minusx x
4
minus
minus minusx
2
1
4
2
minus =
8
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17 Basic integration and its applications 593
Subtracting the two equations before integrating is particularlyuseful when one of the curves is partly below the x-axis If x is always above then the expression we are integrating f g )x minus )x is always positive so we do not have to worryabout the signs of f )x and g )x themselves
Worked example 1710
Find the area bounded by the curves y x minus and y x 2
Sketch the graph to see therelative position of two
curves
Using GDC
= ex minus 5
y = 3 2
y
minus2 2
A
Find the intersection points ndash
use calculatorintersections x = minus2 658
Write down the integralrepresenting the area
rea =minus
minus818
1 658
xminus
= minus
1 6 8
x
Evaluate the integral usingcalculator = 21 6 (3SF)
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594 Topic 6 Calculus
Exercise 17J
1 Find the shaded areas
(a) (i)
y = 2x+ 1
y = (xminus1)2
y
x
(ii) y =x+ 1
y = 4xminusx2minus1
y
x
(b) (i)
y =minusx2minus4x+ 12
y
x
y =x2 + 2x+ 12
(ii)
y =x2minus2x+ 9
y
x
y = 4xminusx2 + 5
(c) (i)y =x2 minusx
y = 2xminusx2
y
x
(ii)y =x2minus7x+ 7
y = 3minusxminusx2
y
x
2 Find the area enclosed between the graphs of y x +x minus and + [6 marks]
3 Find the area enclosed by the curve 2
the y-axisand the line [6 marks]
4 Find the area between the curves1
and in theregion lt lt π [6 marks]
5 Show that the area of the shaded region alongside is2
[6 marks]
y = x2
y
xminus1 2
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17 Basic integration and its applications 595
6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]
7 Find the total area enclosed between the graphs of
)minus 2 and minus2 7 1+ [6 marks]
8 Te area enclosed between the curve y x 2 and the line
y mx is 10 Find the value of m if m 0 [7 marks]
9 Show that the shaded area in the diagram below is [8 marks]
y = 2 minus xy2 = x
y
x
y = cos x y = sin x
y
x
Summary
bull Integration is the reverse process of differentiation
bull Any integral without limits (inde1047297nite) will generate a constant of integration
bull For all rational n ne minus1 int +x c
1
1
bull If minus we get the natural logarithm function int minus =
bull Te integral of the exponential function is int e c
bull Te integrals of the trigonometric functions are
int si minus=
int tan =
bull Te de1047297nite integral has limits f( ) x b
dint is found by evaluating the integrated expression
at b and then subtracting the integrated expression evaluated at a
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596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
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17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
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7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
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574 Topic 6 Calculus
Exercise 17C
1 Find the following integrals
(a) (i) int 9x x (ii) int 12x x
(b) (i) int (ii) int (c) (i) 9 (ii) int
1dx
(d) (i) (ii)
(e) (i) int x x (ii) int 33 x x
(f) (i)2
(ii)3
2 Find the following integrals
(a) (i) int 3 (ii) int 7 z
(b) (i) int q (ii) int r r
In t he in tegra l do no t
forge t t he d x or t he
equi va len t We wi l l
ma ke more use o f i t
la ter T he func tion
you are in tegra ting
is ac tua l l y being
mu l tip lied b y d x
so you cou ld
wri te ques tion
1( f )(ii ) as int2 x
e x a m h i n t
Worked example 172
Find (a) 3
(b) int
( )minusx
2
d
Write the cube root as a powerand use rules of exponents
(a)
int xx
= int x x
Dividing by10
3
7
31+ is the
same as multiplying by3
10
Expand the brackets first thenuse rules of exponents
(b) int +
x
x minus
Dividing by a fraction isthe same as multiplying by
its reciprocal
times10
10
3x c = +10
= minusminus xx 2
= + +5
95 3
minus times c
= ++5 3 1
minus c
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17 Basic integration and its applications 575
(c) (i) 13
(ii) int 57
(d) (i) int 2h (ii) int
d p
p4
3 Find the following integrals (a) (i) int +x minus x 2 d (ii) int x minus x + d
(b) (i) int +1 1
4d (ii) int times minus times
1 15
v d
(c) (i) int x x x d (ii) int 3
(d) (i) int ( 3 (ii) int )2
4 Find int 1
x [4 marks]
17D Integrating x ndash1 and ex
When integrating int +n
1
1 we were careful to exclude
the case n = minus1
In Key point 168 we saw that ( )n Reversing this gives
KEY POINT 174
int minusx c=x
In Key point 167 we saw that x e( We can use this to
integrate the exponential function
KEY POINT 175
e
We will modify
this rule in Section17H
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576 Topic 6 Calculus
Exercise 17D 1 Find the following integrals
(a) (i) (ii)
(b) (i) int 1
x (ii) int
1
x
(c) (i) int minus x 2 1
d (ii) int + x 3
d
(d) (i) int +
2x (ii) int
x x minusx 2
2 Find the following integrals
(a) (i) int ex (ii) int ex x
(b) (i)x
(ii)7 x
(c) (i) int + x
(ii) int +
17E Integrating trigonometric functions
We can expand the set of functions that we can integrate bycontinuing to refer back to work covered in chapter 16
We saw in Key point 166 that (s x which means
that
Similarly as ( then int si minus
KEY POINT 176
Te integrals of trigonometric functions
int s minus s
=
We do not have a function whose derivative is ta and so
have no way (yet) of 1047297nding ta We will meet a method that
enables us to establish this in chapter 19 but for completeness
the result is given here
KEY POINT 176a
tan
See Exercise 19B
for establishing this
result
T he in tegra l o f
tan x is no t gi ven
in t he Formu la
boo k le t and is wor t h
remem bering
e x a m h i n t
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17 Basic integration and its applications 577
Exercise 17E
1 Find the following integrals
(a) (i) int s n x cminus os x d (ii) x s n x d
(b) (i) int t n (ii)
sin
2 3
(c) (i)7
(ii)
(d) (i) int minus x s n x d (ii) cos minus x sminus ncos x
2 Find int s n
cos
x cos
x x d [5 marks]
3 Find int cos x sminus n [5 marks]
17F Finding the equation of a curve
We have seen how we can integrate the function y
to 1047297nd the
equation of the original curve except for the unknown constant
of integration Tis is because the gradientd y
x determines the
shape of the curve but not exactly where it is However if weare also given the coordinates of a point on the curve we can
then determine the constant and hence specify the originalfunction precisely
If we again consider x 2 which we met at the start of this
chapter we know that the original function must have equation y = x 2 + c for some constant value c
If we are also told that the curve passes through the point(1 minus1) we can 1047297nd c and specify which of the family of curvesour function must be
(1minus1)
c = minus6
c = minus4
c = minus2
c = 4
c = 2
c = 0
y
x0
Look back to Worked
example 162 where given the gradient
we could draw many
different curves bychanging the starting
point
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578 Topic 6 Calculus
Te above example illustrates the general procedure for 1047297ndingthe equation of a curve from its gradient function
KEY POINT 177
o 1047297nd the equation for y given the gradientx
and onepoint ( p q) on the curve
1 Integrate remembering +c
2 Find the constant of integration by substitutingx = p y = q
Exercise 17F
1 Find the equation of the original curve if
(a) (i) and the curve passes through (ndash2 7)
(ii) x = 2 and the curve passes through (0 5)
(b) (i)d
d
y
x x
1 and the curve passes through (4 8)
(ii)1
2 and the curve passes through (1 3)
(c) (i) y
= + and the curve passes through (1 1)
(ii)d y
x = e and the curve passes through (ln5 0)
Worked example 173
Te gradient of a curve is given by = + and the curve passes through the point
(1 ndash4) Find the equation of the curve
To find y from ddy x
we need tointegrate
Donrsquot forget + c
= + 5xminus d
= +minus +
The coordinates of the givenpoint must satisfy this
equation so we can find c
When x = = minus so
minus ) + ) += 53 2
c
rArr minus minus + == 4 + 6c rArr
there4 = minusx minus x5+ 6
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17 Basic integration and its applications 579
(d) (i)+
and the curve passes through (e e)
(ii)1
and the curve passes through (e2 5)
(e) (i)d y
x x = x and the curve passesthrough (π 1)
(ii) x 3ta and the curve passes through (0 4)
2 Te derivative of the curve y f )x is1
(a) Find an expression for all possible functions f(x)
(b) If the curve passes through the point (2 7) 1047297nd theequation of the curve [5 marks]
3 Te gradient of a curve is found to be minus2
(a) Find the x -coordinate of the maximum point justifyingthat it is a maximum
(b) Given that the curve passes through the point (0 2)show that the y -coordinate of the maximum point
is minus7 [5 marks]
4 Te gradient of the normal to a curve at any point isequal to the x -coordinate at that point If the curve passesthrough the point (e2 3) 1047297nd the equation of the curvein the form ( ) where is a rationalfunction x gt 0 [6 marks]
17G Definite integration
Until now we have been carrying out a process known as
inde1047297nite integration inde1047297nite in the sense that we have anunknown constant each time for example
1
However there is also a process called de1047297nite integration which yields a numerical answer without the involvementof the constant of integration for example
bb
3 3 3
3 3 minus
int
Here a and b are known as the limits of integration a is the
lower limit and b the upper limit
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580 Topic 6 Calculus
Te square bracket notation means that the integration hastaken place but the limits have not yet been applied o do thiswe simply evaluate the integrated expression at the upper limitand subtract the integrated expression evaluated at the lowerlimit
You may be wondering where the constant of integration hasgone We could write it in as before but we quickly realise thatthis is unnecessary as it will just cancel out at the upper andlower limit each time
b c
bb
3
3
1
1
1
Te value of x is a dummy variable it does not come intothe answer But both a and b can vary and affect the resultChanging x to a different variable does not change the answerFor example
u a x bb
31 1
int
Worked example 174
Find the exact value of1
1+
Integrate and write in squarebrackets x
+ ]xint
Evaluate the integratedexpression at the upper
and lower limits andsubtract the lower from
the upper
=(In (e) + 4 (e)) minus (In (1) + 4 (1))
= + minusminus =
Ma ke sure you
kno w ho w to e va lua te
de fini te in tegra ls on your
ca lcu la tor as e xp lained
on Ca lcu la tor s ki l ls s hee t
1 0 on t he C D - R OM
I t can sa ve you time
and you can e va lua te
in tegra ls you donrsquo t kno w
ho w to do a lge braica l l y
E ven w hen you are
as ked to find t he e xac t
va lue o f t he in tegra l you
can c hec k your ans wer
on t he ca lcu la tor
e x a m h i n t
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17 Basic integration and its applications 581
Exercise 17G
1 Evaluate the following de1047297nite integrals giving exact answers
(a) (i) x x 2
(ii)1
4
1
(b) (i)2π
int (ii) sπ
π2
(c) (i) ex 1
int (ii) 31
1
eminusint
2 Evaluate correct to three signi1047297cant 1047297gures
(a) (i)3
1 4
int (ii)1
int
(b) (i) e1
int (ii) ne
1int 13 Find the exact value of the integral e x int
π [5 marks]
4 Show that the value of the integral12
x k
k
is independentof k [4 marks]
5 If x ) 73
evaluate 13
[4 marks]
6 Solve the equation 1 [5 marks]
y
x
y = f (x)
a b
A
17H Geometrical significance of definiteintegration
Now we have a method that gives a numerical value for anintegral the natural question to ask is what does this numbermean
On Fill-in proof sheet 20 on the CD-ROM Te fundamentaltheorem of calculus we show that as long as f )x is positivethe de1047297nite integral of f )x between the limits a and b is thearea enclosed between the curve the x -axis and the lines x = a and x = b
KEY POINT 178
rea )x b
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582 Topic 6 Calculus
If you are sketching the graph on the calculator you can get it toshade and evaluate the required area see Calculator skillssheet 10 on the CD-ROM You need to show the sketch as a partof your working if it is not already shown in the question
Worked example 175
Find the exact area enclosed between the x -axis the curve y = sin x and the lines x = 0 and x = π
Sketch the graph and identifythe area required = in
3
Integrate and write in squarebrackets
A = ]= minus
0
3ππ
Evaluate the integratedexpression at the upper andlower limit and subtract the
lower from the upper
= minus )cos cminus minus os
= minus minus ( )minus =
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17 Basic integration and its applications 583
The Ancient Greekshad developed ideasof limiting processessimilar to those used
in calculus but it took nearly2000 years for these ideas
to be formalised This wasdone almost simultaneouslyby Isaac Newton andGottried Leibniz in the 17thCentury Is this a coincidenceor is it often the case that along period of slow progressis often necessary to get tothe stage of majorbreakthroughsSupplementary sheet 10
looks at some other peoplewho can claim to haveinvented calculus
In the 17th Century integration was defined as thearea under a curve The area was broken downinto small rectangles each with a height f (x ) and a
width of a small bit of x called ∆x The total areawas approximately the sum of all of these rectangles
x a
x
f x x sum )∆
Isaac Newton one of the pioneers of calculus was also abig fan of writing in English rather than Greek So sigmabecame the English letter lsquoSrsquo and delta became the Englishletter d so that when the limit is taken as the width of therectangles become vanishingly small then the expressionbecomes
x a
)x dint This illustrates another very important interpretation of
integration ndash the infinite sum of infinitesimally small parts
Worked example 176
Find the area A in this graphy
x
y = x(xminus1)(2minusx)
1 2A
0
Write down the integral to beevaluated then use calculator
x x xminusint 1
x = minus y D
The area must be positive there4 =
When the curve is entirely below the x -axis the integral will givea negative value Te modulus of this value is the area
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584 Topic 6 Calculus
Unfortunately the relationship between integrals and areas isnot so simple when there are parts of the curve above and belowthe axis Tose bits above the axis contribute positively to thearea but bits below the axis contribute negatively to the areaWe must separate out the sections above the axis and belowthe axis
Worked example 177
(a) Find x x 1
4
x int 1 d
(b) Find the area enclosed between the x -axis the curve y minus x + and the lines= 1 and 4
Apply standard integration (a) x x
4
xminus1
= ( ( ) ( ( )minus + minus +
= minus =3
4
The value found above canrsquot bethe correct area for (b)
Sketch the curve to see exactlywhich area we are being asked
to find
(b) y
2 minus + 3
1 3
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17 Basic integration and its applications 585
Te fact that the integral was zero in Worked example 177 part(a) means that the area above the axis is exactly cancelled by thearea below the axis
Tis example warns us that when asked to 1047297nd an area we mustalways sketch the graph and identify exactly where each partof the area is If we are evaluating the area on the calculator wecan use the modulus function to ensure that the entire graph isabove the x -axis Using the function from Worked example 177
1
4
int 1y
x
y = |x2 minus 4x + 3|
1 3 4
continued
The area is made up of twoparts so evaluate each of
them separately
x
1
33
xminusint
( ) minus = minus
there4 rea below the axis is
x x3
4
2xminus int
minus ( ) =
4
there4 rea above the axis is
Total area = + =3 3
Transformationsof graphs using the
modulus function
were covered inchapter 7
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586 Topic 6 Calculus
KEY POINT 179
Te area bounded by the curve y f )x the x-axis and the
lines x = a and x = b is given byb
x int
When working without a calculator if the curve crosses
the x -axis between a and b we need to split the area intoseveral parts and 1047297nd each one separately
Te interpretation of integrals as areas causes one inconsistency
with our previous work Consider the integral1
2
1
x dminus
int
Graphically we can see that this area should exist
x
y
minus1minus2
However if we do the integration we 1047297nd that
12
1
2
1
]minus
minus
minus
minus
minus n minus
minusn
1
= ln
1
minus n
Tis is the correct answer (which we could have found usingthe symmetry of the curve) but it goes through a stage wherewe had to take logarithms of negative numbers and this issomething we are not allowed to do We avoid this by rede1047297ning
the integral ofx
as
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17 Basic integration and its applications 587
KEY POINT 174 AGAIN
minusint =
With this de1047297nition we can integrate y =1
over negative
numbers and the integral above becomes
12
1
2
1
x
e oreminus
minusint
n
Notice that the answer is negative since the required area is below
the x -axis We can still not integrate1
with a negative lower and
positive upper limit since the graph has an asymptote at x = 0
You may rightly be alittle uncomfortablewith inserting amodulus function lsquojust
because it worksrsquo Inmathematics do the ends
justify the means
Exercise 17H
1 Find the shaded areas
(a) (i)y =x2
y
x1 2
(ii)y = 1
x2
y
x2 4
(b) (i) y =x2
minus4x+ 3
y
x1 2
(ii)y =x2
minus4
y
x1
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588 Topic 6 Calculus
(c) (i)y = x3 minus x
y
xminus1 2
(ii)y = x2minus3x
y
x
5
2 Te area enclosed by the x -axis the curve andthe line x = k is 18 Find the value of k [6 marks]
3 (a) Find3
int (b) Find the area between the curve minus2 and the
x -axis between x = 0 and x = 3 [5 marks]
4 Between x = 0 and x = 3 the area of the graph y x minus2 below the x -axis equals the area above thex -axis Find the value of k [6 marks]
5 Find the area enclosed by the curve 1x minus minus 0
and the x -axis [7 marks]
lsquo Find t he area
enc losedrsquo means firs t
find a c losed region
bounded b y t he
cur ves men tioned
t hen find i ts area
A s ke tc h is a ver y
use fu l too l
e x a m h i n t
17I The area between a curve andthe y -axis
Consider the diagram alongside How can we 1047297nd the shadedarea A
One possible strategy is to construct a box around the graph todivide up the regions of interest You can integrate to 1047297nd thearea labelled A1 and then by adding and subtracting the areas ofthe blue and red rectangles shown calculate A
y
x
y = f (x)
c
d
A
y
x
y = f (x)
c
d A1
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17 Basic integration and its applications 589
Happily there is a quicker way we can treat x as a functionof y effectively re1047298ecting the whole diagram in the line y = x and then use the same method as in the previous section
KEY POINT 1710
Te area bounded by the curve y the y -axis and the
lines y = c and y = d is given by y d
) dint where is
the expression for x in terms of y
You may have realised that this is related to inverse functions from Section 5E
x
y
x = f (y)
cd
A
Worked example 178
Te curve shown has equation y 1minus Find the shaded area
y
x
y = 2radic x minus 1
10
Express x in terms of y x = 2
rArr = + y
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590 Topic 6 Calculus
Exercise 17I
continued
Find the limits on the y -axisIt may help to label them on the
graph
When x = minusWhen x = 1 = =minus 6
y
y = 2radic minus
1
10
Write down the integral andevaluate using calculator
Area fro GDC= +2
1 2=d
1 Find the shaded areas
(a) (i) y
x
y = x2
1
6
(ii) y
x
y = x3
1
3
(b) (i) y
xy = 1
x2
2
1
(ii) y
x
y =radic x
1
5
(c) (i)y
x
y = ln x
1e
e
(ii) y
x
y = 3x
1 6
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17 Basic integration and its applications 591
2 Te diagram shows the curve If the shadedarea is 504 1047297nd the value of a [6 marks]
y
x
y = radic
x
a
2a
0
3 Find the exact value of the area enclosed by the graphof + the line y = 2 and the y -axis [6 marks]
4 Te diagram shows the graph of
Te shaded area is 39 units Find the value of a [7 marks]y
x
y =radic x
4 a
5 Te diagram shows the graph of 2 where a isin infin Te area of the pink region is equal to the area of the blue
region Give two equations for a in terms of b and hencegive a in exact form and determine the size of theblue area [8 marks]y
x
y = x2
b
1 a
17J The area between two curves
So far we have only looked at areas bounded by a curve and oneof the coordinate axes but we can also 1047297nd areas bounded bytwo curvesTe area A in the diagram can be found by taking the areabounded by f )x and the x -axis and subtracting the areabounded by and the x -axis that is
A
b
minus
y
x
y = f (x)
y = g(x)
a b
A
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592 Topic 6 Calculus
We can do the subtraction before integrating so that we onlyhave to integrate one expression instead of two Tis gives analternative formula for the area
KEY POINT 1711
Te area A between two curves f (x ) and g (x ) is
Ab
(
where a and b are the x-coordinates of the intersectionpoints of the two curves
Worked example 179
Find the area A enclosed between + and minus2 +
First find the x -coordinates ofintersection
For intersection
x x x
rArr =
rArr ) =rArr =
minus
minus
+x
4minusx 0
Make a rough sketch to see therelative positions of the two curves
y
x1 4
A
2x + 1y =
2 minus 3x + 5y = x
Subtract the lower curve from thehigher before integrating
= x
minusx x
4
minus
minus minusx
2
1
4
2
minus =
8
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17 Basic integration and its applications 593
Subtracting the two equations before integrating is particularlyuseful when one of the curves is partly below the x-axis If x is always above then the expression we are integrating f g )x minus )x is always positive so we do not have to worryabout the signs of f )x and g )x themselves
Worked example 1710
Find the area bounded by the curves y x minus and y x 2
Sketch the graph to see therelative position of two
curves
Using GDC
= ex minus 5
y = 3 2
y
minus2 2
A
Find the intersection points ndash
use calculatorintersections x = minus2 658
Write down the integralrepresenting the area
rea =minus
minus818
1 658
xminus
= minus
1 6 8
x
Evaluate the integral usingcalculator = 21 6 (3SF)
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594 Topic 6 Calculus
Exercise 17J
1 Find the shaded areas
(a) (i)
y = 2x+ 1
y = (xminus1)2
y
x
(ii) y =x+ 1
y = 4xminusx2minus1
y
x
(b) (i)
y =minusx2minus4x+ 12
y
x
y =x2 + 2x+ 12
(ii)
y =x2minus2x+ 9
y
x
y = 4xminusx2 + 5
(c) (i)y =x2 minusx
y = 2xminusx2
y
x
(ii)y =x2minus7x+ 7
y = 3minusxminusx2
y
x
2 Find the area enclosed between the graphs of y x +x minus and + [6 marks]
3 Find the area enclosed by the curve 2
the y-axisand the line [6 marks]
4 Find the area between the curves1
and in theregion lt lt π [6 marks]
5 Show that the area of the shaded region alongside is2
[6 marks]
y = x2
y
xminus1 2
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17 Basic integration and its applications 595
6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]
7 Find the total area enclosed between the graphs of
)minus 2 and minus2 7 1+ [6 marks]
8 Te area enclosed between the curve y x 2 and the line
y mx is 10 Find the value of m if m 0 [7 marks]
9 Show that the shaded area in the diagram below is [8 marks]
y = 2 minus xy2 = x
y
x
y = cos x y = sin x
y
x
Summary
bull Integration is the reverse process of differentiation
bull Any integral without limits (inde1047297nite) will generate a constant of integration
bull For all rational n ne minus1 int +x c
1
1
bull If minus we get the natural logarithm function int minus =
bull Te integral of the exponential function is int e c
bull Te integrals of the trigonometric functions are
int si minus=
int tan =
bull Te de1047297nite integral has limits f( ) x b
dint is found by evaluating the integrated expression
at b and then subtracting the integrated expression evaluated at a
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596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
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17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
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7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
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17 Basic integration and its applications 575
(c) (i) 13
(ii) int 57
(d) (i) int 2h (ii) int
d p
p4
3 Find the following integrals (a) (i) int +x minus x 2 d (ii) int x minus x + d
(b) (i) int +1 1
4d (ii) int times minus times
1 15
v d
(c) (i) int x x x d (ii) int 3
(d) (i) int ( 3 (ii) int )2
4 Find int 1
x [4 marks]
17D Integrating x ndash1 and ex
When integrating int +n
1
1 we were careful to exclude
the case n = minus1
In Key point 168 we saw that ( )n Reversing this gives
KEY POINT 174
int minusx c=x
In Key point 167 we saw that x e( We can use this to
integrate the exponential function
KEY POINT 175
e
We will modify
this rule in Section17H
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576 Topic 6 Calculus
Exercise 17D 1 Find the following integrals
(a) (i) (ii)
(b) (i) int 1
x (ii) int
1
x
(c) (i) int minus x 2 1
d (ii) int + x 3
d
(d) (i) int +
2x (ii) int
x x minusx 2
2 Find the following integrals
(a) (i) int ex (ii) int ex x
(b) (i)x
(ii)7 x
(c) (i) int + x
(ii) int +
17E Integrating trigonometric functions
We can expand the set of functions that we can integrate bycontinuing to refer back to work covered in chapter 16
We saw in Key point 166 that (s x which means
that
Similarly as ( then int si minus
KEY POINT 176
Te integrals of trigonometric functions
int s minus s
=
We do not have a function whose derivative is ta and so
have no way (yet) of 1047297nding ta We will meet a method that
enables us to establish this in chapter 19 but for completeness
the result is given here
KEY POINT 176a
tan
See Exercise 19B
for establishing this
result
T he in tegra l o f
tan x is no t gi ven
in t he Formu la
boo k le t and is wor t h
remem bering
e x a m h i n t
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17 Basic integration and its applications 577
Exercise 17E
1 Find the following integrals
(a) (i) int s n x cminus os x d (ii) x s n x d
(b) (i) int t n (ii)
sin
2 3
(c) (i)7
(ii)
(d) (i) int minus x s n x d (ii) cos minus x sminus ncos x
2 Find int s n
cos
x cos
x x d [5 marks]
3 Find int cos x sminus n [5 marks]
17F Finding the equation of a curve
We have seen how we can integrate the function y
to 1047297nd the
equation of the original curve except for the unknown constant
of integration Tis is because the gradientd y
x determines the
shape of the curve but not exactly where it is However if weare also given the coordinates of a point on the curve we can
then determine the constant and hence specify the originalfunction precisely
If we again consider x 2 which we met at the start of this
chapter we know that the original function must have equation y = x 2 + c for some constant value c
If we are also told that the curve passes through the point(1 minus1) we can 1047297nd c and specify which of the family of curvesour function must be
(1minus1)
c = minus6
c = minus4
c = minus2
c = 4
c = 2
c = 0
y
x0
Look back to Worked
example 162 where given the gradient
we could draw many
different curves bychanging the starting
point
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578 Topic 6 Calculus
Te above example illustrates the general procedure for 1047297ndingthe equation of a curve from its gradient function
KEY POINT 177
o 1047297nd the equation for y given the gradientx
and onepoint ( p q) on the curve
1 Integrate remembering +c
2 Find the constant of integration by substitutingx = p y = q
Exercise 17F
1 Find the equation of the original curve if
(a) (i) and the curve passes through (ndash2 7)
(ii) x = 2 and the curve passes through (0 5)
(b) (i)d
d
y
x x
1 and the curve passes through (4 8)
(ii)1
2 and the curve passes through (1 3)
(c) (i) y
= + and the curve passes through (1 1)
(ii)d y
x = e and the curve passes through (ln5 0)
Worked example 173
Te gradient of a curve is given by = + and the curve passes through the point
(1 ndash4) Find the equation of the curve
To find y from ddy x
we need tointegrate
Donrsquot forget + c
= + 5xminus d
= +minus +
The coordinates of the givenpoint must satisfy this
equation so we can find c
When x = = minus so
minus ) + ) += 53 2
c
rArr minus minus + == 4 + 6c rArr
there4 = minusx minus x5+ 6
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17 Basic integration and its applications 579
(d) (i)+
and the curve passes through (e e)
(ii)1
and the curve passes through (e2 5)
(e) (i)d y
x x = x and the curve passesthrough (π 1)
(ii) x 3ta and the curve passes through (0 4)
2 Te derivative of the curve y f )x is1
(a) Find an expression for all possible functions f(x)
(b) If the curve passes through the point (2 7) 1047297nd theequation of the curve [5 marks]
3 Te gradient of a curve is found to be minus2
(a) Find the x -coordinate of the maximum point justifyingthat it is a maximum
(b) Given that the curve passes through the point (0 2)show that the y -coordinate of the maximum point
is minus7 [5 marks]
4 Te gradient of the normal to a curve at any point isequal to the x -coordinate at that point If the curve passesthrough the point (e2 3) 1047297nd the equation of the curvein the form ( ) where is a rationalfunction x gt 0 [6 marks]
17G Definite integration
Until now we have been carrying out a process known as
inde1047297nite integration inde1047297nite in the sense that we have anunknown constant each time for example
1
However there is also a process called de1047297nite integration which yields a numerical answer without the involvementof the constant of integration for example
bb
3 3 3
3 3 minus
int
Here a and b are known as the limits of integration a is the
lower limit and b the upper limit
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580 Topic 6 Calculus
Te square bracket notation means that the integration hastaken place but the limits have not yet been applied o do thiswe simply evaluate the integrated expression at the upper limitand subtract the integrated expression evaluated at the lowerlimit
You may be wondering where the constant of integration hasgone We could write it in as before but we quickly realise thatthis is unnecessary as it will just cancel out at the upper andlower limit each time
b c
bb
3
3
1
1
1
Te value of x is a dummy variable it does not come intothe answer But both a and b can vary and affect the resultChanging x to a different variable does not change the answerFor example
u a x bb
31 1
int
Worked example 174
Find the exact value of1
1+
Integrate and write in squarebrackets x
+ ]xint
Evaluate the integratedexpression at the upper
and lower limits andsubtract the lower from
the upper
=(In (e) + 4 (e)) minus (In (1) + 4 (1))
= + minusminus =
Ma ke sure you
kno w ho w to e va lua te
de fini te in tegra ls on your
ca lcu la tor as e xp lained
on Ca lcu la tor s ki l ls s hee t
1 0 on t he C D - R OM
I t can sa ve you time
and you can e va lua te
in tegra ls you donrsquo t kno w
ho w to do a lge braica l l y
E ven w hen you are
as ked to find t he e xac t
va lue o f t he in tegra l you
can c hec k your ans wer
on t he ca lcu la tor
e x a m h i n t
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17 Basic integration and its applications 581
Exercise 17G
1 Evaluate the following de1047297nite integrals giving exact answers
(a) (i) x x 2
(ii)1
4
1
(b) (i)2π
int (ii) sπ
π2
(c) (i) ex 1
int (ii) 31
1
eminusint
2 Evaluate correct to three signi1047297cant 1047297gures
(a) (i)3
1 4
int (ii)1
int
(b) (i) e1
int (ii) ne
1int 13 Find the exact value of the integral e x int
π [5 marks]
4 Show that the value of the integral12
x k
k
is independentof k [4 marks]
5 If x ) 73
evaluate 13
[4 marks]
6 Solve the equation 1 [5 marks]
y
x
y = f (x)
a b
A
17H Geometrical significance of definiteintegration
Now we have a method that gives a numerical value for anintegral the natural question to ask is what does this numbermean
On Fill-in proof sheet 20 on the CD-ROM Te fundamentaltheorem of calculus we show that as long as f )x is positivethe de1047297nite integral of f )x between the limits a and b is thearea enclosed between the curve the x -axis and the lines x = a and x = b
KEY POINT 178
rea )x b
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582 Topic 6 Calculus
If you are sketching the graph on the calculator you can get it toshade and evaluate the required area see Calculator skillssheet 10 on the CD-ROM You need to show the sketch as a partof your working if it is not already shown in the question
Worked example 175
Find the exact area enclosed between the x -axis the curve y = sin x and the lines x = 0 and x = π
Sketch the graph and identifythe area required = in
3
Integrate and write in squarebrackets
A = ]= minus
0
3ππ
Evaluate the integratedexpression at the upper andlower limit and subtract the
lower from the upper
= minus )cos cminus minus os
= minus minus ( )minus =
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17 Basic integration and its applications 583
The Ancient Greekshad developed ideasof limiting processessimilar to those used
in calculus but it took nearly2000 years for these ideas
to be formalised This wasdone almost simultaneouslyby Isaac Newton andGottried Leibniz in the 17thCentury Is this a coincidenceor is it often the case that along period of slow progressis often necessary to get tothe stage of majorbreakthroughsSupplementary sheet 10
looks at some other peoplewho can claim to haveinvented calculus
In the 17th Century integration was defined as thearea under a curve The area was broken downinto small rectangles each with a height f (x ) and a
width of a small bit of x called ∆x The total areawas approximately the sum of all of these rectangles
x a
x
f x x sum )∆
Isaac Newton one of the pioneers of calculus was also abig fan of writing in English rather than Greek So sigmabecame the English letter lsquoSrsquo and delta became the Englishletter d so that when the limit is taken as the width of therectangles become vanishingly small then the expressionbecomes
x a
)x dint This illustrates another very important interpretation of
integration ndash the infinite sum of infinitesimally small parts
Worked example 176
Find the area A in this graphy
x
y = x(xminus1)(2minusx)
1 2A
0
Write down the integral to beevaluated then use calculator
x x xminusint 1
x = minus y D
The area must be positive there4 =
When the curve is entirely below the x -axis the integral will givea negative value Te modulus of this value is the area
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584 Topic 6 Calculus
Unfortunately the relationship between integrals and areas isnot so simple when there are parts of the curve above and belowthe axis Tose bits above the axis contribute positively to thearea but bits below the axis contribute negatively to the areaWe must separate out the sections above the axis and belowthe axis
Worked example 177
(a) Find x x 1
4
x int 1 d
(b) Find the area enclosed between the x -axis the curve y minus x + and the lines= 1 and 4
Apply standard integration (a) x x
4
xminus1
= ( ( ) ( ( )minus + minus +
= minus =3
4
The value found above canrsquot bethe correct area for (b)
Sketch the curve to see exactlywhich area we are being asked
to find
(b) y
2 minus + 3
1 3
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17 Basic integration and its applications 585
Te fact that the integral was zero in Worked example 177 part(a) means that the area above the axis is exactly cancelled by thearea below the axis
Tis example warns us that when asked to 1047297nd an area we mustalways sketch the graph and identify exactly where each partof the area is If we are evaluating the area on the calculator wecan use the modulus function to ensure that the entire graph isabove the x -axis Using the function from Worked example 177
1
4
int 1y
x
y = |x2 minus 4x + 3|
1 3 4
continued
The area is made up of twoparts so evaluate each of
them separately
x
1
33
xminusint
( ) minus = minus
there4 rea below the axis is
x x3
4
2xminus int
minus ( ) =
4
there4 rea above the axis is
Total area = + =3 3
Transformationsof graphs using the
modulus function
were covered inchapter 7
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586 Topic 6 Calculus
KEY POINT 179
Te area bounded by the curve y f )x the x-axis and the
lines x = a and x = b is given byb
x int
When working without a calculator if the curve crosses
the x -axis between a and b we need to split the area intoseveral parts and 1047297nd each one separately
Te interpretation of integrals as areas causes one inconsistency
with our previous work Consider the integral1
2
1
x dminus
int
Graphically we can see that this area should exist
x
y
minus1minus2
However if we do the integration we 1047297nd that
12
1
2
1
]minus
minus
minus
minus
minus n minus
minusn
1
= ln
1
minus n
Tis is the correct answer (which we could have found usingthe symmetry of the curve) but it goes through a stage wherewe had to take logarithms of negative numbers and this issomething we are not allowed to do We avoid this by rede1047297ning
the integral ofx
as
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17 Basic integration and its applications 587
KEY POINT 174 AGAIN
minusint =
With this de1047297nition we can integrate y =1
over negative
numbers and the integral above becomes
12
1
2
1
x
e oreminus
minusint
n
Notice that the answer is negative since the required area is below
the x -axis We can still not integrate1
with a negative lower and
positive upper limit since the graph has an asymptote at x = 0
You may rightly be alittle uncomfortablewith inserting amodulus function lsquojust
because it worksrsquo Inmathematics do the ends
justify the means
Exercise 17H
1 Find the shaded areas
(a) (i)y =x2
y
x1 2
(ii)y = 1
x2
y
x2 4
(b) (i) y =x2
minus4x+ 3
y
x1 2
(ii)y =x2
minus4
y
x1
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588 Topic 6 Calculus
(c) (i)y = x3 minus x
y
xminus1 2
(ii)y = x2minus3x
y
x
5
2 Te area enclosed by the x -axis the curve andthe line x = k is 18 Find the value of k [6 marks]
3 (a) Find3
int (b) Find the area between the curve minus2 and the
x -axis between x = 0 and x = 3 [5 marks]
4 Between x = 0 and x = 3 the area of the graph y x minus2 below the x -axis equals the area above thex -axis Find the value of k [6 marks]
5 Find the area enclosed by the curve 1x minus minus 0
and the x -axis [7 marks]
lsquo Find t he area
enc losedrsquo means firs t
find a c losed region
bounded b y t he
cur ves men tioned
t hen find i ts area
A s ke tc h is a ver y
use fu l too l
e x a m h i n t
17I The area between a curve andthe y -axis
Consider the diagram alongside How can we 1047297nd the shadedarea A
One possible strategy is to construct a box around the graph todivide up the regions of interest You can integrate to 1047297nd thearea labelled A1 and then by adding and subtracting the areas ofthe blue and red rectangles shown calculate A
y
x
y = f (x)
c
d
A
y
x
y = f (x)
c
d A1
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17 Basic integration and its applications 589
Happily there is a quicker way we can treat x as a functionof y effectively re1047298ecting the whole diagram in the line y = x and then use the same method as in the previous section
KEY POINT 1710
Te area bounded by the curve y the y -axis and the
lines y = c and y = d is given by y d
) dint where is
the expression for x in terms of y
You may have realised that this is related to inverse functions from Section 5E
x
y
x = f (y)
cd
A
Worked example 178
Te curve shown has equation y 1minus Find the shaded area
y
x
y = 2radic x minus 1
10
Express x in terms of y x = 2
rArr = + y
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590 Topic 6 Calculus
Exercise 17I
continued
Find the limits on the y -axisIt may help to label them on the
graph
When x = minusWhen x = 1 = =minus 6
y
y = 2radic minus
1
10
Write down the integral andevaluate using calculator
Area fro GDC= +2
1 2=d
1 Find the shaded areas
(a) (i) y
x
y = x2
1
6
(ii) y
x
y = x3
1
3
(b) (i) y
xy = 1
x2
2
1
(ii) y
x
y =radic x
1
5
(c) (i)y
x
y = ln x
1e
e
(ii) y
x
y = 3x
1 6
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17 Basic integration and its applications 591
2 Te diagram shows the curve If the shadedarea is 504 1047297nd the value of a [6 marks]
y
x
y = radic
x
a
2a
0
3 Find the exact value of the area enclosed by the graphof + the line y = 2 and the y -axis [6 marks]
4 Te diagram shows the graph of
Te shaded area is 39 units Find the value of a [7 marks]y
x
y =radic x
4 a
5 Te diagram shows the graph of 2 where a isin infin Te area of the pink region is equal to the area of the blue
region Give two equations for a in terms of b and hencegive a in exact form and determine the size of theblue area [8 marks]y
x
y = x2
b
1 a
17J The area between two curves
So far we have only looked at areas bounded by a curve and oneof the coordinate axes but we can also 1047297nd areas bounded bytwo curvesTe area A in the diagram can be found by taking the areabounded by f )x and the x -axis and subtracting the areabounded by and the x -axis that is
A
b
minus
y
x
y = f (x)
y = g(x)
a b
A
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592 Topic 6 Calculus
We can do the subtraction before integrating so that we onlyhave to integrate one expression instead of two Tis gives analternative formula for the area
KEY POINT 1711
Te area A between two curves f (x ) and g (x ) is
Ab
(
where a and b are the x-coordinates of the intersectionpoints of the two curves
Worked example 179
Find the area A enclosed between + and minus2 +
First find the x -coordinates ofintersection
For intersection
x x x
rArr =
rArr ) =rArr =
minus
minus
+x
4minusx 0
Make a rough sketch to see therelative positions of the two curves
y
x1 4
A
2x + 1y =
2 minus 3x + 5y = x
Subtract the lower curve from thehigher before integrating
= x
minusx x
4
minus
minus minusx
2
1
4
2
minus =
8
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17 Basic integration and its applications 593
Subtracting the two equations before integrating is particularlyuseful when one of the curves is partly below the x-axis If x is always above then the expression we are integrating f g )x minus )x is always positive so we do not have to worryabout the signs of f )x and g )x themselves
Worked example 1710
Find the area bounded by the curves y x minus and y x 2
Sketch the graph to see therelative position of two
curves
Using GDC
= ex minus 5
y = 3 2
y
minus2 2
A
Find the intersection points ndash
use calculatorintersections x = minus2 658
Write down the integralrepresenting the area
rea =minus
minus818
1 658
xminus
= minus
1 6 8
x
Evaluate the integral usingcalculator = 21 6 (3SF)
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594 Topic 6 Calculus
Exercise 17J
1 Find the shaded areas
(a) (i)
y = 2x+ 1
y = (xminus1)2
y
x
(ii) y =x+ 1
y = 4xminusx2minus1
y
x
(b) (i)
y =minusx2minus4x+ 12
y
x
y =x2 + 2x+ 12
(ii)
y =x2minus2x+ 9
y
x
y = 4xminusx2 + 5
(c) (i)y =x2 minusx
y = 2xminusx2
y
x
(ii)y =x2minus7x+ 7
y = 3minusxminusx2
y
x
2 Find the area enclosed between the graphs of y x +x minus and + [6 marks]
3 Find the area enclosed by the curve 2
the y-axisand the line [6 marks]
4 Find the area between the curves1
and in theregion lt lt π [6 marks]
5 Show that the area of the shaded region alongside is2
[6 marks]
y = x2
y
xminus1 2
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17 Basic integration and its applications 595
6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]
7 Find the total area enclosed between the graphs of
)minus 2 and minus2 7 1+ [6 marks]
8 Te area enclosed between the curve y x 2 and the line
y mx is 10 Find the value of m if m 0 [7 marks]
9 Show that the shaded area in the diagram below is [8 marks]
y = 2 minus xy2 = x
y
x
y = cos x y = sin x
y
x
Summary
bull Integration is the reverse process of differentiation
bull Any integral without limits (inde1047297nite) will generate a constant of integration
bull For all rational n ne minus1 int +x c
1
1
bull If minus we get the natural logarithm function int minus =
bull Te integral of the exponential function is int e c
bull Te integrals of the trigonometric functions are
int si minus=
int tan =
bull Te de1047297nite integral has limits f( ) x b
dint is found by evaluating the integrated expression
at b and then subtracting the integrated expression evaluated at a
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596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
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17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
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7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
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576 Topic 6 Calculus
Exercise 17D 1 Find the following integrals
(a) (i) (ii)
(b) (i) int 1
x (ii) int
1
x
(c) (i) int minus x 2 1
d (ii) int + x 3
d
(d) (i) int +
2x (ii) int
x x minusx 2
2 Find the following integrals
(a) (i) int ex (ii) int ex x
(b) (i)x
(ii)7 x
(c) (i) int + x
(ii) int +
17E Integrating trigonometric functions
We can expand the set of functions that we can integrate bycontinuing to refer back to work covered in chapter 16
We saw in Key point 166 that (s x which means
that
Similarly as ( then int si minus
KEY POINT 176
Te integrals of trigonometric functions
int s minus s
=
We do not have a function whose derivative is ta and so
have no way (yet) of 1047297nding ta We will meet a method that
enables us to establish this in chapter 19 but for completeness
the result is given here
KEY POINT 176a
tan
See Exercise 19B
for establishing this
result
T he in tegra l o f
tan x is no t gi ven
in t he Formu la
boo k le t and is wor t h
remem bering
e x a m h i n t
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17 Basic integration and its applications 577
Exercise 17E
1 Find the following integrals
(a) (i) int s n x cminus os x d (ii) x s n x d
(b) (i) int t n (ii)
sin
2 3
(c) (i)7
(ii)
(d) (i) int minus x s n x d (ii) cos minus x sminus ncos x
2 Find int s n
cos
x cos
x x d [5 marks]
3 Find int cos x sminus n [5 marks]
17F Finding the equation of a curve
We have seen how we can integrate the function y
to 1047297nd the
equation of the original curve except for the unknown constant
of integration Tis is because the gradientd y
x determines the
shape of the curve but not exactly where it is However if weare also given the coordinates of a point on the curve we can
then determine the constant and hence specify the originalfunction precisely
If we again consider x 2 which we met at the start of this
chapter we know that the original function must have equation y = x 2 + c for some constant value c
If we are also told that the curve passes through the point(1 minus1) we can 1047297nd c and specify which of the family of curvesour function must be
(1minus1)
c = minus6
c = minus4
c = minus2
c = 4
c = 2
c = 0
y
x0
Look back to Worked
example 162 where given the gradient
we could draw many
different curves bychanging the starting
point
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578 Topic 6 Calculus
Te above example illustrates the general procedure for 1047297ndingthe equation of a curve from its gradient function
KEY POINT 177
o 1047297nd the equation for y given the gradientx
and onepoint ( p q) on the curve
1 Integrate remembering +c
2 Find the constant of integration by substitutingx = p y = q
Exercise 17F
1 Find the equation of the original curve if
(a) (i) and the curve passes through (ndash2 7)
(ii) x = 2 and the curve passes through (0 5)
(b) (i)d
d
y
x x
1 and the curve passes through (4 8)
(ii)1
2 and the curve passes through (1 3)
(c) (i) y
= + and the curve passes through (1 1)
(ii)d y
x = e and the curve passes through (ln5 0)
Worked example 173
Te gradient of a curve is given by = + and the curve passes through the point
(1 ndash4) Find the equation of the curve
To find y from ddy x
we need tointegrate
Donrsquot forget + c
= + 5xminus d
= +minus +
The coordinates of the givenpoint must satisfy this
equation so we can find c
When x = = minus so
minus ) + ) += 53 2
c
rArr minus minus + == 4 + 6c rArr
there4 = minusx minus x5+ 6
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17 Basic integration and its applications 579
(d) (i)+
and the curve passes through (e e)
(ii)1
and the curve passes through (e2 5)
(e) (i)d y
x x = x and the curve passesthrough (π 1)
(ii) x 3ta and the curve passes through (0 4)
2 Te derivative of the curve y f )x is1
(a) Find an expression for all possible functions f(x)
(b) If the curve passes through the point (2 7) 1047297nd theequation of the curve [5 marks]
3 Te gradient of a curve is found to be minus2
(a) Find the x -coordinate of the maximum point justifyingthat it is a maximum
(b) Given that the curve passes through the point (0 2)show that the y -coordinate of the maximum point
is minus7 [5 marks]
4 Te gradient of the normal to a curve at any point isequal to the x -coordinate at that point If the curve passesthrough the point (e2 3) 1047297nd the equation of the curvein the form ( ) where is a rationalfunction x gt 0 [6 marks]
17G Definite integration
Until now we have been carrying out a process known as
inde1047297nite integration inde1047297nite in the sense that we have anunknown constant each time for example
1
However there is also a process called de1047297nite integration which yields a numerical answer without the involvementof the constant of integration for example
bb
3 3 3
3 3 minus
int
Here a and b are known as the limits of integration a is the
lower limit and b the upper limit
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580 Topic 6 Calculus
Te square bracket notation means that the integration hastaken place but the limits have not yet been applied o do thiswe simply evaluate the integrated expression at the upper limitand subtract the integrated expression evaluated at the lowerlimit
You may be wondering where the constant of integration hasgone We could write it in as before but we quickly realise thatthis is unnecessary as it will just cancel out at the upper andlower limit each time
b c
bb
3
3
1
1
1
Te value of x is a dummy variable it does not come intothe answer But both a and b can vary and affect the resultChanging x to a different variable does not change the answerFor example
u a x bb
31 1
int
Worked example 174
Find the exact value of1
1+
Integrate and write in squarebrackets x
+ ]xint
Evaluate the integratedexpression at the upper
and lower limits andsubtract the lower from
the upper
=(In (e) + 4 (e)) minus (In (1) + 4 (1))
= + minusminus =
Ma ke sure you
kno w ho w to e va lua te
de fini te in tegra ls on your
ca lcu la tor as e xp lained
on Ca lcu la tor s ki l ls s hee t
1 0 on t he C D - R OM
I t can sa ve you time
and you can e va lua te
in tegra ls you donrsquo t kno w
ho w to do a lge braica l l y
E ven w hen you are
as ked to find t he e xac t
va lue o f t he in tegra l you
can c hec k your ans wer
on t he ca lcu la tor
e x a m h i n t
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17 Basic integration and its applications 581
Exercise 17G
1 Evaluate the following de1047297nite integrals giving exact answers
(a) (i) x x 2
(ii)1
4
1
(b) (i)2π
int (ii) sπ
π2
(c) (i) ex 1
int (ii) 31
1
eminusint
2 Evaluate correct to three signi1047297cant 1047297gures
(a) (i)3
1 4
int (ii)1
int
(b) (i) e1
int (ii) ne
1int 13 Find the exact value of the integral e x int
π [5 marks]
4 Show that the value of the integral12
x k
k
is independentof k [4 marks]
5 If x ) 73
evaluate 13
[4 marks]
6 Solve the equation 1 [5 marks]
y
x
y = f (x)
a b
A
17H Geometrical significance of definiteintegration
Now we have a method that gives a numerical value for anintegral the natural question to ask is what does this numbermean
On Fill-in proof sheet 20 on the CD-ROM Te fundamentaltheorem of calculus we show that as long as f )x is positivethe de1047297nite integral of f )x between the limits a and b is thearea enclosed between the curve the x -axis and the lines x = a and x = b
KEY POINT 178
rea )x b
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582 Topic 6 Calculus
If you are sketching the graph on the calculator you can get it toshade and evaluate the required area see Calculator skillssheet 10 on the CD-ROM You need to show the sketch as a partof your working if it is not already shown in the question
Worked example 175
Find the exact area enclosed between the x -axis the curve y = sin x and the lines x = 0 and x = π
Sketch the graph and identifythe area required = in
3
Integrate and write in squarebrackets
A = ]= minus
0
3ππ
Evaluate the integratedexpression at the upper andlower limit and subtract the
lower from the upper
= minus )cos cminus minus os
= minus minus ( )minus =
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17 Basic integration and its applications 583
The Ancient Greekshad developed ideasof limiting processessimilar to those used
in calculus but it took nearly2000 years for these ideas
to be formalised This wasdone almost simultaneouslyby Isaac Newton andGottried Leibniz in the 17thCentury Is this a coincidenceor is it often the case that along period of slow progressis often necessary to get tothe stage of majorbreakthroughsSupplementary sheet 10
looks at some other peoplewho can claim to haveinvented calculus
In the 17th Century integration was defined as thearea under a curve The area was broken downinto small rectangles each with a height f (x ) and a
width of a small bit of x called ∆x The total areawas approximately the sum of all of these rectangles
x a
x
f x x sum )∆
Isaac Newton one of the pioneers of calculus was also abig fan of writing in English rather than Greek So sigmabecame the English letter lsquoSrsquo and delta became the Englishletter d so that when the limit is taken as the width of therectangles become vanishingly small then the expressionbecomes
x a
)x dint This illustrates another very important interpretation of
integration ndash the infinite sum of infinitesimally small parts
Worked example 176
Find the area A in this graphy
x
y = x(xminus1)(2minusx)
1 2A
0
Write down the integral to beevaluated then use calculator
x x xminusint 1
x = minus y D
The area must be positive there4 =
When the curve is entirely below the x -axis the integral will givea negative value Te modulus of this value is the area
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584 Topic 6 Calculus
Unfortunately the relationship between integrals and areas isnot so simple when there are parts of the curve above and belowthe axis Tose bits above the axis contribute positively to thearea but bits below the axis contribute negatively to the areaWe must separate out the sections above the axis and belowthe axis
Worked example 177
(a) Find x x 1
4
x int 1 d
(b) Find the area enclosed between the x -axis the curve y minus x + and the lines= 1 and 4
Apply standard integration (a) x x
4
xminus1
= ( ( ) ( ( )minus + minus +
= minus =3
4
The value found above canrsquot bethe correct area for (b)
Sketch the curve to see exactlywhich area we are being asked
to find
(b) y
2 minus + 3
1 3
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17 Basic integration and its applications 585
Te fact that the integral was zero in Worked example 177 part(a) means that the area above the axis is exactly cancelled by thearea below the axis
Tis example warns us that when asked to 1047297nd an area we mustalways sketch the graph and identify exactly where each partof the area is If we are evaluating the area on the calculator wecan use the modulus function to ensure that the entire graph isabove the x -axis Using the function from Worked example 177
1
4
int 1y
x
y = |x2 minus 4x + 3|
1 3 4
continued
The area is made up of twoparts so evaluate each of
them separately
x
1
33
xminusint
( ) minus = minus
there4 rea below the axis is
x x3
4
2xminus int
minus ( ) =
4
there4 rea above the axis is
Total area = + =3 3
Transformationsof graphs using the
modulus function
were covered inchapter 7
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586 Topic 6 Calculus
KEY POINT 179
Te area bounded by the curve y f )x the x-axis and the
lines x = a and x = b is given byb
x int
When working without a calculator if the curve crosses
the x -axis between a and b we need to split the area intoseveral parts and 1047297nd each one separately
Te interpretation of integrals as areas causes one inconsistency
with our previous work Consider the integral1
2
1
x dminus
int
Graphically we can see that this area should exist
x
y
minus1minus2
However if we do the integration we 1047297nd that
12
1
2
1
]minus
minus
minus
minus
minus n minus
minusn
1
= ln
1
minus n
Tis is the correct answer (which we could have found usingthe symmetry of the curve) but it goes through a stage wherewe had to take logarithms of negative numbers and this issomething we are not allowed to do We avoid this by rede1047297ning
the integral ofx
as
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17 Basic integration and its applications 587
KEY POINT 174 AGAIN
minusint =
With this de1047297nition we can integrate y =1
over negative
numbers and the integral above becomes
12
1
2
1
x
e oreminus
minusint
n
Notice that the answer is negative since the required area is below
the x -axis We can still not integrate1
with a negative lower and
positive upper limit since the graph has an asymptote at x = 0
You may rightly be alittle uncomfortablewith inserting amodulus function lsquojust
because it worksrsquo Inmathematics do the ends
justify the means
Exercise 17H
1 Find the shaded areas
(a) (i)y =x2
y
x1 2
(ii)y = 1
x2
y
x2 4
(b) (i) y =x2
minus4x+ 3
y
x1 2
(ii)y =x2
minus4
y
x1
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588 Topic 6 Calculus
(c) (i)y = x3 minus x
y
xminus1 2
(ii)y = x2minus3x
y
x
5
2 Te area enclosed by the x -axis the curve andthe line x = k is 18 Find the value of k [6 marks]
3 (a) Find3
int (b) Find the area between the curve minus2 and the
x -axis between x = 0 and x = 3 [5 marks]
4 Between x = 0 and x = 3 the area of the graph y x minus2 below the x -axis equals the area above thex -axis Find the value of k [6 marks]
5 Find the area enclosed by the curve 1x minus minus 0
and the x -axis [7 marks]
lsquo Find t he area
enc losedrsquo means firs t
find a c losed region
bounded b y t he
cur ves men tioned
t hen find i ts area
A s ke tc h is a ver y
use fu l too l
e x a m h i n t
17I The area between a curve andthe y -axis
Consider the diagram alongside How can we 1047297nd the shadedarea A
One possible strategy is to construct a box around the graph todivide up the regions of interest You can integrate to 1047297nd thearea labelled A1 and then by adding and subtracting the areas ofthe blue and red rectangles shown calculate A
y
x
y = f (x)
c
d
A
y
x
y = f (x)
c
d A1
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17 Basic integration and its applications 589
Happily there is a quicker way we can treat x as a functionof y effectively re1047298ecting the whole diagram in the line y = x and then use the same method as in the previous section
KEY POINT 1710
Te area bounded by the curve y the y -axis and the
lines y = c and y = d is given by y d
) dint where is
the expression for x in terms of y
You may have realised that this is related to inverse functions from Section 5E
x
y
x = f (y)
cd
A
Worked example 178
Te curve shown has equation y 1minus Find the shaded area
y
x
y = 2radic x minus 1
10
Express x in terms of y x = 2
rArr = + y
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590 Topic 6 Calculus
Exercise 17I
continued
Find the limits on the y -axisIt may help to label them on the
graph
When x = minusWhen x = 1 = =minus 6
y
y = 2radic minus
1
10
Write down the integral andevaluate using calculator
Area fro GDC= +2
1 2=d
1 Find the shaded areas
(a) (i) y
x
y = x2
1
6
(ii) y
x
y = x3
1
3
(b) (i) y
xy = 1
x2
2
1
(ii) y
x
y =radic x
1
5
(c) (i)y
x
y = ln x
1e
e
(ii) y
x
y = 3x
1 6
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17 Basic integration and its applications 591
2 Te diagram shows the curve If the shadedarea is 504 1047297nd the value of a [6 marks]
y
x
y = radic
x
a
2a
0
3 Find the exact value of the area enclosed by the graphof + the line y = 2 and the y -axis [6 marks]
4 Te diagram shows the graph of
Te shaded area is 39 units Find the value of a [7 marks]y
x
y =radic x
4 a
5 Te diagram shows the graph of 2 where a isin infin Te area of the pink region is equal to the area of the blue
region Give two equations for a in terms of b and hencegive a in exact form and determine the size of theblue area [8 marks]y
x
y = x2
b
1 a
17J The area between two curves
So far we have only looked at areas bounded by a curve and oneof the coordinate axes but we can also 1047297nd areas bounded bytwo curvesTe area A in the diagram can be found by taking the areabounded by f )x and the x -axis and subtracting the areabounded by and the x -axis that is
A
b
minus
y
x
y = f (x)
y = g(x)
a b
A
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592 Topic 6 Calculus
We can do the subtraction before integrating so that we onlyhave to integrate one expression instead of two Tis gives analternative formula for the area
KEY POINT 1711
Te area A between two curves f (x ) and g (x ) is
Ab
(
where a and b are the x-coordinates of the intersectionpoints of the two curves
Worked example 179
Find the area A enclosed between + and minus2 +
First find the x -coordinates ofintersection
For intersection
x x x
rArr =
rArr ) =rArr =
minus
minus
+x
4minusx 0
Make a rough sketch to see therelative positions of the two curves
y
x1 4
A
2x + 1y =
2 minus 3x + 5y = x
Subtract the lower curve from thehigher before integrating
= x
minusx x
4
minus
minus minusx
2
1
4
2
minus =
8
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17 Basic integration and its applications 593
Subtracting the two equations before integrating is particularlyuseful when one of the curves is partly below the x-axis If x is always above then the expression we are integrating f g )x minus )x is always positive so we do not have to worryabout the signs of f )x and g )x themselves
Worked example 1710
Find the area bounded by the curves y x minus and y x 2
Sketch the graph to see therelative position of two
curves
Using GDC
= ex minus 5
y = 3 2
y
minus2 2
A
Find the intersection points ndash
use calculatorintersections x = minus2 658
Write down the integralrepresenting the area
rea =minus
minus818
1 658
xminus
= minus
1 6 8
x
Evaluate the integral usingcalculator = 21 6 (3SF)
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594 Topic 6 Calculus
Exercise 17J
1 Find the shaded areas
(a) (i)
y = 2x+ 1
y = (xminus1)2
y
x
(ii) y =x+ 1
y = 4xminusx2minus1
y
x
(b) (i)
y =minusx2minus4x+ 12
y
x
y =x2 + 2x+ 12
(ii)
y =x2minus2x+ 9
y
x
y = 4xminusx2 + 5
(c) (i)y =x2 minusx
y = 2xminusx2
y
x
(ii)y =x2minus7x+ 7
y = 3minusxminusx2
y
x
2 Find the area enclosed between the graphs of y x +x minus and + [6 marks]
3 Find the area enclosed by the curve 2
the y-axisand the line [6 marks]
4 Find the area between the curves1
and in theregion lt lt π [6 marks]
5 Show that the area of the shaded region alongside is2
[6 marks]
y = x2
y
xminus1 2
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17 Basic integration and its applications 595
6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]
7 Find the total area enclosed between the graphs of
)minus 2 and minus2 7 1+ [6 marks]
8 Te area enclosed between the curve y x 2 and the line
y mx is 10 Find the value of m if m 0 [7 marks]
9 Show that the shaded area in the diagram below is [8 marks]
y = 2 minus xy2 = x
y
x
y = cos x y = sin x
y
x
Summary
bull Integration is the reverse process of differentiation
bull Any integral without limits (inde1047297nite) will generate a constant of integration
bull For all rational n ne minus1 int +x c
1
1
bull If minus we get the natural logarithm function int minus =
bull Te integral of the exponential function is int e c
bull Te integrals of the trigonometric functions are
int si minus=
int tan =
bull Te de1047297nite integral has limits f( ) x b
dint is found by evaluating the integrated expression
at b and then subtracting the integrated expression evaluated at a
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596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
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17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
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7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
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17 Basic integration and its applications 577
Exercise 17E
1 Find the following integrals
(a) (i) int s n x cminus os x d (ii) x s n x d
(b) (i) int t n (ii)
sin
2 3
(c) (i)7
(ii)
(d) (i) int minus x s n x d (ii) cos minus x sminus ncos x
2 Find int s n
cos
x cos
x x d [5 marks]
3 Find int cos x sminus n [5 marks]
17F Finding the equation of a curve
We have seen how we can integrate the function y
to 1047297nd the
equation of the original curve except for the unknown constant
of integration Tis is because the gradientd y
x determines the
shape of the curve but not exactly where it is However if weare also given the coordinates of a point on the curve we can
then determine the constant and hence specify the originalfunction precisely
If we again consider x 2 which we met at the start of this
chapter we know that the original function must have equation y = x 2 + c for some constant value c
If we are also told that the curve passes through the point(1 minus1) we can 1047297nd c and specify which of the family of curvesour function must be
(1minus1)
c = minus6
c = minus4
c = minus2
c = 4
c = 2
c = 0
y
x0
Look back to Worked
example 162 where given the gradient
we could draw many
different curves bychanging the starting
point
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578 Topic 6 Calculus
Te above example illustrates the general procedure for 1047297ndingthe equation of a curve from its gradient function
KEY POINT 177
o 1047297nd the equation for y given the gradientx
and onepoint ( p q) on the curve
1 Integrate remembering +c
2 Find the constant of integration by substitutingx = p y = q
Exercise 17F
1 Find the equation of the original curve if
(a) (i) and the curve passes through (ndash2 7)
(ii) x = 2 and the curve passes through (0 5)
(b) (i)d
d
y
x x
1 and the curve passes through (4 8)
(ii)1
2 and the curve passes through (1 3)
(c) (i) y
= + and the curve passes through (1 1)
(ii)d y
x = e and the curve passes through (ln5 0)
Worked example 173
Te gradient of a curve is given by = + and the curve passes through the point
(1 ndash4) Find the equation of the curve
To find y from ddy x
we need tointegrate
Donrsquot forget + c
= + 5xminus d
= +minus +
The coordinates of the givenpoint must satisfy this
equation so we can find c
When x = = minus so
minus ) + ) += 53 2
c
rArr minus minus + == 4 + 6c rArr
there4 = minusx minus x5+ 6
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17 Basic integration and its applications 579
(d) (i)+
and the curve passes through (e e)
(ii)1
and the curve passes through (e2 5)
(e) (i)d y
x x = x and the curve passesthrough (π 1)
(ii) x 3ta and the curve passes through (0 4)
2 Te derivative of the curve y f )x is1
(a) Find an expression for all possible functions f(x)
(b) If the curve passes through the point (2 7) 1047297nd theequation of the curve [5 marks]
3 Te gradient of a curve is found to be minus2
(a) Find the x -coordinate of the maximum point justifyingthat it is a maximum
(b) Given that the curve passes through the point (0 2)show that the y -coordinate of the maximum point
is minus7 [5 marks]
4 Te gradient of the normal to a curve at any point isequal to the x -coordinate at that point If the curve passesthrough the point (e2 3) 1047297nd the equation of the curvein the form ( ) where is a rationalfunction x gt 0 [6 marks]
17G Definite integration
Until now we have been carrying out a process known as
inde1047297nite integration inde1047297nite in the sense that we have anunknown constant each time for example
1
However there is also a process called de1047297nite integration which yields a numerical answer without the involvementof the constant of integration for example
bb
3 3 3
3 3 minus
int
Here a and b are known as the limits of integration a is the
lower limit and b the upper limit
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580 Topic 6 Calculus
Te square bracket notation means that the integration hastaken place but the limits have not yet been applied o do thiswe simply evaluate the integrated expression at the upper limitand subtract the integrated expression evaluated at the lowerlimit
You may be wondering where the constant of integration hasgone We could write it in as before but we quickly realise thatthis is unnecessary as it will just cancel out at the upper andlower limit each time
b c
bb
3
3
1
1
1
Te value of x is a dummy variable it does not come intothe answer But both a and b can vary and affect the resultChanging x to a different variable does not change the answerFor example
u a x bb
31 1
int
Worked example 174
Find the exact value of1
1+
Integrate and write in squarebrackets x
+ ]xint
Evaluate the integratedexpression at the upper
and lower limits andsubtract the lower from
the upper
=(In (e) + 4 (e)) minus (In (1) + 4 (1))
= + minusminus =
Ma ke sure you
kno w ho w to e va lua te
de fini te in tegra ls on your
ca lcu la tor as e xp lained
on Ca lcu la tor s ki l ls s hee t
1 0 on t he C D - R OM
I t can sa ve you time
and you can e va lua te
in tegra ls you donrsquo t kno w
ho w to do a lge braica l l y
E ven w hen you are
as ked to find t he e xac t
va lue o f t he in tegra l you
can c hec k your ans wer
on t he ca lcu la tor
e x a m h i n t
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17 Basic integration and its applications 581
Exercise 17G
1 Evaluate the following de1047297nite integrals giving exact answers
(a) (i) x x 2
(ii)1
4
1
(b) (i)2π
int (ii) sπ
π2
(c) (i) ex 1
int (ii) 31
1
eminusint
2 Evaluate correct to three signi1047297cant 1047297gures
(a) (i)3
1 4
int (ii)1
int
(b) (i) e1
int (ii) ne
1int 13 Find the exact value of the integral e x int
π [5 marks]
4 Show that the value of the integral12
x k
k
is independentof k [4 marks]
5 If x ) 73
evaluate 13
[4 marks]
6 Solve the equation 1 [5 marks]
y
x
y = f (x)
a b
A
17H Geometrical significance of definiteintegration
Now we have a method that gives a numerical value for anintegral the natural question to ask is what does this numbermean
On Fill-in proof sheet 20 on the CD-ROM Te fundamentaltheorem of calculus we show that as long as f )x is positivethe de1047297nite integral of f )x between the limits a and b is thearea enclosed between the curve the x -axis and the lines x = a and x = b
KEY POINT 178
rea )x b
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582 Topic 6 Calculus
If you are sketching the graph on the calculator you can get it toshade and evaluate the required area see Calculator skillssheet 10 on the CD-ROM You need to show the sketch as a partof your working if it is not already shown in the question
Worked example 175
Find the exact area enclosed between the x -axis the curve y = sin x and the lines x = 0 and x = π
Sketch the graph and identifythe area required = in
3
Integrate and write in squarebrackets
A = ]= minus
0
3ππ
Evaluate the integratedexpression at the upper andlower limit and subtract the
lower from the upper
= minus )cos cminus minus os
= minus minus ( )minus =
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17 Basic integration and its applications 583
The Ancient Greekshad developed ideasof limiting processessimilar to those used
in calculus but it took nearly2000 years for these ideas
to be formalised This wasdone almost simultaneouslyby Isaac Newton andGottried Leibniz in the 17thCentury Is this a coincidenceor is it often the case that along period of slow progressis often necessary to get tothe stage of majorbreakthroughsSupplementary sheet 10
looks at some other peoplewho can claim to haveinvented calculus
In the 17th Century integration was defined as thearea under a curve The area was broken downinto small rectangles each with a height f (x ) and a
width of a small bit of x called ∆x The total areawas approximately the sum of all of these rectangles
x a
x
f x x sum )∆
Isaac Newton one of the pioneers of calculus was also abig fan of writing in English rather than Greek So sigmabecame the English letter lsquoSrsquo and delta became the Englishletter d so that when the limit is taken as the width of therectangles become vanishingly small then the expressionbecomes
x a
)x dint This illustrates another very important interpretation of
integration ndash the infinite sum of infinitesimally small parts
Worked example 176
Find the area A in this graphy
x
y = x(xminus1)(2minusx)
1 2A
0
Write down the integral to beevaluated then use calculator
x x xminusint 1
x = minus y D
The area must be positive there4 =
When the curve is entirely below the x -axis the integral will givea negative value Te modulus of this value is the area
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584 Topic 6 Calculus
Unfortunately the relationship between integrals and areas isnot so simple when there are parts of the curve above and belowthe axis Tose bits above the axis contribute positively to thearea but bits below the axis contribute negatively to the areaWe must separate out the sections above the axis and belowthe axis
Worked example 177
(a) Find x x 1
4
x int 1 d
(b) Find the area enclosed between the x -axis the curve y minus x + and the lines= 1 and 4
Apply standard integration (a) x x
4
xminus1
= ( ( ) ( ( )minus + minus +
= minus =3
4
The value found above canrsquot bethe correct area for (b)
Sketch the curve to see exactlywhich area we are being asked
to find
(b) y
2 minus + 3
1 3
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17 Basic integration and its applications 585
Te fact that the integral was zero in Worked example 177 part(a) means that the area above the axis is exactly cancelled by thearea below the axis
Tis example warns us that when asked to 1047297nd an area we mustalways sketch the graph and identify exactly where each partof the area is If we are evaluating the area on the calculator wecan use the modulus function to ensure that the entire graph isabove the x -axis Using the function from Worked example 177
1
4
int 1y
x
y = |x2 minus 4x + 3|
1 3 4
continued
The area is made up of twoparts so evaluate each of
them separately
x
1
33
xminusint
( ) minus = minus
there4 rea below the axis is
x x3
4
2xminus int
minus ( ) =
4
there4 rea above the axis is
Total area = + =3 3
Transformationsof graphs using the
modulus function
were covered inchapter 7
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586 Topic 6 Calculus
KEY POINT 179
Te area bounded by the curve y f )x the x-axis and the
lines x = a and x = b is given byb
x int
When working without a calculator if the curve crosses
the x -axis between a and b we need to split the area intoseveral parts and 1047297nd each one separately
Te interpretation of integrals as areas causes one inconsistency
with our previous work Consider the integral1
2
1
x dminus
int
Graphically we can see that this area should exist
x
y
minus1minus2
However if we do the integration we 1047297nd that
12
1
2
1
]minus
minus
minus
minus
minus n minus
minusn
1
= ln
1
minus n
Tis is the correct answer (which we could have found usingthe symmetry of the curve) but it goes through a stage wherewe had to take logarithms of negative numbers and this issomething we are not allowed to do We avoid this by rede1047297ning
the integral ofx
as
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17 Basic integration and its applications 587
KEY POINT 174 AGAIN
minusint =
With this de1047297nition we can integrate y =1
over negative
numbers and the integral above becomes
12
1
2
1
x
e oreminus
minusint
n
Notice that the answer is negative since the required area is below
the x -axis We can still not integrate1
with a negative lower and
positive upper limit since the graph has an asymptote at x = 0
You may rightly be alittle uncomfortablewith inserting amodulus function lsquojust
because it worksrsquo Inmathematics do the ends
justify the means
Exercise 17H
1 Find the shaded areas
(a) (i)y =x2
y
x1 2
(ii)y = 1
x2
y
x2 4
(b) (i) y =x2
minus4x+ 3
y
x1 2
(ii)y =x2
minus4
y
x1
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588 Topic 6 Calculus
(c) (i)y = x3 minus x
y
xminus1 2
(ii)y = x2minus3x
y
x
5
2 Te area enclosed by the x -axis the curve andthe line x = k is 18 Find the value of k [6 marks]
3 (a) Find3
int (b) Find the area between the curve minus2 and the
x -axis between x = 0 and x = 3 [5 marks]
4 Between x = 0 and x = 3 the area of the graph y x minus2 below the x -axis equals the area above thex -axis Find the value of k [6 marks]
5 Find the area enclosed by the curve 1x minus minus 0
and the x -axis [7 marks]
lsquo Find t he area
enc losedrsquo means firs t
find a c losed region
bounded b y t he
cur ves men tioned
t hen find i ts area
A s ke tc h is a ver y
use fu l too l
e x a m h i n t
17I The area between a curve andthe y -axis
Consider the diagram alongside How can we 1047297nd the shadedarea A
One possible strategy is to construct a box around the graph todivide up the regions of interest You can integrate to 1047297nd thearea labelled A1 and then by adding and subtracting the areas ofthe blue and red rectangles shown calculate A
y
x
y = f (x)
c
d
A
y
x
y = f (x)
c
d A1
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17 Basic integration and its applications 589
Happily there is a quicker way we can treat x as a functionof y effectively re1047298ecting the whole diagram in the line y = x and then use the same method as in the previous section
KEY POINT 1710
Te area bounded by the curve y the y -axis and the
lines y = c and y = d is given by y d
) dint where is
the expression for x in terms of y
You may have realised that this is related to inverse functions from Section 5E
x
y
x = f (y)
cd
A
Worked example 178
Te curve shown has equation y 1minus Find the shaded area
y
x
y = 2radic x minus 1
10
Express x in terms of y x = 2
rArr = + y
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590 Topic 6 Calculus
Exercise 17I
continued
Find the limits on the y -axisIt may help to label them on the
graph
When x = minusWhen x = 1 = =minus 6
y
y = 2radic minus
1
10
Write down the integral andevaluate using calculator
Area fro GDC= +2
1 2=d
1 Find the shaded areas
(a) (i) y
x
y = x2
1
6
(ii) y
x
y = x3
1
3
(b) (i) y
xy = 1
x2
2
1
(ii) y
x
y =radic x
1
5
(c) (i)y
x
y = ln x
1e
e
(ii) y
x
y = 3x
1 6
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17 Basic integration and its applications 591
2 Te diagram shows the curve If the shadedarea is 504 1047297nd the value of a [6 marks]
y
x
y = radic
x
a
2a
0
3 Find the exact value of the area enclosed by the graphof + the line y = 2 and the y -axis [6 marks]
4 Te diagram shows the graph of
Te shaded area is 39 units Find the value of a [7 marks]y
x
y =radic x
4 a
5 Te diagram shows the graph of 2 where a isin infin Te area of the pink region is equal to the area of the blue
region Give two equations for a in terms of b and hencegive a in exact form and determine the size of theblue area [8 marks]y
x
y = x2
b
1 a
17J The area between two curves
So far we have only looked at areas bounded by a curve and oneof the coordinate axes but we can also 1047297nd areas bounded bytwo curvesTe area A in the diagram can be found by taking the areabounded by f )x and the x -axis and subtracting the areabounded by and the x -axis that is
A
b
minus
y
x
y = f (x)
y = g(x)
a b
A
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592 Topic 6 Calculus
We can do the subtraction before integrating so that we onlyhave to integrate one expression instead of two Tis gives analternative formula for the area
KEY POINT 1711
Te area A between two curves f (x ) and g (x ) is
Ab
(
where a and b are the x-coordinates of the intersectionpoints of the two curves
Worked example 179
Find the area A enclosed between + and minus2 +
First find the x -coordinates ofintersection
For intersection
x x x
rArr =
rArr ) =rArr =
minus
minus
+x
4minusx 0
Make a rough sketch to see therelative positions of the two curves
y
x1 4
A
2x + 1y =
2 minus 3x + 5y = x
Subtract the lower curve from thehigher before integrating
= x
minusx x
4
minus
minus minusx
2
1
4
2
minus =
8
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17 Basic integration and its applications 593
Subtracting the two equations before integrating is particularlyuseful when one of the curves is partly below the x-axis If x is always above then the expression we are integrating f g )x minus )x is always positive so we do not have to worryabout the signs of f )x and g )x themselves
Worked example 1710
Find the area bounded by the curves y x minus and y x 2
Sketch the graph to see therelative position of two
curves
Using GDC
= ex minus 5
y = 3 2
y
minus2 2
A
Find the intersection points ndash
use calculatorintersections x = minus2 658
Write down the integralrepresenting the area
rea =minus
minus818
1 658
xminus
= minus
1 6 8
x
Evaluate the integral usingcalculator = 21 6 (3SF)
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594 Topic 6 Calculus
Exercise 17J
1 Find the shaded areas
(a) (i)
y = 2x+ 1
y = (xminus1)2
y
x
(ii) y =x+ 1
y = 4xminusx2minus1
y
x
(b) (i)
y =minusx2minus4x+ 12
y
x
y =x2 + 2x+ 12
(ii)
y =x2minus2x+ 9
y
x
y = 4xminusx2 + 5
(c) (i)y =x2 minusx
y = 2xminusx2
y
x
(ii)y =x2minus7x+ 7
y = 3minusxminusx2
y
x
2 Find the area enclosed between the graphs of y x +x minus and + [6 marks]
3 Find the area enclosed by the curve 2
the y-axisand the line [6 marks]
4 Find the area between the curves1
and in theregion lt lt π [6 marks]
5 Show that the area of the shaded region alongside is2
[6 marks]
y = x2
y
xminus1 2
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17 Basic integration and its applications 595
6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]
7 Find the total area enclosed between the graphs of
)minus 2 and minus2 7 1+ [6 marks]
8 Te area enclosed between the curve y x 2 and the line
y mx is 10 Find the value of m if m 0 [7 marks]
9 Show that the shaded area in the diagram below is [8 marks]
y = 2 minus xy2 = x
y
x
y = cos x y = sin x
y
x
Summary
bull Integration is the reverse process of differentiation
bull Any integral without limits (inde1047297nite) will generate a constant of integration
bull For all rational n ne minus1 int +x c
1
1
bull If minus we get the natural logarithm function int minus =
bull Te integral of the exponential function is int e c
bull Te integrals of the trigonometric functions are
int si minus=
int tan =
bull Te de1047297nite integral has limits f( ) x b
dint is found by evaluating the integrated expression
at b and then subtracting the integrated expression evaluated at a
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596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
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17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
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7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
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578 Topic 6 Calculus
Te above example illustrates the general procedure for 1047297ndingthe equation of a curve from its gradient function
KEY POINT 177
o 1047297nd the equation for y given the gradientx
and onepoint ( p q) on the curve
1 Integrate remembering +c
2 Find the constant of integration by substitutingx = p y = q
Exercise 17F
1 Find the equation of the original curve if
(a) (i) and the curve passes through (ndash2 7)
(ii) x = 2 and the curve passes through (0 5)
(b) (i)d
d
y
x x
1 and the curve passes through (4 8)
(ii)1
2 and the curve passes through (1 3)
(c) (i) y
= + and the curve passes through (1 1)
(ii)d y
x = e and the curve passes through (ln5 0)
Worked example 173
Te gradient of a curve is given by = + and the curve passes through the point
(1 ndash4) Find the equation of the curve
To find y from ddy x
we need tointegrate
Donrsquot forget + c
= + 5xminus d
= +minus +
The coordinates of the givenpoint must satisfy this
equation so we can find c
When x = = minus so
minus ) + ) += 53 2
c
rArr minus minus + == 4 + 6c rArr
there4 = minusx minus x5+ 6
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17 Basic integration and its applications 579
(d) (i)+
and the curve passes through (e e)
(ii)1
and the curve passes through (e2 5)
(e) (i)d y
x x = x and the curve passesthrough (π 1)
(ii) x 3ta and the curve passes through (0 4)
2 Te derivative of the curve y f )x is1
(a) Find an expression for all possible functions f(x)
(b) If the curve passes through the point (2 7) 1047297nd theequation of the curve [5 marks]
3 Te gradient of a curve is found to be minus2
(a) Find the x -coordinate of the maximum point justifyingthat it is a maximum
(b) Given that the curve passes through the point (0 2)show that the y -coordinate of the maximum point
is minus7 [5 marks]
4 Te gradient of the normal to a curve at any point isequal to the x -coordinate at that point If the curve passesthrough the point (e2 3) 1047297nd the equation of the curvein the form ( ) where is a rationalfunction x gt 0 [6 marks]
17G Definite integration
Until now we have been carrying out a process known as
inde1047297nite integration inde1047297nite in the sense that we have anunknown constant each time for example
1
However there is also a process called de1047297nite integration which yields a numerical answer without the involvementof the constant of integration for example
bb
3 3 3
3 3 minus
int
Here a and b are known as the limits of integration a is the
lower limit and b the upper limit
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580 Topic 6 Calculus
Te square bracket notation means that the integration hastaken place but the limits have not yet been applied o do thiswe simply evaluate the integrated expression at the upper limitand subtract the integrated expression evaluated at the lowerlimit
You may be wondering where the constant of integration hasgone We could write it in as before but we quickly realise thatthis is unnecessary as it will just cancel out at the upper andlower limit each time
b c
bb
3
3
1
1
1
Te value of x is a dummy variable it does not come intothe answer But both a and b can vary and affect the resultChanging x to a different variable does not change the answerFor example
u a x bb
31 1
int
Worked example 174
Find the exact value of1
1+
Integrate and write in squarebrackets x
+ ]xint
Evaluate the integratedexpression at the upper
and lower limits andsubtract the lower from
the upper
=(In (e) + 4 (e)) minus (In (1) + 4 (1))
= + minusminus =
Ma ke sure you
kno w ho w to e va lua te
de fini te in tegra ls on your
ca lcu la tor as e xp lained
on Ca lcu la tor s ki l ls s hee t
1 0 on t he C D - R OM
I t can sa ve you time
and you can e va lua te
in tegra ls you donrsquo t kno w
ho w to do a lge braica l l y
E ven w hen you are
as ked to find t he e xac t
va lue o f t he in tegra l you
can c hec k your ans wer
on t he ca lcu la tor
e x a m h i n t
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17 Basic integration and its applications 581
Exercise 17G
1 Evaluate the following de1047297nite integrals giving exact answers
(a) (i) x x 2
(ii)1
4
1
(b) (i)2π
int (ii) sπ
π2
(c) (i) ex 1
int (ii) 31
1
eminusint
2 Evaluate correct to three signi1047297cant 1047297gures
(a) (i)3
1 4
int (ii)1
int
(b) (i) e1
int (ii) ne
1int 13 Find the exact value of the integral e x int
π [5 marks]
4 Show that the value of the integral12
x k
k
is independentof k [4 marks]
5 If x ) 73
evaluate 13
[4 marks]
6 Solve the equation 1 [5 marks]
y
x
y = f (x)
a b
A
17H Geometrical significance of definiteintegration
Now we have a method that gives a numerical value for anintegral the natural question to ask is what does this numbermean
On Fill-in proof sheet 20 on the CD-ROM Te fundamentaltheorem of calculus we show that as long as f )x is positivethe de1047297nite integral of f )x between the limits a and b is thearea enclosed between the curve the x -axis and the lines x = a and x = b
KEY POINT 178
rea )x b
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582 Topic 6 Calculus
If you are sketching the graph on the calculator you can get it toshade and evaluate the required area see Calculator skillssheet 10 on the CD-ROM You need to show the sketch as a partof your working if it is not already shown in the question
Worked example 175
Find the exact area enclosed between the x -axis the curve y = sin x and the lines x = 0 and x = π
Sketch the graph and identifythe area required = in
3
Integrate and write in squarebrackets
A = ]= minus
0
3ππ
Evaluate the integratedexpression at the upper andlower limit and subtract the
lower from the upper
= minus )cos cminus minus os
= minus minus ( )minus =
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17 Basic integration and its applications 583
The Ancient Greekshad developed ideasof limiting processessimilar to those used
in calculus but it took nearly2000 years for these ideas
to be formalised This wasdone almost simultaneouslyby Isaac Newton andGottried Leibniz in the 17thCentury Is this a coincidenceor is it often the case that along period of slow progressis often necessary to get tothe stage of majorbreakthroughsSupplementary sheet 10
looks at some other peoplewho can claim to haveinvented calculus
In the 17th Century integration was defined as thearea under a curve The area was broken downinto small rectangles each with a height f (x ) and a
width of a small bit of x called ∆x The total areawas approximately the sum of all of these rectangles
x a
x
f x x sum )∆
Isaac Newton one of the pioneers of calculus was also abig fan of writing in English rather than Greek So sigmabecame the English letter lsquoSrsquo and delta became the Englishletter d so that when the limit is taken as the width of therectangles become vanishingly small then the expressionbecomes
x a
)x dint This illustrates another very important interpretation of
integration ndash the infinite sum of infinitesimally small parts
Worked example 176
Find the area A in this graphy
x
y = x(xminus1)(2minusx)
1 2A
0
Write down the integral to beevaluated then use calculator
x x xminusint 1
x = minus y D
The area must be positive there4 =
When the curve is entirely below the x -axis the integral will givea negative value Te modulus of this value is the area
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584 Topic 6 Calculus
Unfortunately the relationship between integrals and areas isnot so simple when there are parts of the curve above and belowthe axis Tose bits above the axis contribute positively to thearea but bits below the axis contribute negatively to the areaWe must separate out the sections above the axis and belowthe axis
Worked example 177
(a) Find x x 1
4
x int 1 d
(b) Find the area enclosed between the x -axis the curve y minus x + and the lines= 1 and 4
Apply standard integration (a) x x
4
xminus1
= ( ( ) ( ( )minus + minus +
= minus =3
4
The value found above canrsquot bethe correct area for (b)
Sketch the curve to see exactlywhich area we are being asked
to find
(b) y
2 minus + 3
1 3
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17 Basic integration and its applications 585
Te fact that the integral was zero in Worked example 177 part(a) means that the area above the axis is exactly cancelled by thearea below the axis
Tis example warns us that when asked to 1047297nd an area we mustalways sketch the graph and identify exactly where each partof the area is If we are evaluating the area on the calculator wecan use the modulus function to ensure that the entire graph isabove the x -axis Using the function from Worked example 177
1
4
int 1y
x
y = |x2 minus 4x + 3|
1 3 4
continued
The area is made up of twoparts so evaluate each of
them separately
x
1
33
xminusint
( ) minus = minus
there4 rea below the axis is
x x3
4
2xminus int
minus ( ) =
4
there4 rea above the axis is
Total area = + =3 3
Transformationsof graphs using the
modulus function
were covered inchapter 7
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586 Topic 6 Calculus
KEY POINT 179
Te area bounded by the curve y f )x the x-axis and the
lines x = a and x = b is given byb
x int
When working without a calculator if the curve crosses
the x -axis between a and b we need to split the area intoseveral parts and 1047297nd each one separately
Te interpretation of integrals as areas causes one inconsistency
with our previous work Consider the integral1
2
1
x dminus
int
Graphically we can see that this area should exist
x
y
minus1minus2
However if we do the integration we 1047297nd that
12
1
2
1
]minus
minus
minus
minus
minus n minus
minusn
1
= ln
1
minus n
Tis is the correct answer (which we could have found usingthe symmetry of the curve) but it goes through a stage wherewe had to take logarithms of negative numbers and this issomething we are not allowed to do We avoid this by rede1047297ning
the integral ofx
as
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17 Basic integration and its applications 587
KEY POINT 174 AGAIN
minusint =
With this de1047297nition we can integrate y =1
over negative
numbers and the integral above becomes
12
1
2
1
x
e oreminus
minusint
n
Notice that the answer is negative since the required area is below
the x -axis We can still not integrate1
with a negative lower and
positive upper limit since the graph has an asymptote at x = 0
You may rightly be alittle uncomfortablewith inserting amodulus function lsquojust
because it worksrsquo Inmathematics do the ends
justify the means
Exercise 17H
1 Find the shaded areas
(a) (i)y =x2
y
x1 2
(ii)y = 1
x2
y
x2 4
(b) (i) y =x2
minus4x+ 3
y
x1 2
(ii)y =x2
minus4
y
x1
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588 Topic 6 Calculus
(c) (i)y = x3 minus x
y
xminus1 2
(ii)y = x2minus3x
y
x
5
2 Te area enclosed by the x -axis the curve andthe line x = k is 18 Find the value of k [6 marks]
3 (a) Find3
int (b) Find the area between the curve minus2 and the
x -axis between x = 0 and x = 3 [5 marks]
4 Between x = 0 and x = 3 the area of the graph y x minus2 below the x -axis equals the area above thex -axis Find the value of k [6 marks]
5 Find the area enclosed by the curve 1x minus minus 0
and the x -axis [7 marks]
lsquo Find t he area
enc losedrsquo means firs t
find a c losed region
bounded b y t he
cur ves men tioned
t hen find i ts area
A s ke tc h is a ver y
use fu l too l
e x a m h i n t
17I The area between a curve andthe y -axis
Consider the diagram alongside How can we 1047297nd the shadedarea A
One possible strategy is to construct a box around the graph todivide up the regions of interest You can integrate to 1047297nd thearea labelled A1 and then by adding and subtracting the areas ofthe blue and red rectangles shown calculate A
y
x
y = f (x)
c
d
A
y
x
y = f (x)
c
d A1
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17 Basic integration and its applications 589
Happily there is a quicker way we can treat x as a functionof y effectively re1047298ecting the whole diagram in the line y = x and then use the same method as in the previous section
KEY POINT 1710
Te area bounded by the curve y the y -axis and the
lines y = c and y = d is given by y d
) dint where is
the expression for x in terms of y
You may have realised that this is related to inverse functions from Section 5E
x
y
x = f (y)
cd
A
Worked example 178
Te curve shown has equation y 1minus Find the shaded area
y
x
y = 2radic x minus 1
10
Express x in terms of y x = 2
rArr = + y
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590 Topic 6 Calculus
Exercise 17I
continued
Find the limits on the y -axisIt may help to label them on the
graph
When x = minusWhen x = 1 = =minus 6
y
y = 2radic minus
1
10
Write down the integral andevaluate using calculator
Area fro GDC= +2
1 2=d
1 Find the shaded areas
(a) (i) y
x
y = x2
1
6
(ii) y
x
y = x3
1
3
(b) (i) y
xy = 1
x2
2
1
(ii) y
x
y =radic x
1
5
(c) (i)y
x
y = ln x
1e
e
(ii) y
x
y = 3x
1 6
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17 Basic integration and its applications 591
2 Te diagram shows the curve If the shadedarea is 504 1047297nd the value of a [6 marks]
y
x
y = radic
x
a
2a
0
3 Find the exact value of the area enclosed by the graphof + the line y = 2 and the y -axis [6 marks]
4 Te diagram shows the graph of
Te shaded area is 39 units Find the value of a [7 marks]y
x
y =radic x
4 a
5 Te diagram shows the graph of 2 where a isin infin Te area of the pink region is equal to the area of the blue
region Give two equations for a in terms of b and hencegive a in exact form and determine the size of theblue area [8 marks]y
x
y = x2
b
1 a
17J The area between two curves
So far we have only looked at areas bounded by a curve and oneof the coordinate axes but we can also 1047297nd areas bounded bytwo curvesTe area A in the diagram can be found by taking the areabounded by f )x and the x -axis and subtracting the areabounded by and the x -axis that is
A
b
minus
y
x
y = f (x)
y = g(x)
a b
A
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592 Topic 6 Calculus
We can do the subtraction before integrating so that we onlyhave to integrate one expression instead of two Tis gives analternative formula for the area
KEY POINT 1711
Te area A between two curves f (x ) and g (x ) is
Ab
(
where a and b are the x-coordinates of the intersectionpoints of the two curves
Worked example 179
Find the area A enclosed between + and minus2 +
First find the x -coordinates ofintersection
For intersection
x x x
rArr =
rArr ) =rArr =
minus
minus
+x
4minusx 0
Make a rough sketch to see therelative positions of the two curves
y
x1 4
A
2x + 1y =
2 minus 3x + 5y = x
Subtract the lower curve from thehigher before integrating
= x
minusx x
4
minus
minus minusx
2
1
4
2
minus =
8
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17 Basic integration and its applications 593
Subtracting the two equations before integrating is particularlyuseful when one of the curves is partly below the x-axis If x is always above then the expression we are integrating f g )x minus )x is always positive so we do not have to worryabout the signs of f )x and g )x themselves
Worked example 1710
Find the area bounded by the curves y x minus and y x 2
Sketch the graph to see therelative position of two
curves
Using GDC
= ex minus 5
y = 3 2
y
minus2 2
A
Find the intersection points ndash
use calculatorintersections x = minus2 658
Write down the integralrepresenting the area
rea =minus
minus818
1 658
xminus
= minus
1 6 8
x
Evaluate the integral usingcalculator = 21 6 (3SF)
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594 Topic 6 Calculus
Exercise 17J
1 Find the shaded areas
(a) (i)
y = 2x+ 1
y = (xminus1)2
y
x
(ii) y =x+ 1
y = 4xminusx2minus1
y
x
(b) (i)
y =minusx2minus4x+ 12
y
x
y =x2 + 2x+ 12
(ii)
y =x2minus2x+ 9
y
x
y = 4xminusx2 + 5
(c) (i)y =x2 minusx
y = 2xminusx2
y
x
(ii)y =x2minus7x+ 7
y = 3minusxminusx2
y
x
2 Find the area enclosed between the graphs of y x +x minus and + [6 marks]
3 Find the area enclosed by the curve 2
the y-axisand the line [6 marks]
4 Find the area between the curves1
and in theregion lt lt π [6 marks]
5 Show that the area of the shaded region alongside is2
[6 marks]
y = x2
y
xminus1 2
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17 Basic integration and its applications 595
6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]
7 Find the total area enclosed between the graphs of
)minus 2 and minus2 7 1+ [6 marks]
8 Te area enclosed between the curve y x 2 and the line
y mx is 10 Find the value of m if m 0 [7 marks]
9 Show that the shaded area in the diagram below is [8 marks]
y = 2 minus xy2 = x
y
x
y = cos x y = sin x
y
x
Summary
bull Integration is the reverse process of differentiation
bull Any integral without limits (inde1047297nite) will generate a constant of integration
bull For all rational n ne minus1 int +x c
1
1
bull If minus we get the natural logarithm function int minus =
bull Te integral of the exponential function is int e c
bull Te integrals of the trigonometric functions are
int si minus=
int tan =
bull Te de1047297nite integral has limits f( ) x b
dint is found by evaluating the integrated expression
at b and then subtracting the integrated expression evaluated at a
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596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
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17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
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7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
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17 Basic integration and its applications 579
(d) (i)+
and the curve passes through (e e)
(ii)1
and the curve passes through (e2 5)
(e) (i)d y
x x = x and the curve passesthrough (π 1)
(ii) x 3ta and the curve passes through (0 4)
2 Te derivative of the curve y f )x is1
(a) Find an expression for all possible functions f(x)
(b) If the curve passes through the point (2 7) 1047297nd theequation of the curve [5 marks]
3 Te gradient of a curve is found to be minus2
(a) Find the x -coordinate of the maximum point justifyingthat it is a maximum
(b) Given that the curve passes through the point (0 2)show that the y -coordinate of the maximum point
is minus7 [5 marks]
4 Te gradient of the normal to a curve at any point isequal to the x -coordinate at that point If the curve passesthrough the point (e2 3) 1047297nd the equation of the curvein the form ( ) where is a rationalfunction x gt 0 [6 marks]
17G Definite integration
Until now we have been carrying out a process known as
inde1047297nite integration inde1047297nite in the sense that we have anunknown constant each time for example
1
However there is also a process called de1047297nite integration which yields a numerical answer without the involvementof the constant of integration for example
bb
3 3 3
3 3 minus
int
Here a and b are known as the limits of integration a is the
lower limit and b the upper limit
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580 Topic 6 Calculus
Te square bracket notation means that the integration hastaken place but the limits have not yet been applied o do thiswe simply evaluate the integrated expression at the upper limitand subtract the integrated expression evaluated at the lowerlimit
You may be wondering where the constant of integration hasgone We could write it in as before but we quickly realise thatthis is unnecessary as it will just cancel out at the upper andlower limit each time
b c
bb
3
3
1
1
1
Te value of x is a dummy variable it does not come intothe answer But both a and b can vary and affect the resultChanging x to a different variable does not change the answerFor example
u a x bb
31 1
int
Worked example 174
Find the exact value of1
1+
Integrate and write in squarebrackets x
+ ]xint
Evaluate the integratedexpression at the upper
and lower limits andsubtract the lower from
the upper
=(In (e) + 4 (e)) minus (In (1) + 4 (1))
= + minusminus =
Ma ke sure you
kno w ho w to e va lua te
de fini te in tegra ls on your
ca lcu la tor as e xp lained
on Ca lcu la tor s ki l ls s hee t
1 0 on t he C D - R OM
I t can sa ve you time
and you can e va lua te
in tegra ls you donrsquo t kno w
ho w to do a lge braica l l y
E ven w hen you are
as ked to find t he e xac t
va lue o f t he in tegra l you
can c hec k your ans wer
on t he ca lcu la tor
e x a m h i n t
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17 Basic integration and its applications 581
Exercise 17G
1 Evaluate the following de1047297nite integrals giving exact answers
(a) (i) x x 2
(ii)1
4
1
(b) (i)2π
int (ii) sπ
π2
(c) (i) ex 1
int (ii) 31
1
eminusint
2 Evaluate correct to three signi1047297cant 1047297gures
(a) (i)3
1 4
int (ii)1
int
(b) (i) e1
int (ii) ne
1int 13 Find the exact value of the integral e x int
π [5 marks]
4 Show that the value of the integral12
x k
k
is independentof k [4 marks]
5 If x ) 73
evaluate 13
[4 marks]
6 Solve the equation 1 [5 marks]
y
x
y = f (x)
a b
A
17H Geometrical significance of definiteintegration
Now we have a method that gives a numerical value for anintegral the natural question to ask is what does this numbermean
On Fill-in proof sheet 20 on the CD-ROM Te fundamentaltheorem of calculus we show that as long as f )x is positivethe de1047297nite integral of f )x between the limits a and b is thearea enclosed between the curve the x -axis and the lines x = a and x = b
KEY POINT 178
rea )x b
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582 Topic 6 Calculus
If you are sketching the graph on the calculator you can get it toshade and evaluate the required area see Calculator skillssheet 10 on the CD-ROM You need to show the sketch as a partof your working if it is not already shown in the question
Worked example 175
Find the exact area enclosed between the x -axis the curve y = sin x and the lines x = 0 and x = π
Sketch the graph and identifythe area required = in
3
Integrate and write in squarebrackets
A = ]= minus
0
3ππ
Evaluate the integratedexpression at the upper andlower limit and subtract the
lower from the upper
= minus )cos cminus minus os
= minus minus ( )minus =
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17 Basic integration and its applications 583
The Ancient Greekshad developed ideasof limiting processessimilar to those used
in calculus but it took nearly2000 years for these ideas
to be formalised This wasdone almost simultaneouslyby Isaac Newton andGottried Leibniz in the 17thCentury Is this a coincidenceor is it often the case that along period of slow progressis often necessary to get tothe stage of majorbreakthroughsSupplementary sheet 10
looks at some other peoplewho can claim to haveinvented calculus
In the 17th Century integration was defined as thearea under a curve The area was broken downinto small rectangles each with a height f (x ) and a
width of a small bit of x called ∆x The total areawas approximately the sum of all of these rectangles
x a
x
f x x sum )∆
Isaac Newton one of the pioneers of calculus was also abig fan of writing in English rather than Greek So sigmabecame the English letter lsquoSrsquo and delta became the Englishletter d so that when the limit is taken as the width of therectangles become vanishingly small then the expressionbecomes
x a
)x dint This illustrates another very important interpretation of
integration ndash the infinite sum of infinitesimally small parts
Worked example 176
Find the area A in this graphy
x
y = x(xminus1)(2minusx)
1 2A
0
Write down the integral to beevaluated then use calculator
x x xminusint 1
x = minus y D
The area must be positive there4 =
When the curve is entirely below the x -axis the integral will givea negative value Te modulus of this value is the area
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584 Topic 6 Calculus
Unfortunately the relationship between integrals and areas isnot so simple when there are parts of the curve above and belowthe axis Tose bits above the axis contribute positively to thearea but bits below the axis contribute negatively to the areaWe must separate out the sections above the axis and belowthe axis
Worked example 177
(a) Find x x 1
4
x int 1 d
(b) Find the area enclosed between the x -axis the curve y minus x + and the lines= 1 and 4
Apply standard integration (a) x x
4
xminus1
= ( ( ) ( ( )minus + minus +
= minus =3
4
The value found above canrsquot bethe correct area for (b)
Sketch the curve to see exactlywhich area we are being asked
to find
(b) y
2 minus + 3
1 3
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17 Basic integration and its applications 585
Te fact that the integral was zero in Worked example 177 part(a) means that the area above the axis is exactly cancelled by thearea below the axis
Tis example warns us that when asked to 1047297nd an area we mustalways sketch the graph and identify exactly where each partof the area is If we are evaluating the area on the calculator wecan use the modulus function to ensure that the entire graph isabove the x -axis Using the function from Worked example 177
1
4
int 1y
x
y = |x2 minus 4x + 3|
1 3 4
continued
The area is made up of twoparts so evaluate each of
them separately
x
1
33
xminusint
( ) minus = minus
there4 rea below the axis is
x x3
4
2xminus int
minus ( ) =
4
there4 rea above the axis is
Total area = + =3 3
Transformationsof graphs using the
modulus function
were covered inchapter 7
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586 Topic 6 Calculus
KEY POINT 179
Te area bounded by the curve y f )x the x-axis and the
lines x = a and x = b is given byb
x int
When working without a calculator if the curve crosses
the x -axis between a and b we need to split the area intoseveral parts and 1047297nd each one separately
Te interpretation of integrals as areas causes one inconsistency
with our previous work Consider the integral1
2
1
x dminus
int
Graphically we can see that this area should exist
x
y
minus1minus2
However if we do the integration we 1047297nd that
12
1
2
1
]minus
minus
minus
minus
minus n minus
minusn
1
= ln
1
minus n
Tis is the correct answer (which we could have found usingthe symmetry of the curve) but it goes through a stage wherewe had to take logarithms of negative numbers and this issomething we are not allowed to do We avoid this by rede1047297ning
the integral ofx
as
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17 Basic integration and its applications 587
KEY POINT 174 AGAIN
minusint =
With this de1047297nition we can integrate y =1
over negative
numbers and the integral above becomes
12
1
2
1
x
e oreminus
minusint
n
Notice that the answer is negative since the required area is below
the x -axis We can still not integrate1
with a negative lower and
positive upper limit since the graph has an asymptote at x = 0
You may rightly be alittle uncomfortablewith inserting amodulus function lsquojust
because it worksrsquo Inmathematics do the ends
justify the means
Exercise 17H
1 Find the shaded areas
(a) (i)y =x2
y
x1 2
(ii)y = 1
x2
y
x2 4
(b) (i) y =x2
minus4x+ 3
y
x1 2
(ii)y =x2
minus4
y
x1
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588 Topic 6 Calculus
(c) (i)y = x3 minus x
y
xminus1 2
(ii)y = x2minus3x
y
x
5
2 Te area enclosed by the x -axis the curve andthe line x = k is 18 Find the value of k [6 marks]
3 (a) Find3
int (b) Find the area between the curve minus2 and the
x -axis between x = 0 and x = 3 [5 marks]
4 Between x = 0 and x = 3 the area of the graph y x minus2 below the x -axis equals the area above thex -axis Find the value of k [6 marks]
5 Find the area enclosed by the curve 1x minus minus 0
and the x -axis [7 marks]
lsquo Find t he area
enc losedrsquo means firs t
find a c losed region
bounded b y t he
cur ves men tioned
t hen find i ts area
A s ke tc h is a ver y
use fu l too l
e x a m h i n t
17I The area between a curve andthe y -axis
Consider the diagram alongside How can we 1047297nd the shadedarea A
One possible strategy is to construct a box around the graph todivide up the regions of interest You can integrate to 1047297nd thearea labelled A1 and then by adding and subtracting the areas ofthe blue and red rectangles shown calculate A
y
x
y = f (x)
c
d
A
y
x
y = f (x)
c
d A1
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17 Basic integration and its applications 589
Happily there is a quicker way we can treat x as a functionof y effectively re1047298ecting the whole diagram in the line y = x and then use the same method as in the previous section
KEY POINT 1710
Te area bounded by the curve y the y -axis and the
lines y = c and y = d is given by y d
) dint where is
the expression for x in terms of y
You may have realised that this is related to inverse functions from Section 5E
x
y
x = f (y)
cd
A
Worked example 178
Te curve shown has equation y 1minus Find the shaded area
y
x
y = 2radic x minus 1
10
Express x in terms of y x = 2
rArr = + y
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590 Topic 6 Calculus
Exercise 17I
continued
Find the limits on the y -axisIt may help to label them on the
graph
When x = minusWhen x = 1 = =minus 6
y
y = 2radic minus
1
10
Write down the integral andevaluate using calculator
Area fro GDC= +2
1 2=d
1 Find the shaded areas
(a) (i) y
x
y = x2
1
6
(ii) y
x
y = x3
1
3
(b) (i) y
xy = 1
x2
2
1
(ii) y
x
y =radic x
1
5
(c) (i)y
x
y = ln x
1e
e
(ii) y
x
y = 3x
1 6
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17 Basic integration and its applications 591
2 Te diagram shows the curve If the shadedarea is 504 1047297nd the value of a [6 marks]
y
x
y = radic
x
a
2a
0
3 Find the exact value of the area enclosed by the graphof + the line y = 2 and the y -axis [6 marks]
4 Te diagram shows the graph of
Te shaded area is 39 units Find the value of a [7 marks]y
x
y =radic x
4 a
5 Te diagram shows the graph of 2 where a isin infin Te area of the pink region is equal to the area of the blue
region Give two equations for a in terms of b and hencegive a in exact form and determine the size of theblue area [8 marks]y
x
y = x2
b
1 a
17J The area between two curves
So far we have only looked at areas bounded by a curve and oneof the coordinate axes but we can also 1047297nd areas bounded bytwo curvesTe area A in the diagram can be found by taking the areabounded by f )x and the x -axis and subtracting the areabounded by and the x -axis that is
A
b
minus
y
x
y = f (x)
y = g(x)
a b
A
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592 Topic 6 Calculus
We can do the subtraction before integrating so that we onlyhave to integrate one expression instead of two Tis gives analternative formula for the area
KEY POINT 1711
Te area A between two curves f (x ) and g (x ) is
Ab
(
where a and b are the x-coordinates of the intersectionpoints of the two curves
Worked example 179
Find the area A enclosed between + and minus2 +
First find the x -coordinates ofintersection
For intersection
x x x
rArr =
rArr ) =rArr =
minus
minus
+x
4minusx 0
Make a rough sketch to see therelative positions of the two curves
y
x1 4
A
2x + 1y =
2 minus 3x + 5y = x
Subtract the lower curve from thehigher before integrating
= x
minusx x
4
minus
minus minusx
2
1
4
2
minus =
8
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17 Basic integration and its applications 593
Subtracting the two equations before integrating is particularlyuseful when one of the curves is partly below the x-axis If x is always above then the expression we are integrating f g )x minus )x is always positive so we do not have to worryabout the signs of f )x and g )x themselves
Worked example 1710
Find the area bounded by the curves y x minus and y x 2
Sketch the graph to see therelative position of two
curves
Using GDC
= ex minus 5
y = 3 2
y
minus2 2
A
Find the intersection points ndash
use calculatorintersections x = minus2 658
Write down the integralrepresenting the area
rea =minus
minus818
1 658
xminus
= minus
1 6 8
x
Evaluate the integral usingcalculator = 21 6 (3SF)
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594 Topic 6 Calculus
Exercise 17J
1 Find the shaded areas
(a) (i)
y = 2x+ 1
y = (xminus1)2
y
x
(ii) y =x+ 1
y = 4xminusx2minus1
y
x
(b) (i)
y =minusx2minus4x+ 12
y
x
y =x2 + 2x+ 12
(ii)
y =x2minus2x+ 9
y
x
y = 4xminusx2 + 5
(c) (i)y =x2 minusx
y = 2xminusx2
y
x
(ii)y =x2minus7x+ 7
y = 3minusxminusx2
y
x
2 Find the area enclosed between the graphs of y x +x minus and + [6 marks]
3 Find the area enclosed by the curve 2
the y-axisand the line [6 marks]
4 Find the area between the curves1
and in theregion lt lt π [6 marks]
5 Show that the area of the shaded region alongside is2
[6 marks]
y = x2
y
xminus1 2
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17 Basic integration and its applications 595
6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]
7 Find the total area enclosed between the graphs of
)minus 2 and minus2 7 1+ [6 marks]
8 Te area enclosed between the curve y x 2 and the line
y mx is 10 Find the value of m if m 0 [7 marks]
9 Show that the shaded area in the diagram below is [8 marks]
y = 2 minus xy2 = x
y
x
y = cos x y = sin x
y
x
Summary
bull Integration is the reverse process of differentiation
bull Any integral without limits (inde1047297nite) will generate a constant of integration
bull For all rational n ne minus1 int +x c
1
1
bull If minus we get the natural logarithm function int minus =
bull Te integral of the exponential function is int e c
bull Te integrals of the trigonometric functions are
int si minus=
int tan =
bull Te de1047297nite integral has limits f( ) x b
dint is found by evaluating the integrated expression
at b and then subtracting the integrated expression evaluated at a
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596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
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17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
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7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
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580 Topic 6 Calculus
Te square bracket notation means that the integration hastaken place but the limits have not yet been applied o do thiswe simply evaluate the integrated expression at the upper limitand subtract the integrated expression evaluated at the lowerlimit
You may be wondering where the constant of integration hasgone We could write it in as before but we quickly realise thatthis is unnecessary as it will just cancel out at the upper andlower limit each time
b c
bb
3
3
1
1
1
Te value of x is a dummy variable it does not come intothe answer But both a and b can vary and affect the resultChanging x to a different variable does not change the answerFor example
u a x bb
31 1
int
Worked example 174
Find the exact value of1
1+
Integrate and write in squarebrackets x
+ ]xint
Evaluate the integratedexpression at the upper
and lower limits andsubtract the lower from
the upper
=(In (e) + 4 (e)) minus (In (1) + 4 (1))
= + minusminus =
Ma ke sure you
kno w ho w to e va lua te
de fini te in tegra ls on your
ca lcu la tor as e xp lained
on Ca lcu la tor s ki l ls s hee t
1 0 on t he C D - R OM
I t can sa ve you time
and you can e va lua te
in tegra ls you donrsquo t kno w
ho w to do a lge braica l l y
E ven w hen you are
as ked to find t he e xac t
va lue o f t he in tegra l you
can c hec k your ans wer
on t he ca lcu la tor
e x a m h i n t
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17 Basic integration and its applications 581
Exercise 17G
1 Evaluate the following de1047297nite integrals giving exact answers
(a) (i) x x 2
(ii)1
4
1
(b) (i)2π
int (ii) sπ
π2
(c) (i) ex 1
int (ii) 31
1
eminusint
2 Evaluate correct to three signi1047297cant 1047297gures
(a) (i)3
1 4
int (ii)1
int
(b) (i) e1
int (ii) ne
1int 13 Find the exact value of the integral e x int
π [5 marks]
4 Show that the value of the integral12
x k
k
is independentof k [4 marks]
5 If x ) 73
evaluate 13
[4 marks]
6 Solve the equation 1 [5 marks]
y
x
y = f (x)
a b
A
17H Geometrical significance of definiteintegration
Now we have a method that gives a numerical value for anintegral the natural question to ask is what does this numbermean
On Fill-in proof sheet 20 on the CD-ROM Te fundamentaltheorem of calculus we show that as long as f )x is positivethe de1047297nite integral of f )x between the limits a and b is thearea enclosed between the curve the x -axis and the lines x = a and x = b
KEY POINT 178
rea )x b
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582 Topic 6 Calculus
If you are sketching the graph on the calculator you can get it toshade and evaluate the required area see Calculator skillssheet 10 on the CD-ROM You need to show the sketch as a partof your working if it is not already shown in the question
Worked example 175
Find the exact area enclosed between the x -axis the curve y = sin x and the lines x = 0 and x = π
Sketch the graph and identifythe area required = in
3
Integrate and write in squarebrackets
A = ]= minus
0
3ππ
Evaluate the integratedexpression at the upper andlower limit and subtract the
lower from the upper
= minus )cos cminus minus os
= minus minus ( )minus =
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17 Basic integration and its applications 583
The Ancient Greekshad developed ideasof limiting processessimilar to those used
in calculus but it took nearly2000 years for these ideas
to be formalised This wasdone almost simultaneouslyby Isaac Newton andGottried Leibniz in the 17thCentury Is this a coincidenceor is it often the case that along period of slow progressis often necessary to get tothe stage of majorbreakthroughsSupplementary sheet 10
looks at some other peoplewho can claim to haveinvented calculus
In the 17th Century integration was defined as thearea under a curve The area was broken downinto small rectangles each with a height f (x ) and a
width of a small bit of x called ∆x The total areawas approximately the sum of all of these rectangles
x a
x
f x x sum )∆
Isaac Newton one of the pioneers of calculus was also abig fan of writing in English rather than Greek So sigmabecame the English letter lsquoSrsquo and delta became the Englishletter d so that when the limit is taken as the width of therectangles become vanishingly small then the expressionbecomes
x a
)x dint This illustrates another very important interpretation of
integration ndash the infinite sum of infinitesimally small parts
Worked example 176
Find the area A in this graphy
x
y = x(xminus1)(2minusx)
1 2A
0
Write down the integral to beevaluated then use calculator
x x xminusint 1
x = minus y D
The area must be positive there4 =
When the curve is entirely below the x -axis the integral will givea negative value Te modulus of this value is the area
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584 Topic 6 Calculus
Unfortunately the relationship between integrals and areas isnot so simple when there are parts of the curve above and belowthe axis Tose bits above the axis contribute positively to thearea but bits below the axis contribute negatively to the areaWe must separate out the sections above the axis and belowthe axis
Worked example 177
(a) Find x x 1
4
x int 1 d
(b) Find the area enclosed between the x -axis the curve y minus x + and the lines= 1 and 4
Apply standard integration (a) x x
4
xminus1
= ( ( ) ( ( )minus + minus +
= minus =3
4
The value found above canrsquot bethe correct area for (b)
Sketch the curve to see exactlywhich area we are being asked
to find
(b) y
2 minus + 3
1 3
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17 Basic integration and its applications 585
Te fact that the integral was zero in Worked example 177 part(a) means that the area above the axis is exactly cancelled by thearea below the axis
Tis example warns us that when asked to 1047297nd an area we mustalways sketch the graph and identify exactly where each partof the area is If we are evaluating the area on the calculator wecan use the modulus function to ensure that the entire graph isabove the x -axis Using the function from Worked example 177
1
4
int 1y
x
y = |x2 minus 4x + 3|
1 3 4
continued
The area is made up of twoparts so evaluate each of
them separately
x
1
33
xminusint
( ) minus = minus
there4 rea below the axis is
x x3
4
2xminus int
minus ( ) =
4
there4 rea above the axis is
Total area = + =3 3
Transformationsof graphs using the
modulus function
were covered inchapter 7
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586 Topic 6 Calculus
KEY POINT 179
Te area bounded by the curve y f )x the x-axis and the
lines x = a and x = b is given byb
x int
When working without a calculator if the curve crosses
the x -axis between a and b we need to split the area intoseveral parts and 1047297nd each one separately
Te interpretation of integrals as areas causes one inconsistency
with our previous work Consider the integral1
2
1
x dminus
int
Graphically we can see that this area should exist
x
y
minus1minus2
However if we do the integration we 1047297nd that
12
1
2
1
]minus
minus
minus
minus
minus n minus
minusn
1
= ln
1
minus n
Tis is the correct answer (which we could have found usingthe symmetry of the curve) but it goes through a stage wherewe had to take logarithms of negative numbers and this issomething we are not allowed to do We avoid this by rede1047297ning
the integral ofx
as
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17 Basic integration and its applications 587
KEY POINT 174 AGAIN
minusint =
With this de1047297nition we can integrate y =1
over negative
numbers and the integral above becomes
12
1
2
1
x
e oreminus
minusint
n
Notice that the answer is negative since the required area is below
the x -axis We can still not integrate1
with a negative lower and
positive upper limit since the graph has an asymptote at x = 0
You may rightly be alittle uncomfortablewith inserting amodulus function lsquojust
because it worksrsquo Inmathematics do the ends
justify the means
Exercise 17H
1 Find the shaded areas
(a) (i)y =x2
y
x1 2
(ii)y = 1
x2
y
x2 4
(b) (i) y =x2
minus4x+ 3
y
x1 2
(ii)y =x2
minus4
y
x1
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588 Topic 6 Calculus
(c) (i)y = x3 minus x
y
xminus1 2
(ii)y = x2minus3x
y
x
5
2 Te area enclosed by the x -axis the curve andthe line x = k is 18 Find the value of k [6 marks]
3 (a) Find3
int (b) Find the area between the curve minus2 and the
x -axis between x = 0 and x = 3 [5 marks]
4 Between x = 0 and x = 3 the area of the graph y x minus2 below the x -axis equals the area above thex -axis Find the value of k [6 marks]
5 Find the area enclosed by the curve 1x minus minus 0
and the x -axis [7 marks]
lsquo Find t he area
enc losedrsquo means firs t
find a c losed region
bounded b y t he
cur ves men tioned
t hen find i ts area
A s ke tc h is a ver y
use fu l too l
e x a m h i n t
17I The area between a curve andthe y -axis
Consider the diagram alongside How can we 1047297nd the shadedarea A
One possible strategy is to construct a box around the graph todivide up the regions of interest You can integrate to 1047297nd thearea labelled A1 and then by adding and subtracting the areas ofthe blue and red rectangles shown calculate A
y
x
y = f (x)
c
d
A
y
x
y = f (x)
c
d A1
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17 Basic integration and its applications 589
Happily there is a quicker way we can treat x as a functionof y effectively re1047298ecting the whole diagram in the line y = x and then use the same method as in the previous section
KEY POINT 1710
Te area bounded by the curve y the y -axis and the
lines y = c and y = d is given by y d
) dint where is
the expression for x in terms of y
You may have realised that this is related to inverse functions from Section 5E
x
y
x = f (y)
cd
A
Worked example 178
Te curve shown has equation y 1minus Find the shaded area
y
x
y = 2radic x minus 1
10
Express x in terms of y x = 2
rArr = + y
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590 Topic 6 Calculus
Exercise 17I
continued
Find the limits on the y -axisIt may help to label them on the
graph
When x = minusWhen x = 1 = =minus 6
y
y = 2radic minus
1
10
Write down the integral andevaluate using calculator
Area fro GDC= +2
1 2=d
1 Find the shaded areas
(a) (i) y
x
y = x2
1
6
(ii) y
x
y = x3
1
3
(b) (i) y
xy = 1
x2
2
1
(ii) y
x
y =radic x
1
5
(c) (i)y
x
y = ln x
1e
e
(ii) y
x
y = 3x
1 6
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17 Basic integration and its applications 591
2 Te diagram shows the curve If the shadedarea is 504 1047297nd the value of a [6 marks]
y
x
y = radic
x
a
2a
0
3 Find the exact value of the area enclosed by the graphof + the line y = 2 and the y -axis [6 marks]
4 Te diagram shows the graph of
Te shaded area is 39 units Find the value of a [7 marks]y
x
y =radic x
4 a
5 Te diagram shows the graph of 2 where a isin infin Te area of the pink region is equal to the area of the blue
region Give two equations for a in terms of b and hencegive a in exact form and determine the size of theblue area [8 marks]y
x
y = x2
b
1 a
17J The area between two curves
So far we have only looked at areas bounded by a curve and oneof the coordinate axes but we can also 1047297nd areas bounded bytwo curvesTe area A in the diagram can be found by taking the areabounded by f )x and the x -axis and subtracting the areabounded by and the x -axis that is
A
b
minus
y
x
y = f (x)
y = g(x)
a b
A
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592 Topic 6 Calculus
We can do the subtraction before integrating so that we onlyhave to integrate one expression instead of two Tis gives analternative formula for the area
KEY POINT 1711
Te area A between two curves f (x ) and g (x ) is
Ab
(
where a and b are the x-coordinates of the intersectionpoints of the two curves
Worked example 179
Find the area A enclosed between + and minus2 +
First find the x -coordinates ofintersection
For intersection
x x x
rArr =
rArr ) =rArr =
minus
minus
+x
4minusx 0
Make a rough sketch to see therelative positions of the two curves
y
x1 4
A
2x + 1y =
2 minus 3x + 5y = x
Subtract the lower curve from thehigher before integrating
= x
minusx x
4
minus
minus minusx
2
1
4
2
minus =
8
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17 Basic integration and its applications 593
Subtracting the two equations before integrating is particularlyuseful when one of the curves is partly below the x-axis If x is always above then the expression we are integrating f g )x minus )x is always positive so we do not have to worryabout the signs of f )x and g )x themselves
Worked example 1710
Find the area bounded by the curves y x minus and y x 2
Sketch the graph to see therelative position of two
curves
Using GDC
= ex minus 5
y = 3 2
y
minus2 2
A
Find the intersection points ndash
use calculatorintersections x = minus2 658
Write down the integralrepresenting the area
rea =minus
minus818
1 658
xminus
= minus
1 6 8
x
Evaluate the integral usingcalculator = 21 6 (3SF)
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594 Topic 6 Calculus
Exercise 17J
1 Find the shaded areas
(a) (i)
y = 2x+ 1
y = (xminus1)2
y
x
(ii) y =x+ 1
y = 4xminusx2minus1
y
x
(b) (i)
y =minusx2minus4x+ 12
y
x
y =x2 + 2x+ 12
(ii)
y =x2minus2x+ 9
y
x
y = 4xminusx2 + 5
(c) (i)y =x2 minusx
y = 2xminusx2
y
x
(ii)y =x2minus7x+ 7
y = 3minusxminusx2
y
x
2 Find the area enclosed between the graphs of y x +x minus and + [6 marks]
3 Find the area enclosed by the curve 2
the y-axisand the line [6 marks]
4 Find the area between the curves1
and in theregion lt lt π [6 marks]
5 Show that the area of the shaded region alongside is2
[6 marks]
y = x2
y
xminus1 2
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17 Basic integration and its applications 595
6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]
7 Find the total area enclosed between the graphs of
)minus 2 and minus2 7 1+ [6 marks]
8 Te area enclosed between the curve y x 2 and the line
y mx is 10 Find the value of m if m 0 [7 marks]
9 Show that the shaded area in the diagram below is [8 marks]
y = 2 minus xy2 = x
y
x
y = cos x y = sin x
y
x
Summary
bull Integration is the reverse process of differentiation
bull Any integral without limits (inde1047297nite) will generate a constant of integration
bull For all rational n ne minus1 int +x c
1
1
bull If minus we get the natural logarithm function int minus =
bull Te integral of the exponential function is int e c
bull Te integrals of the trigonometric functions are
int si minus=
int tan =
bull Te de1047297nite integral has limits f( ) x b
dint is found by evaluating the integrated expression
at b and then subtracting the integrated expression evaluated at a
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7242019 Chapter 17 IB Maths HL Cambridge Textbook
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596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
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17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
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7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
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17 Basic integration and its applications 581
Exercise 17G
1 Evaluate the following de1047297nite integrals giving exact answers
(a) (i) x x 2
(ii)1
4
1
(b) (i)2π
int (ii) sπ
π2
(c) (i) ex 1
int (ii) 31
1
eminusint
2 Evaluate correct to three signi1047297cant 1047297gures
(a) (i)3
1 4
int (ii)1
int
(b) (i) e1
int (ii) ne
1int 13 Find the exact value of the integral e x int
π [5 marks]
4 Show that the value of the integral12
x k
k
is independentof k [4 marks]
5 If x ) 73
evaluate 13
[4 marks]
6 Solve the equation 1 [5 marks]
y
x
y = f (x)
a b
A
17H Geometrical significance of definiteintegration
Now we have a method that gives a numerical value for anintegral the natural question to ask is what does this numbermean
On Fill-in proof sheet 20 on the CD-ROM Te fundamentaltheorem of calculus we show that as long as f )x is positivethe de1047297nite integral of f )x between the limits a and b is thearea enclosed between the curve the x -axis and the lines x = a and x = b
KEY POINT 178
rea )x b
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582 Topic 6 Calculus
If you are sketching the graph on the calculator you can get it toshade and evaluate the required area see Calculator skillssheet 10 on the CD-ROM You need to show the sketch as a partof your working if it is not already shown in the question
Worked example 175
Find the exact area enclosed between the x -axis the curve y = sin x and the lines x = 0 and x = π
Sketch the graph and identifythe area required = in
3
Integrate and write in squarebrackets
A = ]= minus
0
3ππ
Evaluate the integratedexpression at the upper andlower limit and subtract the
lower from the upper
= minus )cos cminus minus os
= minus minus ( )minus =
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17 Basic integration and its applications 583
The Ancient Greekshad developed ideasof limiting processessimilar to those used
in calculus but it took nearly2000 years for these ideas
to be formalised This wasdone almost simultaneouslyby Isaac Newton andGottried Leibniz in the 17thCentury Is this a coincidenceor is it often the case that along period of slow progressis often necessary to get tothe stage of majorbreakthroughsSupplementary sheet 10
looks at some other peoplewho can claim to haveinvented calculus
In the 17th Century integration was defined as thearea under a curve The area was broken downinto small rectangles each with a height f (x ) and a
width of a small bit of x called ∆x The total areawas approximately the sum of all of these rectangles
x a
x
f x x sum )∆
Isaac Newton one of the pioneers of calculus was also abig fan of writing in English rather than Greek So sigmabecame the English letter lsquoSrsquo and delta became the Englishletter d so that when the limit is taken as the width of therectangles become vanishingly small then the expressionbecomes
x a
)x dint This illustrates another very important interpretation of
integration ndash the infinite sum of infinitesimally small parts
Worked example 176
Find the area A in this graphy
x
y = x(xminus1)(2minusx)
1 2A
0
Write down the integral to beevaluated then use calculator
x x xminusint 1
x = minus y D
The area must be positive there4 =
When the curve is entirely below the x -axis the integral will givea negative value Te modulus of this value is the area
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584 Topic 6 Calculus
Unfortunately the relationship between integrals and areas isnot so simple when there are parts of the curve above and belowthe axis Tose bits above the axis contribute positively to thearea but bits below the axis contribute negatively to the areaWe must separate out the sections above the axis and belowthe axis
Worked example 177
(a) Find x x 1
4
x int 1 d
(b) Find the area enclosed between the x -axis the curve y minus x + and the lines= 1 and 4
Apply standard integration (a) x x
4
xminus1
= ( ( ) ( ( )minus + minus +
= minus =3
4
The value found above canrsquot bethe correct area for (b)
Sketch the curve to see exactlywhich area we are being asked
to find
(b) y
2 minus + 3
1 3
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17 Basic integration and its applications 585
Te fact that the integral was zero in Worked example 177 part(a) means that the area above the axis is exactly cancelled by thearea below the axis
Tis example warns us that when asked to 1047297nd an area we mustalways sketch the graph and identify exactly where each partof the area is If we are evaluating the area on the calculator wecan use the modulus function to ensure that the entire graph isabove the x -axis Using the function from Worked example 177
1
4
int 1y
x
y = |x2 minus 4x + 3|
1 3 4
continued
The area is made up of twoparts so evaluate each of
them separately
x
1
33
xminusint
( ) minus = minus
there4 rea below the axis is
x x3
4
2xminus int
minus ( ) =
4
there4 rea above the axis is
Total area = + =3 3
Transformationsof graphs using the
modulus function
were covered inchapter 7
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586 Topic 6 Calculus
KEY POINT 179
Te area bounded by the curve y f )x the x-axis and the
lines x = a and x = b is given byb
x int
When working without a calculator if the curve crosses
the x -axis between a and b we need to split the area intoseveral parts and 1047297nd each one separately
Te interpretation of integrals as areas causes one inconsistency
with our previous work Consider the integral1
2
1
x dminus
int
Graphically we can see that this area should exist
x
y
minus1minus2
However if we do the integration we 1047297nd that
12
1
2
1
]minus
minus
minus
minus
minus n minus
minusn
1
= ln
1
minus n
Tis is the correct answer (which we could have found usingthe symmetry of the curve) but it goes through a stage wherewe had to take logarithms of negative numbers and this issomething we are not allowed to do We avoid this by rede1047297ning
the integral ofx
as
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17 Basic integration and its applications 587
KEY POINT 174 AGAIN
minusint =
With this de1047297nition we can integrate y =1
over negative
numbers and the integral above becomes
12
1
2
1
x
e oreminus
minusint
n
Notice that the answer is negative since the required area is below
the x -axis We can still not integrate1
with a negative lower and
positive upper limit since the graph has an asymptote at x = 0
You may rightly be alittle uncomfortablewith inserting amodulus function lsquojust
because it worksrsquo Inmathematics do the ends
justify the means
Exercise 17H
1 Find the shaded areas
(a) (i)y =x2
y
x1 2
(ii)y = 1
x2
y
x2 4
(b) (i) y =x2
minus4x+ 3
y
x1 2
(ii)y =x2
minus4
y
x1
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588 Topic 6 Calculus
(c) (i)y = x3 minus x
y
xminus1 2
(ii)y = x2minus3x
y
x
5
2 Te area enclosed by the x -axis the curve andthe line x = k is 18 Find the value of k [6 marks]
3 (a) Find3
int (b) Find the area between the curve minus2 and the
x -axis between x = 0 and x = 3 [5 marks]
4 Between x = 0 and x = 3 the area of the graph y x minus2 below the x -axis equals the area above thex -axis Find the value of k [6 marks]
5 Find the area enclosed by the curve 1x minus minus 0
and the x -axis [7 marks]
lsquo Find t he area
enc losedrsquo means firs t
find a c losed region
bounded b y t he
cur ves men tioned
t hen find i ts area
A s ke tc h is a ver y
use fu l too l
e x a m h i n t
17I The area between a curve andthe y -axis
Consider the diagram alongside How can we 1047297nd the shadedarea A
One possible strategy is to construct a box around the graph todivide up the regions of interest You can integrate to 1047297nd thearea labelled A1 and then by adding and subtracting the areas ofthe blue and red rectangles shown calculate A
y
x
y = f (x)
c
d
A
y
x
y = f (x)
c
d A1
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17 Basic integration and its applications 589
Happily there is a quicker way we can treat x as a functionof y effectively re1047298ecting the whole diagram in the line y = x and then use the same method as in the previous section
KEY POINT 1710
Te area bounded by the curve y the y -axis and the
lines y = c and y = d is given by y d
) dint where is
the expression for x in terms of y
You may have realised that this is related to inverse functions from Section 5E
x
y
x = f (y)
cd
A
Worked example 178
Te curve shown has equation y 1minus Find the shaded area
y
x
y = 2radic x minus 1
10
Express x in terms of y x = 2
rArr = + y
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590 Topic 6 Calculus
Exercise 17I
continued
Find the limits on the y -axisIt may help to label them on the
graph
When x = minusWhen x = 1 = =minus 6
y
y = 2radic minus
1
10
Write down the integral andevaluate using calculator
Area fro GDC= +2
1 2=d
1 Find the shaded areas
(a) (i) y
x
y = x2
1
6
(ii) y
x
y = x3
1
3
(b) (i) y
xy = 1
x2
2
1
(ii) y
x
y =radic x
1
5
(c) (i)y
x
y = ln x
1e
e
(ii) y
x
y = 3x
1 6
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17 Basic integration and its applications 591
2 Te diagram shows the curve If the shadedarea is 504 1047297nd the value of a [6 marks]
y
x
y = radic
x
a
2a
0
3 Find the exact value of the area enclosed by the graphof + the line y = 2 and the y -axis [6 marks]
4 Te diagram shows the graph of
Te shaded area is 39 units Find the value of a [7 marks]y
x
y =radic x
4 a
5 Te diagram shows the graph of 2 where a isin infin Te area of the pink region is equal to the area of the blue
region Give two equations for a in terms of b and hencegive a in exact form and determine the size of theblue area [8 marks]y
x
y = x2
b
1 a
17J The area between two curves
So far we have only looked at areas bounded by a curve and oneof the coordinate axes but we can also 1047297nd areas bounded bytwo curvesTe area A in the diagram can be found by taking the areabounded by f )x and the x -axis and subtracting the areabounded by and the x -axis that is
A
b
minus
y
x
y = f (x)
y = g(x)
a b
A
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592 Topic 6 Calculus
We can do the subtraction before integrating so that we onlyhave to integrate one expression instead of two Tis gives analternative formula for the area
KEY POINT 1711
Te area A between two curves f (x ) and g (x ) is
Ab
(
where a and b are the x-coordinates of the intersectionpoints of the two curves
Worked example 179
Find the area A enclosed between + and minus2 +
First find the x -coordinates ofintersection
For intersection
x x x
rArr =
rArr ) =rArr =
minus
minus
+x
4minusx 0
Make a rough sketch to see therelative positions of the two curves
y
x1 4
A
2x + 1y =
2 minus 3x + 5y = x
Subtract the lower curve from thehigher before integrating
= x
minusx x
4
minus
minus minusx
2
1
4
2
minus =
8
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17 Basic integration and its applications 593
Subtracting the two equations before integrating is particularlyuseful when one of the curves is partly below the x-axis If x is always above then the expression we are integrating f g )x minus )x is always positive so we do not have to worryabout the signs of f )x and g )x themselves
Worked example 1710
Find the area bounded by the curves y x minus and y x 2
Sketch the graph to see therelative position of two
curves
Using GDC
= ex minus 5
y = 3 2
y
minus2 2
A
Find the intersection points ndash
use calculatorintersections x = minus2 658
Write down the integralrepresenting the area
rea =minus
minus818
1 658
xminus
= minus
1 6 8
x
Evaluate the integral usingcalculator = 21 6 (3SF)
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594 Topic 6 Calculus
Exercise 17J
1 Find the shaded areas
(a) (i)
y = 2x+ 1
y = (xminus1)2
y
x
(ii) y =x+ 1
y = 4xminusx2minus1
y
x
(b) (i)
y =minusx2minus4x+ 12
y
x
y =x2 + 2x+ 12
(ii)
y =x2minus2x+ 9
y
x
y = 4xminusx2 + 5
(c) (i)y =x2 minusx
y = 2xminusx2
y
x
(ii)y =x2minus7x+ 7
y = 3minusxminusx2
y
x
2 Find the area enclosed between the graphs of y x +x minus and + [6 marks]
3 Find the area enclosed by the curve 2
the y-axisand the line [6 marks]
4 Find the area between the curves1
and in theregion lt lt π [6 marks]
5 Show that the area of the shaded region alongside is2
[6 marks]
y = x2
y
xminus1 2
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Not for printing sharing or distribution
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17 Basic integration and its applications 595
6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]
7 Find the total area enclosed between the graphs of
)minus 2 and minus2 7 1+ [6 marks]
8 Te area enclosed between the curve y x 2 and the line
y mx is 10 Find the value of m if m 0 [7 marks]
9 Show that the shaded area in the diagram below is [8 marks]
y = 2 minus xy2 = x
y
x
y = cos x y = sin x
y
x
Summary
bull Integration is the reverse process of differentiation
bull Any integral without limits (inde1047297nite) will generate a constant of integration
bull For all rational n ne minus1 int +x c
1
1
bull If minus we get the natural logarithm function int minus =
bull Te integral of the exponential function is int e c
bull Te integrals of the trigonometric functions are
int si minus=
int tan =
bull Te de1047297nite integral has limits f( ) x b
dint is found by evaluating the integrated expression
at b and then subtracting the integrated expression evaluated at a
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596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
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Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
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7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
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582 Topic 6 Calculus
If you are sketching the graph on the calculator you can get it toshade and evaluate the required area see Calculator skillssheet 10 on the CD-ROM You need to show the sketch as a partof your working if it is not already shown in the question
Worked example 175
Find the exact area enclosed between the x -axis the curve y = sin x and the lines x = 0 and x = π
Sketch the graph and identifythe area required = in
3
Integrate and write in squarebrackets
A = ]= minus
0
3ππ
Evaluate the integratedexpression at the upper andlower limit and subtract the
lower from the upper
= minus )cos cminus minus os
= minus minus ( )minus =
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Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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17 Basic integration and its applications 583
The Ancient Greekshad developed ideasof limiting processessimilar to those used
in calculus but it took nearly2000 years for these ideas
to be formalised This wasdone almost simultaneouslyby Isaac Newton andGottried Leibniz in the 17thCentury Is this a coincidenceor is it often the case that along period of slow progressis often necessary to get tothe stage of majorbreakthroughsSupplementary sheet 10
looks at some other peoplewho can claim to haveinvented calculus
In the 17th Century integration was defined as thearea under a curve The area was broken downinto small rectangles each with a height f (x ) and a
width of a small bit of x called ∆x The total areawas approximately the sum of all of these rectangles
x a
x
f x x sum )∆
Isaac Newton one of the pioneers of calculus was also abig fan of writing in English rather than Greek So sigmabecame the English letter lsquoSrsquo and delta became the Englishletter d so that when the limit is taken as the width of therectangles become vanishingly small then the expressionbecomes
x a
)x dint This illustrates another very important interpretation of
integration ndash the infinite sum of infinitesimally small parts
Worked example 176
Find the area A in this graphy
x
y = x(xminus1)(2minusx)
1 2A
0
Write down the integral to beevaluated then use calculator
x x xminusint 1
x = minus y D
The area must be positive there4 =
When the curve is entirely below the x -axis the integral will givea negative value Te modulus of this value is the area
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Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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584 Topic 6 Calculus
Unfortunately the relationship between integrals and areas isnot so simple when there are parts of the curve above and belowthe axis Tose bits above the axis contribute positively to thearea but bits below the axis contribute negatively to the areaWe must separate out the sections above the axis and belowthe axis
Worked example 177
(a) Find x x 1
4
x int 1 d
(b) Find the area enclosed between the x -axis the curve y minus x + and the lines= 1 and 4
Apply standard integration (a) x x
4
xminus1
= ( ( ) ( ( )minus + minus +
= minus =3
4
The value found above canrsquot bethe correct area for (b)
Sketch the curve to see exactlywhich area we are being asked
to find
(b) y
2 minus + 3
1 3
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Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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17 Basic integration and its applications 585
Te fact that the integral was zero in Worked example 177 part(a) means that the area above the axis is exactly cancelled by thearea below the axis
Tis example warns us that when asked to 1047297nd an area we mustalways sketch the graph and identify exactly where each partof the area is If we are evaluating the area on the calculator wecan use the modulus function to ensure that the entire graph isabove the x -axis Using the function from Worked example 177
1
4
int 1y
x
y = |x2 minus 4x + 3|
1 3 4
continued
The area is made up of twoparts so evaluate each of
them separately
x
1
33
xminusint
( ) minus = minus
there4 rea below the axis is
x x3
4
2xminus int
minus ( ) =
4
there4 rea above the axis is
Total area = + =3 3
Transformationsof graphs using the
modulus function
were covered inchapter 7
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7242019 Chapter 17 IB Maths HL Cambridge Textbook
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586 Topic 6 Calculus
KEY POINT 179
Te area bounded by the curve y f )x the x-axis and the
lines x = a and x = b is given byb
x int
When working without a calculator if the curve crosses
the x -axis between a and b we need to split the area intoseveral parts and 1047297nd each one separately
Te interpretation of integrals as areas causes one inconsistency
with our previous work Consider the integral1
2
1
x dminus
int
Graphically we can see that this area should exist
x
y
minus1minus2
However if we do the integration we 1047297nd that
12
1
2
1
]minus
minus
minus
minus
minus n minus
minusn
1
= ln
1
minus n
Tis is the correct answer (which we could have found usingthe symmetry of the curve) but it goes through a stage wherewe had to take logarithms of negative numbers and this issomething we are not allowed to do We avoid this by rede1047297ning
the integral ofx
as
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17 Basic integration and its applications 587
KEY POINT 174 AGAIN
minusint =
With this de1047297nition we can integrate y =1
over negative
numbers and the integral above becomes
12
1
2
1
x
e oreminus
minusint
n
Notice that the answer is negative since the required area is below
the x -axis We can still not integrate1
with a negative lower and
positive upper limit since the graph has an asymptote at x = 0
You may rightly be alittle uncomfortablewith inserting amodulus function lsquojust
because it worksrsquo Inmathematics do the ends
justify the means
Exercise 17H
1 Find the shaded areas
(a) (i)y =x2
y
x1 2
(ii)y = 1
x2
y
x2 4
(b) (i) y =x2
minus4x+ 3
y
x1 2
(ii)y =x2
minus4
y
x1
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588 Topic 6 Calculus
(c) (i)y = x3 minus x
y
xminus1 2
(ii)y = x2minus3x
y
x
5
2 Te area enclosed by the x -axis the curve andthe line x = k is 18 Find the value of k [6 marks]
3 (a) Find3
int (b) Find the area between the curve minus2 and the
x -axis between x = 0 and x = 3 [5 marks]
4 Between x = 0 and x = 3 the area of the graph y x minus2 below the x -axis equals the area above thex -axis Find the value of k [6 marks]
5 Find the area enclosed by the curve 1x minus minus 0
and the x -axis [7 marks]
lsquo Find t he area
enc losedrsquo means firs t
find a c losed region
bounded b y t he
cur ves men tioned
t hen find i ts area
A s ke tc h is a ver y
use fu l too l
e x a m h i n t
17I The area between a curve andthe y -axis
Consider the diagram alongside How can we 1047297nd the shadedarea A
One possible strategy is to construct a box around the graph todivide up the regions of interest You can integrate to 1047297nd thearea labelled A1 and then by adding and subtracting the areas ofthe blue and red rectangles shown calculate A
y
x
y = f (x)
c
d
A
y
x
y = f (x)
c
d A1
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17 Basic integration and its applications 589
Happily there is a quicker way we can treat x as a functionof y effectively re1047298ecting the whole diagram in the line y = x and then use the same method as in the previous section
KEY POINT 1710
Te area bounded by the curve y the y -axis and the
lines y = c and y = d is given by y d
) dint where is
the expression for x in terms of y
You may have realised that this is related to inverse functions from Section 5E
x
y
x = f (y)
cd
A
Worked example 178
Te curve shown has equation y 1minus Find the shaded area
y
x
y = 2radic x minus 1
10
Express x in terms of y x = 2
rArr = + y
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590 Topic 6 Calculus
Exercise 17I
continued
Find the limits on the y -axisIt may help to label them on the
graph
When x = minusWhen x = 1 = =minus 6
y
y = 2radic minus
1
10
Write down the integral andevaluate using calculator
Area fro GDC= +2
1 2=d
1 Find the shaded areas
(a) (i) y
x
y = x2
1
6
(ii) y
x
y = x3
1
3
(b) (i) y
xy = 1
x2
2
1
(ii) y
x
y =radic x
1
5
(c) (i)y
x
y = ln x
1e
e
(ii) y
x
y = 3x
1 6
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17 Basic integration and its applications 591
2 Te diagram shows the curve If the shadedarea is 504 1047297nd the value of a [6 marks]
y
x
y = radic
x
a
2a
0
3 Find the exact value of the area enclosed by the graphof + the line y = 2 and the y -axis [6 marks]
4 Te diagram shows the graph of
Te shaded area is 39 units Find the value of a [7 marks]y
x
y =radic x
4 a
5 Te diagram shows the graph of 2 where a isin infin Te area of the pink region is equal to the area of the blue
region Give two equations for a in terms of b and hencegive a in exact form and determine the size of theblue area [8 marks]y
x
y = x2
b
1 a
17J The area between two curves
So far we have only looked at areas bounded by a curve and oneof the coordinate axes but we can also 1047297nd areas bounded bytwo curvesTe area A in the diagram can be found by taking the areabounded by f )x and the x -axis and subtracting the areabounded by and the x -axis that is
A
b
minus
y
x
y = f (x)
y = g(x)
a b
A
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592 Topic 6 Calculus
We can do the subtraction before integrating so that we onlyhave to integrate one expression instead of two Tis gives analternative formula for the area
KEY POINT 1711
Te area A between two curves f (x ) and g (x ) is
Ab
(
where a and b are the x-coordinates of the intersectionpoints of the two curves
Worked example 179
Find the area A enclosed between + and minus2 +
First find the x -coordinates ofintersection
For intersection
x x x
rArr =
rArr ) =rArr =
minus
minus
+x
4minusx 0
Make a rough sketch to see therelative positions of the two curves
y
x1 4
A
2x + 1y =
2 minus 3x + 5y = x
Subtract the lower curve from thehigher before integrating
= x
minusx x
4
minus
minus minusx
2
1
4
2
minus =
8
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17 Basic integration and its applications 593
Subtracting the two equations before integrating is particularlyuseful when one of the curves is partly below the x-axis If x is always above then the expression we are integrating f g )x minus )x is always positive so we do not have to worryabout the signs of f )x and g )x themselves
Worked example 1710
Find the area bounded by the curves y x minus and y x 2
Sketch the graph to see therelative position of two
curves
Using GDC
= ex minus 5
y = 3 2
y
minus2 2
A
Find the intersection points ndash
use calculatorintersections x = minus2 658
Write down the integralrepresenting the area
rea =minus
minus818
1 658
xminus
= minus
1 6 8
x
Evaluate the integral usingcalculator = 21 6 (3SF)
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594 Topic 6 Calculus
Exercise 17J
1 Find the shaded areas
(a) (i)
y = 2x+ 1
y = (xminus1)2
y
x
(ii) y =x+ 1
y = 4xminusx2minus1
y
x
(b) (i)
y =minusx2minus4x+ 12
y
x
y =x2 + 2x+ 12
(ii)
y =x2minus2x+ 9
y
x
y = 4xminusx2 + 5
(c) (i)y =x2 minusx
y = 2xminusx2
y
x
(ii)y =x2minus7x+ 7
y = 3minusxminusx2
y
x
2 Find the area enclosed between the graphs of y x +x minus and + [6 marks]
3 Find the area enclosed by the curve 2
the y-axisand the line [6 marks]
4 Find the area between the curves1
and in theregion lt lt π [6 marks]
5 Show that the area of the shaded region alongside is2
[6 marks]
y = x2
y
xminus1 2
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17 Basic integration and its applications 595
6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]
7 Find the total area enclosed between the graphs of
)minus 2 and minus2 7 1+ [6 marks]
8 Te area enclosed between the curve y x 2 and the line
y mx is 10 Find the value of m if m 0 [7 marks]
9 Show that the shaded area in the diagram below is [8 marks]
y = 2 minus xy2 = x
y
x
y = cos x y = sin x
y
x
Summary
bull Integration is the reverse process of differentiation
bull Any integral without limits (inde1047297nite) will generate a constant of integration
bull For all rational n ne minus1 int +x c
1
1
bull If minus we get the natural logarithm function int minus =
bull Te integral of the exponential function is int e c
bull Te integrals of the trigonometric functions are
int si minus=
int tan =
bull Te de1047297nite integral has limits f( ) x b
dint is found by evaluating the integrated expression
at b and then subtracting the integrated expression evaluated at a
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596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
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17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
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7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
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17 Basic integration and its applications 583
The Ancient Greekshad developed ideasof limiting processessimilar to those used
in calculus but it took nearly2000 years for these ideas
to be formalised This wasdone almost simultaneouslyby Isaac Newton andGottried Leibniz in the 17thCentury Is this a coincidenceor is it often the case that along period of slow progressis often necessary to get tothe stage of majorbreakthroughsSupplementary sheet 10
looks at some other peoplewho can claim to haveinvented calculus
In the 17th Century integration was defined as thearea under a curve The area was broken downinto small rectangles each with a height f (x ) and a
width of a small bit of x called ∆x The total areawas approximately the sum of all of these rectangles
x a
x
f x x sum )∆
Isaac Newton one of the pioneers of calculus was also abig fan of writing in English rather than Greek So sigmabecame the English letter lsquoSrsquo and delta became the Englishletter d so that when the limit is taken as the width of therectangles become vanishingly small then the expressionbecomes
x a
)x dint This illustrates another very important interpretation of
integration ndash the infinite sum of infinitesimally small parts
Worked example 176
Find the area A in this graphy
x
y = x(xminus1)(2minusx)
1 2A
0
Write down the integral to beevaluated then use calculator
x x xminusint 1
x = minus y D
The area must be positive there4 =
When the curve is entirely below the x -axis the integral will givea negative value Te modulus of this value is the area
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584 Topic 6 Calculus
Unfortunately the relationship between integrals and areas isnot so simple when there are parts of the curve above and belowthe axis Tose bits above the axis contribute positively to thearea but bits below the axis contribute negatively to the areaWe must separate out the sections above the axis and belowthe axis
Worked example 177
(a) Find x x 1
4
x int 1 d
(b) Find the area enclosed between the x -axis the curve y minus x + and the lines= 1 and 4
Apply standard integration (a) x x
4
xminus1
= ( ( ) ( ( )minus + minus +
= minus =3
4
The value found above canrsquot bethe correct area for (b)
Sketch the curve to see exactlywhich area we are being asked
to find
(b) y
2 minus + 3
1 3
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17 Basic integration and its applications 585
Te fact that the integral was zero in Worked example 177 part(a) means that the area above the axis is exactly cancelled by thearea below the axis
Tis example warns us that when asked to 1047297nd an area we mustalways sketch the graph and identify exactly where each partof the area is If we are evaluating the area on the calculator wecan use the modulus function to ensure that the entire graph isabove the x -axis Using the function from Worked example 177
1
4
int 1y
x
y = |x2 minus 4x + 3|
1 3 4
continued
The area is made up of twoparts so evaluate each of
them separately
x
1
33
xminusint
( ) minus = minus
there4 rea below the axis is
x x3
4
2xminus int
minus ( ) =
4
there4 rea above the axis is
Total area = + =3 3
Transformationsof graphs using the
modulus function
were covered inchapter 7
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586 Topic 6 Calculus
KEY POINT 179
Te area bounded by the curve y f )x the x-axis and the
lines x = a and x = b is given byb
x int
When working without a calculator if the curve crosses
the x -axis between a and b we need to split the area intoseveral parts and 1047297nd each one separately
Te interpretation of integrals as areas causes one inconsistency
with our previous work Consider the integral1
2
1
x dminus
int
Graphically we can see that this area should exist
x
y
minus1minus2
However if we do the integration we 1047297nd that
12
1
2
1
]minus
minus
minus
minus
minus n minus
minusn
1
= ln
1
minus n
Tis is the correct answer (which we could have found usingthe symmetry of the curve) but it goes through a stage wherewe had to take logarithms of negative numbers and this issomething we are not allowed to do We avoid this by rede1047297ning
the integral ofx
as
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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17 Basic integration and its applications 587
KEY POINT 174 AGAIN
minusint =
With this de1047297nition we can integrate y =1
over negative
numbers and the integral above becomes
12
1
2
1
x
e oreminus
minusint
n
Notice that the answer is negative since the required area is below
the x -axis We can still not integrate1
with a negative lower and
positive upper limit since the graph has an asymptote at x = 0
You may rightly be alittle uncomfortablewith inserting amodulus function lsquojust
because it worksrsquo Inmathematics do the ends
justify the means
Exercise 17H
1 Find the shaded areas
(a) (i)y =x2
y
x1 2
(ii)y = 1
x2
y
x2 4
(b) (i) y =x2
minus4x+ 3
y
x1 2
(ii)y =x2
minus4
y
x1
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588 Topic 6 Calculus
(c) (i)y = x3 minus x
y
xminus1 2
(ii)y = x2minus3x
y
x
5
2 Te area enclosed by the x -axis the curve andthe line x = k is 18 Find the value of k [6 marks]
3 (a) Find3
int (b) Find the area between the curve minus2 and the
x -axis between x = 0 and x = 3 [5 marks]
4 Between x = 0 and x = 3 the area of the graph y x minus2 below the x -axis equals the area above thex -axis Find the value of k [6 marks]
5 Find the area enclosed by the curve 1x minus minus 0
and the x -axis [7 marks]
lsquo Find t he area
enc losedrsquo means firs t
find a c losed region
bounded b y t he
cur ves men tioned
t hen find i ts area
A s ke tc h is a ver y
use fu l too l
e x a m h i n t
17I The area between a curve andthe y -axis
Consider the diagram alongside How can we 1047297nd the shadedarea A
One possible strategy is to construct a box around the graph todivide up the regions of interest You can integrate to 1047297nd thearea labelled A1 and then by adding and subtracting the areas ofthe blue and red rectangles shown calculate A
y
x
y = f (x)
c
d
A
y
x
y = f (x)
c
d A1
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2130
17 Basic integration and its applications 589
Happily there is a quicker way we can treat x as a functionof y effectively re1047298ecting the whole diagram in the line y = x and then use the same method as in the previous section
KEY POINT 1710
Te area bounded by the curve y the y -axis and the
lines y = c and y = d is given by y d
) dint where is
the expression for x in terms of y
You may have realised that this is related to inverse functions from Section 5E
x
y
x = f (y)
cd
A
Worked example 178
Te curve shown has equation y 1minus Find the shaded area
y
x
y = 2radic x minus 1
10
Express x in terms of y x = 2
rArr = + y
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590 Topic 6 Calculus
Exercise 17I
continued
Find the limits on the y -axisIt may help to label them on the
graph
When x = minusWhen x = 1 = =minus 6
y
y = 2radic minus
1
10
Write down the integral andevaluate using calculator
Area fro GDC= +2
1 2=d
1 Find the shaded areas
(a) (i) y
x
y = x2
1
6
(ii) y
x
y = x3
1
3
(b) (i) y
xy = 1
x2
2
1
(ii) y
x
y =radic x
1
5
(c) (i)y
x
y = ln x
1e
e
(ii) y
x
y = 3x
1 6
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Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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17 Basic integration and its applications 591
2 Te diagram shows the curve If the shadedarea is 504 1047297nd the value of a [6 marks]
y
x
y = radic
x
a
2a
0
3 Find the exact value of the area enclosed by the graphof + the line y = 2 and the y -axis [6 marks]
4 Te diagram shows the graph of
Te shaded area is 39 units Find the value of a [7 marks]y
x
y =radic x
4 a
5 Te diagram shows the graph of 2 where a isin infin Te area of the pink region is equal to the area of the blue
region Give two equations for a in terms of b and hencegive a in exact form and determine the size of theblue area [8 marks]y
x
y = x2
b
1 a
17J The area between two curves
So far we have only looked at areas bounded by a curve and oneof the coordinate axes but we can also 1047297nd areas bounded bytwo curvesTe area A in the diagram can be found by taking the areabounded by f )x and the x -axis and subtracting the areabounded by and the x -axis that is
A
b
minus
y
x
y = f (x)
y = g(x)
a b
A
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Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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592 Topic 6 Calculus
We can do the subtraction before integrating so that we onlyhave to integrate one expression instead of two Tis gives analternative formula for the area
KEY POINT 1711
Te area A between two curves f (x ) and g (x ) is
Ab
(
where a and b are the x-coordinates of the intersectionpoints of the two curves
Worked example 179
Find the area A enclosed between + and minus2 +
First find the x -coordinates ofintersection
For intersection
x x x
rArr =
rArr ) =rArr =
minus
minus
+x
4minusx 0
Make a rough sketch to see therelative positions of the two curves
y
x1 4
A
2x + 1y =
2 minus 3x + 5y = x
Subtract the lower curve from thehigher before integrating
= x
minusx x
4
minus
minus minusx
2
1
4
2
minus =
8
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Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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17 Basic integration and its applications 593
Subtracting the two equations before integrating is particularlyuseful when one of the curves is partly below the x-axis If x is always above then the expression we are integrating f g )x minus )x is always positive so we do not have to worryabout the signs of f )x and g )x themselves
Worked example 1710
Find the area bounded by the curves y x minus and y x 2
Sketch the graph to see therelative position of two
curves
Using GDC
= ex minus 5
y = 3 2
y
minus2 2
A
Find the intersection points ndash
use calculatorintersections x = minus2 658
Write down the integralrepresenting the area
rea =minus
minus818
1 658
xminus
= minus
1 6 8
x
Evaluate the integral usingcalculator = 21 6 (3SF)
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Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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594 Topic 6 Calculus
Exercise 17J
1 Find the shaded areas
(a) (i)
y = 2x+ 1
y = (xminus1)2
y
x
(ii) y =x+ 1
y = 4xminusx2minus1
y
x
(b) (i)
y =minusx2minus4x+ 12
y
x
y =x2 + 2x+ 12
(ii)
y =x2minus2x+ 9
y
x
y = 4xminusx2 + 5
(c) (i)y =x2 minusx
y = 2xminusx2
y
x
(ii)y =x2minus7x+ 7
y = 3minusxminusx2
y
x
2 Find the area enclosed between the graphs of y x +x minus and + [6 marks]
3 Find the area enclosed by the curve 2
the y-axisand the line [6 marks]
4 Find the area between the curves1
and in theregion lt lt π [6 marks]
5 Show that the area of the shaded region alongside is2
[6 marks]
y = x2
y
xminus1 2
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17 Basic integration and its applications 595
6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]
7 Find the total area enclosed between the graphs of
)minus 2 and minus2 7 1+ [6 marks]
8 Te area enclosed between the curve y x 2 and the line
y mx is 10 Find the value of m if m 0 [7 marks]
9 Show that the shaded area in the diagram below is [8 marks]
y = 2 minus xy2 = x
y
x
y = cos x y = sin x
y
x
Summary
bull Integration is the reverse process of differentiation
bull Any integral without limits (inde1047297nite) will generate a constant of integration
bull For all rational n ne minus1 int +x c
1
1
bull If minus we get the natural logarithm function int minus =
bull Te integral of the exponential function is int e c
bull Te integrals of the trigonometric functions are
int si minus=
int tan =
bull Te de1047297nite integral has limits f( ) x b
dint is found by evaluating the integrated expression
at b and then subtracting the integrated expression evaluated at a
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596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
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17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
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7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
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584 Topic 6 Calculus
Unfortunately the relationship between integrals and areas isnot so simple when there are parts of the curve above and belowthe axis Tose bits above the axis contribute positively to thearea but bits below the axis contribute negatively to the areaWe must separate out the sections above the axis and belowthe axis
Worked example 177
(a) Find x x 1
4
x int 1 d
(b) Find the area enclosed between the x -axis the curve y minus x + and the lines= 1 and 4
Apply standard integration (a) x x
4
xminus1
= ( ( ) ( ( )minus + minus +
= minus =3
4
The value found above canrsquot bethe correct area for (b)
Sketch the curve to see exactlywhich area we are being asked
to find
(b) y
2 minus + 3
1 3
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17 Basic integration and its applications 585
Te fact that the integral was zero in Worked example 177 part(a) means that the area above the axis is exactly cancelled by thearea below the axis
Tis example warns us that when asked to 1047297nd an area we mustalways sketch the graph and identify exactly where each partof the area is If we are evaluating the area on the calculator wecan use the modulus function to ensure that the entire graph isabove the x -axis Using the function from Worked example 177
1
4
int 1y
x
y = |x2 minus 4x + 3|
1 3 4
continued
The area is made up of twoparts so evaluate each of
them separately
x
1
33
xminusint
( ) minus = minus
there4 rea below the axis is
x x3
4
2xminus int
minus ( ) =
4
there4 rea above the axis is
Total area = + =3 3
Transformationsof graphs using the
modulus function
were covered inchapter 7
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586 Topic 6 Calculus
KEY POINT 179
Te area bounded by the curve y f )x the x-axis and the
lines x = a and x = b is given byb
x int
When working without a calculator if the curve crosses
the x -axis between a and b we need to split the area intoseveral parts and 1047297nd each one separately
Te interpretation of integrals as areas causes one inconsistency
with our previous work Consider the integral1
2
1
x dminus
int
Graphically we can see that this area should exist
x
y
minus1minus2
However if we do the integration we 1047297nd that
12
1
2
1
]minus
minus
minus
minus
minus n minus
minusn
1
= ln
1
minus n
Tis is the correct answer (which we could have found usingthe symmetry of the curve) but it goes through a stage wherewe had to take logarithms of negative numbers and this issomething we are not allowed to do We avoid this by rede1047297ning
the integral ofx
as
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17 Basic integration and its applications 587
KEY POINT 174 AGAIN
minusint =
With this de1047297nition we can integrate y =1
over negative
numbers and the integral above becomes
12
1
2
1
x
e oreminus
minusint
n
Notice that the answer is negative since the required area is below
the x -axis We can still not integrate1
with a negative lower and
positive upper limit since the graph has an asymptote at x = 0
You may rightly be alittle uncomfortablewith inserting amodulus function lsquojust
because it worksrsquo Inmathematics do the ends
justify the means
Exercise 17H
1 Find the shaded areas
(a) (i)y =x2
y
x1 2
(ii)y = 1
x2
y
x2 4
(b) (i) y =x2
minus4x+ 3
y
x1 2
(ii)y =x2
minus4
y
x1
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588 Topic 6 Calculus
(c) (i)y = x3 minus x
y
xminus1 2
(ii)y = x2minus3x
y
x
5
2 Te area enclosed by the x -axis the curve andthe line x = k is 18 Find the value of k [6 marks]
3 (a) Find3
int (b) Find the area between the curve minus2 and the
x -axis between x = 0 and x = 3 [5 marks]
4 Between x = 0 and x = 3 the area of the graph y x minus2 below the x -axis equals the area above thex -axis Find the value of k [6 marks]
5 Find the area enclosed by the curve 1x minus minus 0
and the x -axis [7 marks]
lsquo Find t he area
enc losedrsquo means firs t
find a c losed region
bounded b y t he
cur ves men tioned
t hen find i ts area
A s ke tc h is a ver y
use fu l too l
e x a m h i n t
17I The area between a curve andthe y -axis
Consider the diagram alongside How can we 1047297nd the shadedarea A
One possible strategy is to construct a box around the graph todivide up the regions of interest You can integrate to 1047297nd thearea labelled A1 and then by adding and subtracting the areas ofthe blue and red rectangles shown calculate A
y
x
y = f (x)
c
d
A
y
x
y = f (x)
c
d A1
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17 Basic integration and its applications 589
Happily there is a quicker way we can treat x as a functionof y effectively re1047298ecting the whole diagram in the line y = x and then use the same method as in the previous section
KEY POINT 1710
Te area bounded by the curve y the y -axis and the
lines y = c and y = d is given by y d
) dint where is
the expression for x in terms of y
You may have realised that this is related to inverse functions from Section 5E
x
y
x = f (y)
cd
A
Worked example 178
Te curve shown has equation y 1minus Find the shaded area
y
x
y = 2radic x minus 1
10
Express x in terms of y x = 2
rArr = + y
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590 Topic 6 Calculus
Exercise 17I
continued
Find the limits on the y -axisIt may help to label them on the
graph
When x = minusWhen x = 1 = =minus 6
y
y = 2radic minus
1
10
Write down the integral andevaluate using calculator
Area fro GDC= +2
1 2=d
1 Find the shaded areas
(a) (i) y
x
y = x2
1
6
(ii) y
x
y = x3
1
3
(b) (i) y
xy = 1
x2
2
1
(ii) y
x
y =radic x
1
5
(c) (i)y
x
y = ln x
1e
e
(ii) y
x
y = 3x
1 6
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17 Basic integration and its applications 591
2 Te diagram shows the curve If the shadedarea is 504 1047297nd the value of a [6 marks]
y
x
y = radic
x
a
2a
0
3 Find the exact value of the area enclosed by the graphof + the line y = 2 and the y -axis [6 marks]
4 Te diagram shows the graph of
Te shaded area is 39 units Find the value of a [7 marks]y
x
y =radic x
4 a
5 Te diagram shows the graph of 2 where a isin infin Te area of the pink region is equal to the area of the blue
region Give two equations for a in terms of b and hencegive a in exact form and determine the size of theblue area [8 marks]y
x
y = x2
b
1 a
17J The area between two curves
So far we have only looked at areas bounded by a curve and oneof the coordinate axes but we can also 1047297nd areas bounded bytwo curvesTe area A in the diagram can be found by taking the areabounded by f )x and the x -axis and subtracting the areabounded by and the x -axis that is
A
b
minus
y
x
y = f (x)
y = g(x)
a b
A
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592 Topic 6 Calculus
We can do the subtraction before integrating so that we onlyhave to integrate one expression instead of two Tis gives analternative formula for the area
KEY POINT 1711
Te area A between two curves f (x ) and g (x ) is
Ab
(
where a and b are the x-coordinates of the intersectionpoints of the two curves
Worked example 179
Find the area A enclosed between + and minus2 +
First find the x -coordinates ofintersection
For intersection
x x x
rArr =
rArr ) =rArr =
minus
minus
+x
4minusx 0
Make a rough sketch to see therelative positions of the two curves
y
x1 4
A
2x + 1y =
2 minus 3x + 5y = x
Subtract the lower curve from thehigher before integrating
= x
minusx x
4
minus
minus minusx
2
1
4
2
minus =
8
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17 Basic integration and its applications 593
Subtracting the two equations before integrating is particularlyuseful when one of the curves is partly below the x-axis If x is always above then the expression we are integrating f g )x minus )x is always positive so we do not have to worryabout the signs of f )x and g )x themselves
Worked example 1710
Find the area bounded by the curves y x minus and y x 2
Sketch the graph to see therelative position of two
curves
Using GDC
= ex minus 5
y = 3 2
y
minus2 2
A
Find the intersection points ndash
use calculatorintersections x = minus2 658
Write down the integralrepresenting the area
rea =minus
minus818
1 658
xminus
= minus
1 6 8
x
Evaluate the integral usingcalculator = 21 6 (3SF)
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594 Topic 6 Calculus
Exercise 17J
1 Find the shaded areas
(a) (i)
y = 2x+ 1
y = (xminus1)2
y
x
(ii) y =x+ 1
y = 4xminusx2minus1
y
x
(b) (i)
y =minusx2minus4x+ 12
y
x
y =x2 + 2x+ 12
(ii)
y =x2minus2x+ 9
y
x
y = 4xminusx2 + 5
(c) (i)y =x2 minusx
y = 2xminusx2
y
x
(ii)y =x2minus7x+ 7
y = 3minusxminusx2
y
x
2 Find the area enclosed between the graphs of y x +x minus and + [6 marks]
3 Find the area enclosed by the curve 2
the y-axisand the line [6 marks]
4 Find the area between the curves1
and in theregion lt lt π [6 marks]
5 Show that the area of the shaded region alongside is2
[6 marks]
y = x2
y
xminus1 2
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Not for printing sharing or distribution
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17 Basic integration and its applications 595
6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]
7 Find the total area enclosed between the graphs of
)minus 2 and minus2 7 1+ [6 marks]
8 Te area enclosed between the curve y x 2 and the line
y mx is 10 Find the value of m if m 0 [7 marks]
9 Show that the shaded area in the diagram below is [8 marks]
y = 2 minus xy2 = x
y
x
y = cos x y = sin x
y
x
Summary
bull Integration is the reverse process of differentiation
bull Any integral without limits (inde1047297nite) will generate a constant of integration
bull For all rational n ne minus1 int +x c
1
1
bull If minus we get the natural logarithm function int minus =
bull Te integral of the exponential function is int e c
bull Te integrals of the trigonometric functions are
int si minus=
int tan =
bull Te de1047297nite integral has limits f( ) x b
dint is found by evaluating the integrated expression
at b and then subtracting the integrated expression evaluated at a
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Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
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Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
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7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
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17 Basic integration and its applications 585
Te fact that the integral was zero in Worked example 177 part(a) means that the area above the axis is exactly cancelled by thearea below the axis
Tis example warns us that when asked to 1047297nd an area we mustalways sketch the graph and identify exactly where each partof the area is If we are evaluating the area on the calculator wecan use the modulus function to ensure that the entire graph isabove the x -axis Using the function from Worked example 177
1
4
int 1y
x
y = |x2 minus 4x + 3|
1 3 4
continued
The area is made up of twoparts so evaluate each of
them separately
x
1
33
xminusint
( ) minus = minus
there4 rea below the axis is
x x3
4
2xminus int
minus ( ) =
4
there4 rea above the axis is
Total area = + =3 3
Transformationsof graphs using the
modulus function
were covered inchapter 7
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Not for printing sharing or distribution
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586 Topic 6 Calculus
KEY POINT 179
Te area bounded by the curve y f )x the x-axis and the
lines x = a and x = b is given byb
x int
When working without a calculator if the curve crosses
the x -axis between a and b we need to split the area intoseveral parts and 1047297nd each one separately
Te interpretation of integrals as areas causes one inconsistency
with our previous work Consider the integral1
2
1
x dminus
int
Graphically we can see that this area should exist
x
y
minus1minus2
However if we do the integration we 1047297nd that
12
1
2
1
]minus
minus
minus
minus
minus n minus
minusn
1
= ln
1
minus n
Tis is the correct answer (which we could have found usingthe symmetry of the curve) but it goes through a stage wherewe had to take logarithms of negative numbers and this issomething we are not allowed to do We avoid this by rede1047297ning
the integral ofx
as
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 1930
17 Basic integration and its applications 587
KEY POINT 174 AGAIN
minusint =
With this de1047297nition we can integrate y =1
over negative
numbers and the integral above becomes
12
1
2
1
x
e oreminus
minusint
n
Notice that the answer is negative since the required area is below
the x -axis We can still not integrate1
with a negative lower and
positive upper limit since the graph has an asymptote at x = 0
You may rightly be alittle uncomfortablewith inserting amodulus function lsquojust
because it worksrsquo Inmathematics do the ends
justify the means
Exercise 17H
1 Find the shaded areas
(a) (i)y =x2
y
x1 2
(ii)y = 1
x2
y
x2 4
(b) (i) y =x2
minus4x+ 3
y
x1 2
(ii)y =x2
minus4
y
x1
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Not for printing sharing or distribution
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588 Topic 6 Calculus
(c) (i)y = x3 minus x
y
xminus1 2
(ii)y = x2minus3x
y
x
5
2 Te area enclosed by the x -axis the curve andthe line x = k is 18 Find the value of k [6 marks]
3 (a) Find3
int (b) Find the area between the curve minus2 and the
x -axis between x = 0 and x = 3 [5 marks]
4 Between x = 0 and x = 3 the area of the graph y x minus2 below the x -axis equals the area above thex -axis Find the value of k [6 marks]
5 Find the area enclosed by the curve 1x minus minus 0
and the x -axis [7 marks]
lsquo Find t he area
enc losedrsquo means firs t
find a c losed region
bounded b y t he
cur ves men tioned
t hen find i ts area
A s ke tc h is a ver y
use fu l too l
e x a m h i n t
17I The area between a curve andthe y -axis
Consider the diagram alongside How can we 1047297nd the shadedarea A
One possible strategy is to construct a box around the graph todivide up the regions of interest You can integrate to 1047297nd thearea labelled A1 and then by adding and subtracting the areas ofthe blue and red rectangles shown calculate A
y
x
y = f (x)
c
d
A
y
x
y = f (x)
c
d A1
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2130
17 Basic integration and its applications 589
Happily there is a quicker way we can treat x as a functionof y effectively re1047298ecting the whole diagram in the line y = x and then use the same method as in the previous section
KEY POINT 1710
Te area bounded by the curve y the y -axis and the
lines y = c and y = d is given by y d
) dint where is
the expression for x in terms of y
You may have realised that this is related to inverse functions from Section 5E
x
y
x = f (y)
cd
A
Worked example 178
Te curve shown has equation y 1minus Find the shaded area
y
x
y = 2radic x minus 1
10
Express x in terms of y x = 2
rArr = + y
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7242019 Chapter 17 IB Maths HL Cambridge Textbook
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590 Topic 6 Calculus
Exercise 17I
continued
Find the limits on the y -axisIt may help to label them on the
graph
When x = minusWhen x = 1 = =minus 6
y
y = 2radic minus
1
10
Write down the integral andevaluate using calculator
Area fro GDC= +2
1 2=d
1 Find the shaded areas
(a) (i) y
x
y = x2
1
6
(ii) y
x
y = x3
1
3
(b) (i) y
xy = 1
x2
2
1
(ii) y
x
y =radic x
1
5
(c) (i)y
x
y = ln x
1e
e
(ii) y
x
y = 3x
1 6
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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17 Basic integration and its applications 591
2 Te diagram shows the curve If the shadedarea is 504 1047297nd the value of a [6 marks]
y
x
y = radic
x
a
2a
0
3 Find the exact value of the area enclosed by the graphof + the line y = 2 and the y -axis [6 marks]
4 Te diagram shows the graph of
Te shaded area is 39 units Find the value of a [7 marks]y
x
y =radic x
4 a
5 Te diagram shows the graph of 2 where a isin infin Te area of the pink region is equal to the area of the blue
region Give two equations for a in terms of b and hencegive a in exact form and determine the size of theblue area [8 marks]y
x
y = x2
b
1 a
17J The area between two curves
So far we have only looked at areas bounded by a curve and oneof the coordinate axes but we can also 1047297nd areas bounded bytwo curvesTe area A in the diagram can be found by taking the areabounded by f )x and the x -axis and subtracting the areabounded by and the x -axis that is
A
b
minus
y
x
y = f (x)
y = g(x)
a b
A
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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592 Topic 6 Calculus
We can do the subtraction before integrating so that we onlyhave to integrate one expression instead of two Tis gives analternative formula for the area
KEY POINT 1711
Te area A between two curves f (x ) and g (x ) is
Ab
(
where a and b are the x-coordinates of the intersectionpoints of the two curves
Worked example 179
Find the area A enclosed between + and minus2 +
First find the x -coordinates ofintersection
For intersection
x x x
rArr =
rArr ) =rArr =
minus
minus
+x
4minusx 0
Make a rough sketch to see therelative positions of the two curves
y
x1 4
A
2x + 1y =
2 minus 3x + 5y = x
Subtract the lower curve from thehigher before integrating
= x
minusx x
4
minus
minus minusx
2
1
4
2
minus =
8
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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17 Basic integration and its applications 593
Subtracting the two equations before integrating is particularlyuseful when one of the curves is partly below the x-axis If x is always above then the expression we are integrating f g )x minus )x is always positive so we do not have to worryabout the signs of f )x and g )x themselves
Worked example 1710
Find the area bounded by the curves y x minus and y x 2
Sketch the graph to see therelative position of two
curves
Using GDC
= ex minus 5
y = 3 2
y
minus2 2
A
Find the intersection points ndash
use calculatorintersections x = minus2 658
Write down the integralrepresenting the area
rea =minus
minus818
1 658
xminus
= minus
1 6 8
x
Evaluate the integral usingcalculator = 21 6 (3SF)
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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594 Topic 6 Calculus
Exercise 17J
1 Find the shaded areas
(a) (i)
y = 2x+ 1
y = (xminus1)2
y
x
(ii) y =x+ 1
y = 4xminusx2minus1
y
x
(b) (i)
y =minusx2minus4x+ 12
y
x
y =x2 + 2x+ 12
(ii)
y =x2minus2x+ 9
y
x
y = 4xminusx2 + 5
(c) (i)y =x2 minusx
y = 2xminusx2
y
x
(ii)y =x2minus7x+ 7
y = 3minusxminusx2
y
x
2 Find the area enclosed between the graphs of y x +x minus and + [6 marks]
3 Find the area enclosed by the curve 2
the y-axisand the line [6 marks]
4 Find the area between the curves1
and in theregion lt lt π [6 marks]
5 Show that the area of the shaded region alongside is2
[6 marks]
y = x2
y
xminus1 2
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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17 Basic integration and its applications 595
6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]
7 Find the total area enclosed between the graphs of
)minus 2 and minus2 7 1+ [6 marks]
8 Te area enclosed between the curve y x 2 and the line
y mx is 10 Find the value of m if m 0 [7 marks]
9 Show that the shaded area in the diagram below is [8 marks]
y = 2 minus xy2 = x
y
x
y = cos x y = sin x
y
x
Summary
bull Integration is the reverse process of differentiation
bull Any integral without limits (inde1047297nite) will generate a constant of integration
bull For all rational n ne minus1 int +x c
1
1
bull If minus we get the natural logarithm function int minus =
bull Te integral of the exponential function is int e c
bull Te integrals of the trigonometric functions are
int si minus=
int tan =
bull Te de1047297nite integral has limits f( ) x b
dint is found by evaluating the integrated expression
at b and then subtracting the integrated expression evaluated at a
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2830
596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2930
17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
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Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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586 Topic 6 Calculus
KEY POINT 179
Te area bounded by the curve y f )x the x-axis and the
lines x = a and x = b is given byb
x int
When working without a calculator if the curve crosses
the x -axis between a and b we need to split the area intoseveral parts and 1047297nd each one separately
Te interpretation of integrals as areas causes one inconsistency
with our previous work Consider the integral1
2
1
x dminus
int
Graphically we can see that this area should exist
x
y
minus1minus2
However if we do the integration we 1047297nd that
12
1
2
1
]minus
minus
minus
minus
minus n minus
minusn
1
= ln
1
minus n
Tis is the correct answer (which we could have found usingthe symmetry of the curve) but it goes through a stage wherewe had to take logarithms of negative numbers and this issomething we are not allowed to do We avoid this by rede1047297ning
the integral ofx
as
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 1930
17 Basic integration and its applications 587
KEY POINT 174 AGAIN
minusint =
With this de1047297nition we can integrate y =1
over negative
numbers and the integral above becomes
12
1
2
1
x
e oreminus
minusint
n
Notice that the answer is negative since the required area is below
the x -axis We can still not integrate1
with a negative lower and
positive upper limit since the graph has an asymptote at x = 0
You may rightly be alittle uncomfortablewith inserting amodulus function lsquojust
because it worksrsquo Inmathematics do the ends
justify the means
Exercise 17H
1 Find the shaded areas
(a) (i)y =x2
y
x1 2
(ii)y = 1
x2
y
x2 4
(b) (i) y =x2
minus4x+ 3
y
x1 2
(ii)y =x2
minus4
y
x1
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2030
588 Topic 6 Calculus
(c) (i)y = x3 minus x
y
xminus1 2
(ii)y = x2minus3x
y
x
5
2 Te area enclosed by the x -axis the curve andthe line x = k is 18 Find the value of k [6 marks]
3 (a) Find3
int (b) Find the area between the curve minus2 and the
x -axis between x = 0 and x = 3 [5 marks]
4 Between x = 0 and x = 3 the area of the graph y x minus2 below the x -axis equals the area above thex -axis Find the value of k [6 marks]
5 Find the area enclosed by the curve 1x minus minus 0
and the x -axis [7 marks]
lsquo Find t he area
enc losedrsquo means firs t
find a c losed region
bounded b y t he
cur ves men tioned
t hen find i ts area
A s ke tc h is a ver y
use fu l too l
e x a m h i n t
17I The area between a curve andthe y -axis
Consider the diagram alongside How can we 1047297nd the shadedarea A
One possible strategy is to construct a box around the graph todivide up the regions of interest You can integrate to 1047297nd thearea labelled A1 and then by adding and subtracting the areas ofthe blue and red rectangles shown calculate A
y
x
y = f (x)
c
d
A
y
x
y = f (x)
c
d A1
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2130
17 Basic integration and its applications 589
Happily there is a quicker way we can treat x as a functionof y effectively re1047298ecting the whole diagram in the line y = x and then use the same method as in the previous section
KEY POINT 1710
Te area bounded by the curve y the y -axis and the
lines y = c and y = d is given by y d
) dint where is
the expression for x in terms of y
You may have realised that this is related to inverse functions from Section 5E
x
y
x = f (y)
cd
A
Worked example 178
Te curve shown has equation y 1minus Find the shaded area
y
x
y = 2radic x minus 1
10
Express x in terms of y x = 2
rArr = + y
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Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2230
590 Topic 6 Calculus
Exercise 17I
continued
Find the limits on the y -axisIt may help to label them on the
graph
When x = minusWhen x = 1 = =minus 6
y
y = 2radic minus
1
10
Write down the integral andevaluate using calculator
Area fro GDC= +2
1 2=d
1 Find the shaded areas
(a) (i) y
x
y = x2
1
6
(ii) y
x
y = x3
1
3
(b) (i) y
xy = 1
x2
2
1
(ii) y
x
y =radic x
1
5
(c) (i)y
x
y = ln x
1e
e
(ii) y
x
y = 3x
1 6
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2330
17 Basic integration and its applications 591
2 Te diagram shows the curve If the shadedarea is 504 1047297nd the value of a [6 marks]
y
x
y = radic
x
a
2a
0
3 Find the exact value of the area enclosed by the graphof + the line y = 2 and the y -axis [6 marks]
4 Te diagram shows the graph of
Te shaded area is 39 units Find the value of a [7 marks]y
x
y =radic x
4 a
5 Te diagram shows the graph of 2 where a isin infin Te area of the pink region is equal to the area of the blue
region Give two equations for a in terms of b and hencegive a in exact form and determine the size of theblue area [8 marks]y
x
y = x2
b
1 a
17J The area between two curves
So far we have only looked at areas bounded by a curve and oneof the coordinate axes but we can also 1047297nd areas bounded bytwo curvesTe area A in the diagram can be found by taking the areabounded by f )x and the x -axis and subtracting the areabounded by and the x -axis that is
A
b
minus
y
x
y = f (x)
y = g(x)
a b
A
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2430
592 Topic 6 Calculus
We can do the subtraction before integrating so that we onlyhave to integrate one expression instead of two Tis gives analternative formula for the area
KEY POINT 1711
Te area A between two curves f (x ) and g (x ) is
Ab
(
where a and b are the x-coordinates of the intersectionpoints of the two curves
Worked example 179
Find the area A enclosed between + and minus2 +
First find the x -coordinates ofintersection
For intersection
x x x
rArr =
rArr ) =rArr =
minus
minus
+x
4minusx 0
Make a rough sketch to see therelative positions of the two curves
y
x1 4
A
2x + 1y =
2 minus 3x + 5y = x
Subtract the lower curve from thehigher before integrating
= x
minusx x
4
minus
minus minusx
2
1
4
2
minus =
8
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2530
17 Basic integration and its applications 593
Subtracting the two equations before integrating is particularlyuseful when one of the curves is partly below the x-axis If x is always above then the expression we are integrating f g )x minus )x is always positive so we do not have to worryabout the signs of f )x and g )x themselves
Worked example 1710
Find the area bounded by the curves y x minus and y x 2
Sketch the graph to see therelative position of two
curves
Using GDC
= ex minus 5
y = 3 2
y
minus2 2
A
Find the intersection points ndash
use calculatorintersections x = minus2 658
Write down the integralrepresenting the area
rea =minus
minus818
1 658
xminus
= minus
1 6 8
x
Evaluate the integral usingcalculator = 21 6 (3SF)
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2630
594 Topic 6 Calculus
Exercise 17J
1 Find the shaded areas
(a) (i)
y = 2x+ 1
y = (xminus1)2
y
x
(ii) y =x+ 1
y = 4xminusx2minus1
y
x
(b) (i)
y =minusx2minus4x+ 12
y
x
y =x2 + 2x+ 12
(ii)
y =x2minus2x+ 9
y
x
y = 4xminusx2 + 5
(c) (i)y =x2 minusx
y = 2xminusx2
y
x
(ii)y =x2minus7x+ 7
y = 3minusxminusx2
y
x
2 Find the area enclosed between the graphs of y x +x minus and + [6 marks]
3 Find the area enclosed by the curve 2
the y-axisand the line [6 marks]
4 Find the area between the curves1
and in theregion lt lt π [6 marks]
5 Show that the area of the shaded region alongside is2
[6 marks]
y = x2
y
xminus1 2
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2730
17 Basic integration and its applications 595
6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]
7 Find the total area enclosed between the graphs of
)minus 2 and minus2 7 1+ [6 marks]
8 Te area enclosed between the curve y x 2 and the line
y mx is 10 Find the value of m if m 0 [7 marks]
9 Show that the shaded area in the diagram below is [8 marks]
y = 2 minus xy2 = x
y
x
y = cos x y = sin x
y
x
Summary
bull Integration is the reverse process of differentiation
bull Any integral without limits (inde1047297nite) will generate a constant of integration
bull For all rational n ne minus1 int +x c
1
1
bull If minus we get the natural logarithm function int minus =
bull Te integral of the exponential function is int e c
bull Te integrals of the trigonometric functions are
int si minus=
int tan =
bull Te de1047297nite integral has limits f( ) x b
dint is found by evaluating the integrated expression
at b and then subtracting the integrated expression evaluated at a
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2830
596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2930
17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 3030
7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 1930
17 Basic integration and its applications 587
KEY POINT 174 AGAIN
minusint =
With this de1047297nition we can integrate y =1
over negative
numbers and the integral above becomes
12
1
2
1
x
e oreminus
minusint
n
Notice that the answer is negative since the required area is below
the x -axis We can still not integrate1
with a negative lower and
positive upper limit since the graph has an asymptote at x = 0
You may rightly be alittle uncomfortablewith inserting amodulus function lsquojust
because it worksrsquo Inmathematics do the ends
justify the means
Exercise 17H
1 Find the shaded areas
(a) (i)y =x2
y
x1 2
(ii)y = 1
x2
y
x2 4
(b) (i) y =x2
minus4x+ 3
y
x1 2
(ii)y =x2
minus4
y
x1
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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588 Topic 6 Calculus
(c) (i)y = x3 minus x
y
xminus1 2
(ii)y = x2minus3x
y
x
5
2 Te area enclosed by the x -axis the curve andthe line x = k is 18 Find the value of k [6 marks]
3 (a) Find3
int (b) Find the area between the curve minus2 and the
x -axis between x = 0 and x = 3 [5 marks]
4 Between x = 0 and x = 3 the area of the graph y x minus2 below the x -axis equals the area above thex -axis Find the value of k [6 marks]
5 Find the area enclosed by the curve 1x minus minus 0
and the x -axis [7 marks]
lsquo Find t he area
enc losedrsquo means firs t
find a c losed region
bounded b y t he
cur ves men tioned
t hen find i ts area
A s ke tc h is a ver y
use fu l too l
e x a m h i n t
17I The area between a curve andthe y -axis
Consider the diagram alongside How can we 1047297nd the shadedarea A
One possible strategy is to construct a box around the graph todivide up the regions of interest You can integrate to 1047297nd thearea labelled A1 and then by adding and subtracting the areas ofthe blue and red rectangles shown calculate A
y
x
y = f (x)
c
d
A
y
x
y = f (x)
c
d A1
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2130
17 Basic integration and its applications 589
Happily there is a quicker way we can treat x as a functionof y effectively re1047298ecting the whole diagram in the line y = x and then use the same method as in the previous section
KEY POINT 1710
Te area bounded by the curve y the y -axis and the
lines y = c and y = d is given by y d
) dint where is
the expression for x in terms of y
You may have realised that this is related to inverse functions from Section 5E
x
y
x = f (y)
cd
A
Worked example 178
Te curve shown has equation y 1minus Find the shaded area
y
x
y = 2radic x minus 1
10
Express x in terms of y x = 2
rArr = + y
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2230
590 Topic 6 Calculus
Exercise 17I
continued
Find the limits on the y -axisIt may help to label them on the
graph
When x = minusWhen x = 1 = =minus 6
y
y = 2radic minus
1
10
Write down the integral andevaluate using calculator
Area fro GDC= +2
1 2=d
1 Find the shaded areas
(a) (i) y
x
y = x2
1
6
(ii) y
x
y = x3
1
3
(b) (i) y
xy = 1
x2
2
1
(ii) y
x
y =radic x
1
5
(c) (i)y
x
y = ln x
1e
e
(ii) y
x
y = 3x
1 6
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17 Basic integration and its applications 591
2 Te diagram shows the curve If the shadedarea is 504 1047297nd the value of a [6 marks]
y
x
y = radic
x
a
2a
0
3 Find the exact value of the area enclosed by the graphof + the line y = 2 and the y -axis [6 marks]
4 Te diagram shows the graph of
Te shaded area is 39 units Find the value of a [7 marks]y
x
y =radic x
4 a
5 Te diagram shows the graph of 2 where a isin infin Te area of the pink region is equal to the area of the blue
region Give two equations for a in terms of b and hencegive a in exact form and determine the size of theblue area [8 marks]y
x
y = x2
b
1 a
17J The area between two curves
So far we have only looked at areas bounded by a curve and oneof the coordinate axes but we can also 1047297nd areas bounded bytwo curvesTe area A in the diagram can be found by taking the areabounded by f )x and the x -axis and subtracting the areabounded by and the x -axis that is
A
b
minus
y
x
y = f (x)
y = g(x)
a b
A
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592 Topic 6 Calculus
We can do the subtraction before integrating so that we onlyhave to integrate one expression instead of two Tis gives analternative formula for the area
KEY POINT 1711
Te area A between two curves f (x ) and g (x ) is
Ab
(
where a and b are the x-coordinates of the intersectionpoints of the two curves
Worked example 179
Find the area A enclosed between + and minus2 +
First find the x -coordinates ofintersection
For intersection
x x x
rArr =
rArr ) =rArr =
minus
minus
+x
4minusx 0
Make a rough sketch to see therelative positions of the two curves
y
x1 4
A
2x + 1y =
2 minus 3x + 5y = x
Subtract the lower curve from thehigher before integrating
= x
minusx x
4
minus
minus minusx
2
1
4
2
minus =
8
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17 Basic integration and its applications 593
Subtracting the two equations before integrating is particularlyuseful when one of the curves is partly below the x-axis If x is always above then the expression we are integrating f g )x minus )x is always positive so we do not have to worryabout the signs of f )x and g )x themselves
Worked example 1710
Find the area bounded by the curves y x minus and y x 2
Sketch the graph to see therelative position of two
curves
Using GDC
= ex minus 5
y = 3 2
y
minus2 2
A
Find the intersection points ndash
use calculatorintersections x = minus2 658
Write down the integralrepresenting the area
rea =minus
minus818
1 658
xminus
= minus
1 6 8
x
Evaluate the integral usingcalculator = 21 6 (3SF)
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594 Topic 6 Calculus
Exercise 17J
1 Find the shaded areas
(a) (i)
y = 2x+ 1
y = (xminus1)2
y
x
(ii) y =x+ 1
y = 4xminusx2minus1
y
x
(b) (i)
y =minusx2minus4x+ 12
y
x
y =x2 + 2x+ 12
(ii)
y =x2minus2x+ 9
y
x
y = 4xminusx2 + 5
(c) (i)y =x2 minusx
y = 2xminusx2
y
x
(ii)y =x2minus7x+ 7
y = 3minusxminusx2
y
x
2 Find the area enclosed between the graphs of y x +x minus and + [6 marks]
3 Find the area enclosed by the curve 2
the y-axisand the line [6 marks]
4 Find the area between the curves1
and in theregion lt lt π [6 marks]
5 Show that the area of the shaded region alongside is2
[6 marks]
y = x2
y
xminus1 2
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17 Basic integration and its applications 595
6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]
7 Find the total area enclosed between the graphs of
)minus 2 and minus2 7 1+ [6 marks]
8 Te area enclosed between the curve y x 2 and the line
y mx is 10 Find the value of m if m 0 [7 marks]
9 Show that the shaded area in the diagram below is [8 marks]
y = 2 minus xy2 = x
y
x
y = cos x y = sin x
y
x
Summary
bull Integration is the reverse process of differentiation
bull Any integral without limits (inde1047297nite) will generate a constant of integration
bull For all rational n ne minus1 int +x c
1
1
bull If minus we get the natural logarithm function int minus =
bull Te integral of the exponential function is int e c
bull Te integrals of the trigonometric functions are
int si minus=
int tan =
bull Te de1047297nite integral has limits f( ) x b
dint is found by evaluating the integrated expression
at b and then subtracting the integrated expression evaluated at a
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596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
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17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
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7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
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588 Topic 6 Calculus
(c) (i)y = x3 minus x
y
xminus1 2
(ii)y = x2minus3x
y
x
5
2 Te area enclosed by the x -axis the curve andthe line x = k is 18 Find the value of k [6 marks]
3 (a) Find3
int (b) Find the area between the curve minus2 and the
x -axis between x = 0 and x = 3 [5 marks]
4 Between x = 0 and x = 3 the area of the graph y x minus2 below the x -axis equals the area above thex -axis Find the value of k [6 marks]
5 Find the area enclosed by the curve 1x minus minus 0
and the x -axis [7 marks]
lsquo Find t he area
enc losedrsquo means firs t
find a c losed region
bounded b y t he
cur ves men tioned
t hen find i ts area
A s ke tc h is a ver y
use fu l too l
e x a m h i n t
17I The area between a curve andthe y -axis
Consider the diagram alongside How can we 1047297nd the shadedarea A
One possible strategy is to construct a box around the graph todivide up the regions of interest You can integrate to 1047297nd thearea labelled A1 and then by adding and subtracting the areas ofthe blue and red rectangles shown calculate A
y
x
y = f (x)
c
d
A
y
x
y = f (x)
c
d A1
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17 Basic integration and its applications 589
Happily there is a quicker way we can treat x as a functionof y effectively re1047298ecting the whole diagram in the line y = x and then use the same method as in the previous section
KEY POINT 1710
Te area bounded by the curve y the y -axis and the
lines y = c and y = d is given by y d
) dint where is
the expression for x in terms of y
You may have realised that this is related to inverse functions from Section 5E
x
y
x = f (y)
cd
A
Worked example 178
Te curve shown has equation y 1minus Find the shaded area
y
x
y = 2radic x minus 1
10
Express x in terms of y x = 2
rArr = + y
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590 Topic 6 Calculus
Exercise 17I
continued
Find the limits on the y -axisIt may help to label them on the
graph
When x = minusWhen x = 1 = =minus 6
y
y = 2radic minus
1
10
Write down the integral andevaluate using calculator
Area fro GDC= +2
1 2=d
1 Find the shaded areas
(a) (i) y
x
y = x2
1
6
(ii) y
x
y = x3
1
3
(b) (i) y
xy = 1
x2
2
1
(ii) y
x
y =radic x
1
5
(c) (i)y
x
y = ln x
1e
e
(ii) y
x
y = 3x
1 6
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17 Basic integration and its applications 591
2 Te diagram shows the curve If the shadedarea is 504 1047297nd the value of a [6 marks]
y
x
y = radic
x
a
2a
0
3 Find the exact value of the area enclosed by the graphof + the line y = 2 and the y -axis [6 marks]
4 Te diagram shows the graph of
Te shaded area is 39 units Find the value of a [7 marks]y
x
y =radic x
4 a
5 Te diagram shows the graph of 2 where a isin infin Te area of the pink region is equal to the area of the blue
region Give two equations for a in terms of b and hencegive a in exact form and determine the size of theblue area [8 marks]y
x
y = x2
b
1 a
17J The area between two curves
So far we have only looked at areas bounded by a curve and oneof the coordinate axes but we can also 1047297nd areas bounded bytwo curvesTe area A in the diagram can be found by taking the areabounded by f )x and the x -axis and subtracting the areabounded by and the x -axis that is
A
b
minus
y
x
y = f (x)
y = g(x)
a b
A
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592 Topic 6 Calculus
We can do the subtraction before integrating so that we onlyhave to integrate one expression instead of two Tis gives analternative formula for the area
KEY POINT 1711
Te area A between two curves f (x ) and g (x ) is
Ab
(
where a and b are the x-coordinates of the intersectionpoints of the two curves
Worked example 179
Find the area A enclosed between + and minus2 +
First find the x -coordinates ofintersection
For intersection
x x x
rArr =
rArr ) =rArr =
minus
minus
+x
4minusx 0
Make a rough sketch to see therelative positions of the two curves
y
x1 4
A
2x + 1y =
2 minus 3x + 5y = x
Subtract the lower curve from thehigher before integrating
= x
minusx x
4
minus
minus minusx
2
1
4
2
minus =
8
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7242019 Chapter 17 IB Maths HL Cambridge Textbook
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17 Basic integration and its applications 593
Subtracting the two equations before integrating is particularlyuseful when one of the curves is partly below the x-axis If x is always above then the expression we are integrating f g )x minus )x is always positive so we do not have to worryabout the signs of f )x and g )x themselves
Worked example 1710
Find the area bounded by the curves y x minus and y x 2
Sketch the graph to see therelative position of two
curves
Using GDC
= ex minus 5
y = 3 2
y
minus2 2
A
Find the intersection points ndash
use calculatorintersections x = minus2 658
Write down the integralrepresenting the area
rea =minus
minus818
1 658
xminus
= minus
1 6 8
x
Evaluate the integral usingcalculator = 21 6 (3SF)
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594 Topic 6 Calculus
Exercise 17J
1 Find the shaded areas
(a) (i)
y = 2x+ 1
y = (xminus1)2
y
x
(ii) y =x+ 1
y = 4xminusx2minus1
y
x
(b) (i)
y =minusx2minus4x+ 12
y
x
y =x2 + 2x+ 12
(ii)
y =x2minus2x+ 9
y
x
y = 4xminusx2 + 5
(c) (i)y =x2 minusx
y = 2xminusx2
y
x
(ii)y =x2minus7x+ 7
y = 3minusxminusx2
y
x
2 Find the area enclosed between the graphs of y x +x minus and + [6 marks]
3 Find the area enclosed by the curve 2
the y-axisand the line [6 marks]
4 Find the area between the curves1
and in theregion lt lt π [6 marks]
5 Show that the area of the shaded region alongside is2
[6 marks]
y = x2
y
xminus1 2
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17 Basic integration and its applications 595
6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]
7 Find the total area enclosed between the graphs of
)minus 2 and minus2 7 1+ [6 marks]
8 Te area enclosed between the curve y x 2 and the line
y mx is 10 Find the value of m if m 0 [7 marks]
9 Show that the shaded area in the diagram below is [8 marks]
y = 2 minus xy2 = x
y
x
y = cos x y = sin x
y
x
Summary
bull Integration is the reverse process of differentiation
bull Any integral without limits (inde1047297nite) will generate a constant of integration
bull For all rational n ne minus1 int +x c
1
1
bull If minus we get the natural logarithm function int minus =
bull Te integral of the exponential function is int e c
bull Te integrals of the trigonometric functions are
int si minus=
int tan =
bull Te de1047297nite integral has limits f( ) x b
dint is found by evaluating the integrated expression
at b and then subtracting the integrated expression evaluated at a
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Not for printing sharing or distribution
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596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
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Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
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7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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17 Basic integration and its applications 589
Happily there is a quicker way we can treat x as a functionof y effectively re1047298ecting the whole diagram in the line y = x and then use the same method as in the previous section
KEY POINT 1710
Te area bounded by the curve y the y -axis and the
lines y = c and y = d is given by y d
) dint where is
the expression for x in terms of y
You may have realised that this is related to inverse functions from Section 5E
x
y
x = f (y)
cd
A
Worked example 178
Te curve shown has equation y 1minus Find the shaded area
y
x
y = 2radic x minus 1
10
Express x in terms of y x = 2
rArr = + y
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590 Topic 6 Calculus
Exercise 17I
continued
Find the limits on the y -axisIt may help to label them on the
graph
When x = minusWhen x = 1 = =minus 6
y
y = 2radic minus
1
10
Write down the integral andevaluate using calculator
Area fro GDC= +2
1 2=d
1 Find the shaded areas
(a) (i) y
x
y = x2
1
6
(ii) y
x
y = x3
1
3
(b) (i) y
xy = 1
x2
2
1
(ii) y
x
y =radic x
1
5
(c) (i)y
x
y = ln x
1e
e
(ii) y
x
y = 3x
1 6
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17 Basic integration and its applications 591
2 Te diagram shows the curve If the shadedarea is 504 1047297nd the value of a [6 marks]
y
x
y = radic
x
a
2a
0
3 Find the exact value of the area enclosed by the graphof + the line y = 2 and the y -axis [6 marks]
4 Te diagram shows the graph of
Te shaded area is 39 units Find the value of a [7 marks]y
x
y =radic x
4 a
5 Te diagram shows the graph of 2 where a isin infin Te area of the pink region is equal to the area of the blue
region Give two equations for a in terms of b and hencegive a in exact form and determine the size of theblue area [8 marks]y
x
y = x2
b
1 a
17J The area between two curves
So far we have only looked at areas bounded by a curve and oneof the coordinate axes but we can also 1047297nd areas bounded bytwo curvesTe area A in the diagram can be found by taking the areabounded by f )x and the x -axis and subtracting the areabounded by and the x -axis that is
A
b
minus
y
x
y = f (x)
y = g(x)
a b
A
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592 Topic 6 Calculus
We can do the subtraction before integrating so that we onlyhave to integrate one expression instead of two Tis gives analternative formula for the area
KEY POINT 1711
Te area A between two curves f (x ) and g (x ) is
Ab
(
where a and b are the x-coordinates of the intersectionpoints of the two curves
Worked example 179
Find the area A enclosed between + and minus2 +
First find the x -coordinates ofintersection
For intersection
x x x
rArr =
rArr ) =rArr =
minus
minus
+x
4minusx 0
Make a rough sketch to see therelative positions of the two curves
y
x1 4
A
2x + 1y =
2 minus 3x + 5y = x
Subtract the lower curve from thehigher before integrating
= x
minusx x
4
minus
minus minusx
2
1
4
2
minus =
8
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7242019 Chapter 17 IB Maths HL Cambridge Textbook
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17 Basic integration and its applications 593
Subtracting the two equations before integrating is particularlyuseful when one of the curves is partly below the x-axis If x is always above then the expression we are integrating f g )x minus )x is always positive so we do not have to worryabout the signs of f )x and g )x themselves
Worked example 1710
Find the area bounded by the curves y x minus and y x 2
Sketch the graph to see therelative position of two
curves
Using GDC
= ex minus 5
y = 3 2
y
minus2 2
A
Find the intersection points ndash
use calculatorintersections x = minus2 658
Write down the integralrepresenting the area
rea =minus
minus818
1 658
xminus
= minus
1 6 8
x
Evaluate the integral usingcalculator = 21 6 (3SF)
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594 Topic 6 Calculus
Exercise 17J
1 Find the shaded areas
(a) (i)
y = 2x+ 1
y = (xminus1)2
y
x
(ii) y =x+ 1
y = 4xminusx2minus1
y
x
(b) (i)
y =minusx2minus4x+ 12
y
x
y =x2 + 2x+ 12
(ii)
y =x2minus2x+ 9
y
x
y = 4xminusx2 + 5
(c) (i)y =x2 minusx
y = 2xminusx2
y
x
(ii)y =x2minus7x+ 7
y = 3minusxminusx2
y
x
2 Find the area enclosed between the graphs of y x +x minus and + [6 marks]
3 Find the area enclosed by the curve 2
the y-axisand the line [6 marks]
4 Find the area between the curves1
and in theregion lt lt π [6 marks]
5 Show that the area of the shaded region alongside is2
[6 marks]
y = x2
y
xminus1 2
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7242019 Chapter 17 IB Maths HL Cambridge Textbook
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17 Basic integration and its applications 595
6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]
7 Find the total area enclosed between the graphs of
)minus 2 and minus2 7 1+ [6 marks]
8 Te area enclosed between the curve y x 2 and the line
y mx is 10 Find the value of m if m 0 [7 marks]
9 Show that the shaded area in the diagram below is [8 marks]
y = 2 minus xy2 = x
y
x
y = cos x y = sin x
y
x
Summary
bull Integration is the reverse process of differentiation
bull Any integral without limits (inde1047297nite) will generate a constant of integration
bull For all rational n ne minus1 int +x c
1
1
bull If minus we get the natural logarithm function int minus =
bull Te integral of the exponential function is int e c
bull Te integrals of the trigonometric functions are
int si minus=
int tan =
bull Te de1047297nite integral has limits f( ) x b
dint is found by evaluating the integrated expression
at b and then subtracting the integrated expression evaluated at a
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Not for printing sharing or distribution
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596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
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Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
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7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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590 Topic 6 Calculus
Exercise 17I
continued
Find the limits on the y -axisIt may help to label them on the
graph
When x = minusWhen x = 1 = =minus 6
y
y = 2radic minus
1
10
Write down the integral andevaluate using calculator
Area fro GDC= +2
1 2=d
1 Find the shaded areas
(a) (i) y
x
y = x2
1
6
(ii) y
x
y = x3
1
3
(b) (i) y
xy = 1
x2
2
1
(ii) y
x
y =radic x
1
5
(c) (i)y
x
y = ln x
1e
e
(ii) y
x
y = 3x
1 6
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17 Basic integration and its applications 591
2 Te diagram shows the curve If the shadedarea is 504 1047297nd the value of a [6 marks]
y
x
y = radic
x
a
2a
0
3 Find the exact value of the area enclosed by the graphof + the line y = 2 and the y -axis [6 marks]
4 Te diagram shows the graph of
Te shaded area is 39 units Find the value of a [7 marks]y
x
y =radic x
4 a
5 Te diagram shows the graph of 2 where a isin infin Te area of the pink region is equal to the area of the blue
region Give two equations for a in terms of b and hencegive a in exact form and determine the size of theblue area [8 marks]y
x
y = x2
b
1 a
17J The area between two curves
So far we have only looked at areas bounded by a curve and oneof the coordinate axes but we can also 1047297nd areas bounded bytwo curvesTe area A in the diagram can be found by taking the areabounded by f )x and the x -axis and subtracting the areabounded by and the x -axis that is
A
b
minus
y
x
y = f (x)
y = g(x)
a b
A
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7242019 Chapter 17 IB Maths HL Cambridge Textbook
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592 Topic 6 Calculus
We can do the subtraction before integrating so that we onlyhave to integrate one expression instead of two Tis gives analternative formula for the area
KEY POINT 1711
Te area A between two curves f (x ) and g (x ) is
Ab
(
where a and b are the x-coordinates of the intersectionpoints of the two curves
Worked example 179
Find the area A enclosed between + and minus2 +
First find the x -coordinates ofintersection
For intersection
x x x
rArr =
rArr ) =rArr =
minus
minus
+x
4minusx 0
Make a rough sketch to see therelative positions of the two curves
y
x1 4
A
2x + 1y =
2 minus 3x + 5y = x
Subtract the lower curve from thehigher before integrating
= x
minusx x
4
minus
minus minusx
2
1
4
2
minus =
8
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7242019 Chapter 17 IB Maths HL Cambridge Textbook
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17 Basic integration and its applications 593
Subtracting the two equations before integrating is particularlyuseful when one of the curves is partly below the x-axis If x is always above then the expression we are integrating f g )x minus )x is always positive so we do not have to worryabout the signs of f )x and g )x themselves
Worked example 1710
Find the area bounded by the curves y x minus and y x 2
Sketch the graph to see therelative position of two
curves
Using GDC
= ex minus 5
y = 3 2
y
minus2 2
A
Find the intersection points ndash
use calculatorintersections x = minus2 658
Write down the integralrepresenting the area
rea =minus
minus818
1 658
xminus
= minus
1 6 8
x
Evaluate the integral usingcalculator = 21 6 (3SF)
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7242019 Chapter 17 IB Maths HL Cambridge Textbook
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594 Topic 6 Calculus
Exercise 17J
1 Find the shaded areas
(a) (i)
y = 2x+ 1
y = (xminus1)2
y
x
(ii) y =x+ 1
y = 4xminusx2minus1
y
x
(b) (i)
y =minusx2minus4x+ 12
y
x
y =x2 + 2x+ 12
(ii)
y =x2minus2x+ 9
y
x
y = 4xminusx2 + 5
(c) (i)y =x2 minusx
y = 2xminusx2
y
x
(ii)y =x2minus7x+ 7
y = 3minusxminusx2
y
x
2 Find the area enclosed between the graphs of y x +x minus and + [6 marks]
3 Find the area enclosed by the curve 2
the y-axisand the line [6 marks]
4 Find the area between the curves1
and in theregion lt lt π [6 marks]
5 Show that the area of the shaded region alongside is2
[6 marks]
y = x2
y
xminus1 2
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7242019 Chapter 17 IB Maths HL Cambridge Textbook
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17 Basic integration and its applications 595
6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]
7 Find the total area enclosed between the graphs of
)minus 2 and minus2 7 1+ [6 marks]
8 Te area enclosed between the curve y x 2 and the line
y mx is 10 Find the value of m if m 0 [7 marks]
9 Show that the shaded area in the diagram below is [8 marks]
y = 2 minus xy2 = x
y
x
y = cos x y = sin x
y
x
Summary
bull Integration is the reverse process of differentiation
bull Any integral without limits (inde1047297nite) will generate a constant of integration
bull For all rational n ne minus1 int +x c
1
1
bull If minus we get the natural logarithm function int minus =
bull Te integral of the exponential function is int e c
bull Te integrals of the trigonometric functions are
int si minus=
int tan =
bull Te de1047297nite integral has limits f( ) x b
dint is found by evaluating the integrated expression
at b and then subtracting the integrated expression evaluated at a
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2930
17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
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7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2330
17 Basic integration and its applications 591
2 Te diagram shows the curve If the shadedarea is 504 1047297nd the value of a [6 marks]
y
x
y = radic
x
a
2a
0
3 Find the exact value of the area enclosed by the graphof + the line y = 2 and the y -axis [6 marks]
4 Te diagram shows the graph of
Te shaded area is 39 units Find the value of a [7 marks]y
x
y =radic x
4 a
5 Te diagram shows the graph of 2 where a isin infin Te area of the pink region is equal to the area of the blue
region Give two equations for a in terms of b and hencegive a in exact form and determine the size of theblue area [8 marks]y
x
y = x2
b
1 a
17J The area between two curves
So far we have only looked at areas bounded by a curve and oneof the coordinate axes but we can also 1047297nd areas bounded bytwo curvesTe area A in the diagram can be found by taking the areabounded by f )x and the x -axis and subtracting the areabounded by and the x -axis that is
A
b
minus
y
x
y = f (x)
y = g(x)
a b
A
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7242019 Chapter 17 IB Maths HL Cambridge Textbook
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592 Topic 6 Calculus
We can do the subtraction before integrating so that we onlyhave to integrate one expression instead of two Tis gives analternative formula for the area
KEY POINT 1711
Te area A between two curves f (x ) and g (x ) is
Ab
(
where a and b are the x-coordinates of the intersectionpoints of the two curves
Worked example 179
Find the area A enclosed between + and minus2 +
First find the x -coordinates ofintersection
For intersection
x x x
rArr =
rArr ) =rArr =
minus
minus
+x
4minusx 0
Make a rough sketch to see therelative positions of the two curves
y
x1 4
A
2x + 1y =
2 minus 3x + 5y = x
Subtract the lower curve from thehigher before integrating
= x
minusx x
4
minus
minus minusx
2
1
4
2
minus =
8
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7242019 Chapter 17 IB Maths HL Cambridge Textbook
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17 Basic integration and its applications 593
Subtracting the two equations before integrating is particularlyuseful when one of the curves is partly below the x-axis If x is always above then the expression we are integrating f g )x minus )x is always positive so we do not have to worryabout the signs of f )x and g )x themselves
Worked example 1710
Find the area bounded by the curves y x minus and y x 2
Sketch the graph to see therelative position of two
curves
Using GDC
= ex minus 5
y = 3 2
y
minus2 2
A
Find the intersection points ndash
use calculatorintersections x = minus2 658
Write down the integralrepresenting the area
rea =minus
minus818
1 658
xminus
= minus
1 6 8
x
Evaluate the integral usingcalculator = 21 6 (3SF)
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7242019 Chapter 17 IB Maths HL Cambridge Textbook
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594 Topic 6 Calculus
Exercise 17J
1 Find the shaded areas
(a) (i)
y = 2x+ 1
y = (xminus1)2
y
x
(ii) y =x+ 1
y = 4xminusx2minus1
y
x
(b) (i)
y =minusx2minus4x+ 12
y
x
y =x2 + 2x+ 12
(ii)
y =x2minus2x+ 9
y
x
y = 4xminusx2 + 5
(c) (i)y =x2 minusx
y = 2xminusx2
y
x
(ii)y =x2minus7x+ 7
y = 3minusxminusx2
y
x
2 Find the area enclosed between the graphs of y x +x minus and + [6 marks]
3 Find the area enclosed by the curve 2
the y-axisand the line [6 marks]
4 Find the area between the curves1
and in theregion lt lt π [6 marks]
5 Show that the area of the shaded region alongside is2
[6 marks]
y = x2
y
xminus1 2
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Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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17 Basic integration and its applications 595
6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]
7 Find the total area enclosed between the graphs of
)minus 2 and minus2 7 1+ [6 marks]
8 Te area enclosed between the curve y x 2 and the line
y mx is 10 Find the value of m if m 0 [7 marks]
9 Show that the shaded area in the diagram below is [8 marks]
y = 2 minus xy2 = x
y
x
y = cos x y = sin x
y
x
Summary
bull Integration is the reverse process of differentiation
bull Any integral without limits (inde1047297nite) will generate a constant of integration
bull For all rational n ne minus1 int +x c
1
1
bull If minus we get the natural logarithm function int minus =
bull Te integral of the exponential function is int e c
bull Te integrals of the trigonometric functions are
int si minus=
int tan =
bull Te de1047297nite integral has limits f( ) x b
dint is found by evaluating the integrated expression
at b and then subtracting the integrated expression evaluated at a
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2930
17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
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7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2430
592 Topic 6 Calculus
We can do the subtraction before integrating so that we onlyhave to integrate one expression instead of two Tis gives analternative formula for the area
KEY POINT 1711
Te area A between two curves f (x ) and g (x ) is
Ab
(
where a and b are the x-coordinates of the intersectionpoints of the two curves
Worked example 179
Find the area A enclosed between + and minus2 +
First find the x -coordinates ofintersection
For intersection
x x x
rArr =
rArr ) =rArr =
minus
minus
+x
4minusx 0
Make a rough sketch to see therelative positions of the two curves
y
x1 4
A
2x + 1y =
2 minus 3x + 5y = x
Subtract the lower curve from thehigher before integrating
= x
minusx x
4
minus
minus minusx
2
1
4
2
minus =
8
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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17 Basic integration and its applications 593
Subtracting the two equations before integrating is particularlyuseful when one of the curves is partly below the x-axis If x is always above then the expression we are integrating f g )x minus )x is always positive so we do not have to worryabout the signs of f )x and g )x themselves
Worked example 1710
Find the area bounded by the curves y x minus and y x 2
Sketch the graph to see therelative position of two
curves
Using GDC
= ex minus 5
y = 3 2
y
minus2 2
A
Find the intersection points ndash
use calculatorintersections x = minus2 658
Write down the integralrepresenting the area
rea =minus
minus818
1 658
xminus
= minus
1 6 8
x
Evaluate the integral usingcalculator = 21 6 (3SF)
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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594 Topic 6 Calculus
Exercise 17J
1 Find the shaded areas
(a) (i)
y = 2x+ 1
y = (xminus1)2
y
x
(ii) y =x+ 1
y = 4xminusx2minus1
y
x
(b) (i)
y =minusx2minus4x+ 12
y
x
y =x2 + 2x+ 12
(ii)
y =x2minus2x+ 9
y
x
y = 4xminusx2 + 5
(c) (i)y =x2 minusx
y = 2xminusx2
y
x
(ii)y =x2minus7x+ 7
y = 3minusxminusx2
y
x
2 Find the area enclosed between the graphs of y x +x minus and + [6 marks]
3 Find the area enclosed by the curve 2
the y-axisand the line [6 marks]
4 Find the area between the curves1
and in theregion lt lt π [6 marks]
5 Show that the area of the shaded region alongside is2
[6 marks]
y = x2
y
xminus1 2
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2730
17 Basic integration and its applications 595
6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]
7 Find the total area enclosed between the graphs of
)minus 2 and minus2 7 1+ [6 marks]
8 Te area enclosed between the curve y x 2 and the line
y mx is 10 Find the value of m if m 0 [7 marks]
9 Show that the shaded area in the diagram below is [8 marks]
y = 2 minus xy2 = x
y
x
y = cos x y = sin x
y
x
Summary
bull Integration is the reverse process of differentiation
bull Any integral without limits (inde1047297nite) will generate a constant of integration
bull For all rational n ne minus1 int +x c
1
1
bull If minus we get the natural logarithm function int minus =
bull Te integral of the exponential function is int e c
bull Te integrals of the trigonometric functions are
int si minus=
int tan =
bull Te de1047297nite integral has limits f( ) x b
dint is found by evaluating the integrated expression
at b and then subtracting the integrated expression evaluated at a
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2830
596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2930
17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2530
17 Basic integration and its applications 593
Subtracting the two equations before integrating is particularlyuseful when one of the curves is partly below the x-axis If x is always above then the expression we are integrating f g )x minus )x is always positive so we do not have to worryabout the signs of f )x and g )x themselves
Worked example 1710
Find the area bounded by the curves y x minus and y x 2
Sketch the graph to see therelative position of two
curves
Using GDC
= ex minus 5
y = 3 2
y
minus2 2
A
Find the intersection points ndash
use calculatorintersections x = minus2 658
Write down the integralrepresenting the area
rea =minus
minus818
1 658
xminus
= minus
1 6 8
x
Evaluate the integral usingcalculator = 21 6 (3SF)
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7242019 Chapter 17 IB Maths HL Cambridge Textbook
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594 Topic 6 Calculus
Exercise 17J
1 Find the shaded areas
(a) (i)
y = 2x+ 1
y = (xminus1)2
y
x
(ii) y =x+ 1
y = 4xminusx2minus1
y
x
(b) (i)
y =minusx2minus4x+ 12
y
x
y =x2 + 2x+ 12
(ii)
y =x2minus2x+ 9
y
x
y = 4xminusx2 + 5
(c) (i)y =x2 minusx
y = 2xminusx2
y
x
(ii)y =x2minus7x+ 7
y = 3minusxminusx2
y
x
2 Find the area enclosed between the graphs of y x +x minus and + [6 marks]
3 Find the area enclosed by the curve 2
the y-axisand the line [6 marks]
4 Find the area between the curves1
and in theregion lt lt π [6 marks]
5 Show that the area of the shaded region alongside is2
[6 marks]
y = x2
y
xminus1 2
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
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17 Basic integration and its applications 595
6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]
7 Find the total area enclosed between the graphs of
)minus 2 and minus2 7 1+ [6 marks]
8 Te area enclosed between the curve y x 2 and the line
y mx is 10 Find the value of m if m 0 [7 marks]
9 Show that the shaded area in the diagram below is [8 marks]
y = 2 minus xy2 = x
y
x
y = cos x y = sin x
y
x
Summary
bull Integration is the reverse process of differentiation
bull Any integral without limits (inde1047297nite) will generate a constant of integration
bull For all rational n ne minus1 int +x c
1
1
bull If minus we get the natural logarithm function int minus =
bull Te integral of the exponential function is int e c
bull Te integrals of the trigonometric functions are
int si minus=
int tan =
bull Te de1047297nite integral has limits f( ) x b
dint is found by evaluating the integrated expression
at b and then subtracting the integrated expression evaluated at a
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2830
596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2930
17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 3030
7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2630
594 Topic 6 Calculus
Exercise 17J
1 Find the shaded areas
(a) (i)
y = 2x+ 1
y = (xminus1)2
y
x
(ii) y =x+ 1
y = 4xminusx2minus1
y
x
(b) (i)
y =minusx2minus4x+ 12
y
x
y =x2 + 2x+ 12
(ii)
y =x2minus2x+ 9
y
x
y = 4xminusx2 + 5
(c) (i)y =x2 minusx
y = 2xminusx2
y
x
(ii)y =x2minus7x+ 7
y = 3minusxminusx2
y
x
2 Find the area enclosed between the graphs of y x +x minus and + [6 marks]
3 Find the area enclosed by the curve 2
the y-axisand the line [6 marks]
4 Find the area between the curves1
and in theregion lt lt π [6 marks]
5 Show that the area of the shaded region alongside is2
[6 marks]
y = x2
y
xminus1 2
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2730
17 Basic integration and its applications 595
6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]
7 Find the total area enclosed between the graphs of
)minus 2 and minus2 7 1+ [6 marks]
8 Te area enclosed between the curve y x 2 and the line
y mx is 10 Find the value of m if m 0 [7 marks]
9 Show that the shaded area in the diagram below is [8 marks]
y = 2 minus xy2 = x
y
x
y = cos x y = sin x
y
x
Summary
bull Integration is the reverse process of differentiation
bull Any integral without limits (inde1047297nite) will generate a constant of integration
bull For all rational n ne minus1 int +x c
1
1
bull If minus we get the natural logarithm function int minus =
bull Te integral of the exponential function is int e c
bull Te integrals of the trigonometric functions are
int si minus=
int tan =
bull Te de1047297nite integral has limits f( ) x b
dint is found by evaluating the integrated expression
at b and then subtracting the integrated expression evaluated at a
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2830
596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2930
17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 3030
7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2730
17 Basic integration and its applications 595
6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]
7 Find the total area enclosed between the graphs of
)minus 2 and minus2 7 1+ [6 marks]
8 Te area enclosed between the curve y x 2 and the line
y mx is 10 Find the value of m if m 0 [7 marks]
9 Show that the shaded area in the diagram below is [8 marks]
y = 2 minus xy2 = x
y
x
y = cos x y = sin x
y
x
Summary
bull Integration is the reverse process of differentiation
bull Any integral without limits (inde1047297nite) will generate a constant of integration
bull For all rational n ne minus1 int +x c
1
1
bull If minus we get the natural logarithm function int minus =
bull Te integral of the exponential function is int e c
bull Te integrals of the trigonometric functions are
int si minus=
int tan =
bull Te de1047297nite integral has limits f( ) x b
dint is found by evaluating the integrated expression
at b and then subtracting the integrated expression evaluated at a
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2830
596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2930
17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 3030
7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2830
596 Topic 6 Calculus
bull Te area between the curve y = f (x ) the x -axis and lines and x is given by
Ab
If the curve goes below the x-axis the value of this integral will be negative
bull On the calculator we can use the modulus function to ensure we are always integrating apositive function
bull Te area between the curve the y -axis and lines y c and y = d is given by A g y
bull Te area between two curves is given by
Ab
(
where x a and x b are the intersection points
Introductory problem revisited
Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used
Te area under the curve of I against t is given by sin cosint 0
ππ For a rectangle of
width π to have the same area the height must beπ
t
l
I = sin t
π
t
l
k
π
You can look at integration as a quite sophisticated way of finding an average value ofa function
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2930
17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 3030
7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 2930
17 Basic integration and its applications 597
Short questions
1 If ) = f prime x s n and
1047297nd x [4 marks]
2 Calculate the area enclosed by the curves and
[6 marks]
[copy IB Organization 2003]
3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]
4 Te diagram shows the graph of n for 1
y
xa b
0
Te red area is three times larger than the blue area Find the value of n [6 marks]
5 Find the inde1047297nite integral
int [5 marks]
6 (a) Solve the equation
a
int
(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]
Mixed examination practice 17
copy Cambridge University Press 2012
Not for printing sharing or distribution
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 3030
7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]
7242019 Chapter 17 IB Maths HL Cambridge Textbook
httpslidepdfcomreaderfullchapter-17-ib-maths-hl-cambridge-textbook 3030
7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]
8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +
What kind of stationary point is ( )1 [5 marks]
(b) Find f )x
Long questions
1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2
(b) Find the area enclosed between these two graphs
(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]
2 (a) Use the identity 1= to show that cos x arcs n x 2
(b) Te diagram below shows part of the curve
y
x
a
P
y = sinx
Write down the x -coordinate of the point P in terms of a
(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions
(d) By considering the blue shaded area 1047297nd arcsin x x int for 1
[12 marks]