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Edexcel Core

Mathematics 4 Integration

Edited by: K V Kumaran


Integration

It might appear to be a bit obvious but you must remember all of your C3

work on differentiation if you are to succeed with this unit.

Content

By the end of this unit you should be able to:

Integrate trigonometric, exponential functions and x

1

Find volumes of revolution

Find integrals using partial fractions and integrating by parts.

Find areas under curves and volumes generated from functions given in parametric

form.


Standard Integrals

Function (f(x)) Integral f(x)dx

xn n 1xc

n 1

1

x

ln x c

(ax + b)n n 1(ax b)c

a(n 1)

1

ax b

1ln ax b c

a

ex ex + c

Sin x -Cos x + c

Cos x Sin x + c

Tan x ln|(Sec x)| + c

Cot x ln|(Sin x)| + c

Cosec x -ln|(Cosec x + Cot x)| + c

Sec x ln|(Sec x + Tan x)| + c

Sec2x Tanx + c

Sec x Tan x Sec x + c

-Cosec2x Cot x + c

eax+b ax bec

a

Cos (ax + b) 1Sin(ax b) c

a

SinnxCosx n 11Sin x c

n 1

CosnxSinx n 11Cos x c

n 1

Sin2x cxSin

x

2

2

2

1

Cos2x cxxSin

2

2

2

1


Integration of trigonometric functions.

To succeed with these types of questions you need to have a very good grasp of

the trig identities from C3 and obviously the differentiation from the same unit.

Integration questions involving trigonometry are either of a substitution type or

areas and volumes from parametrics. The following examples give a flavour but

can’t cover every eventuality.

Example 1

By using the substitution u = sinx or otherwise, find

28sin xsin2xdx

Giving your answer in terms of x.

From C3 you should remember that:

sin2x = 2sinx cosx

Hence:

2 38sin xsin2xdx 16sin xcosxdx

If u = sinx du = cosdx (part of the integral!)

Therefore 3 316sin xcosxdx 16u du

4 44u c 4sin x c


Example 2

Use the substitution x = sin θ to find the exact value of

1

230

2 2

4dx

(1 x )

If x = sin θ then the denominator becomes (1 – sin2θ) which is cos2θ. When this is

raised to the power of 1.5 the result is cos3θ as a denominator. The x’s have now

been substituted but you can’t forget about the dx.

If x = sin θ then dx = cosθdθ. Therefore the integral becomes:

1

23 30

2 22 2

3 2

4 4dx cos d

(1 x ) (1 sin )

4 4cos d d

cos cos

The next thing to spot is that 22

44sec

cos

. Therefore the question becomes:

24sec d 4tan

The question isn’t quite finished as the original limits were given in terms of x. So

for the lower limit of x = 0, sin θ = 0, hence θ = 0. For the upper limit x = 0.5, so

sin θ = 0.5, hence θ =3

. The question asks for the exact value so the integral is

finally.

23 300

4sec d 4tan

4 4 3

33

There are a few tricks in this question but practice makes perfect and you must

remember all of your differentiation and trig work from C3 and C4.

Use ideas outlined in the examples above to carry out the next question.

Use the substitution u2 = 1 – cos2θ and integration to find

2

0Sin2 1 Cos2 d

.

(You will need to use implicit differentiation when dealing with the u2.)


Integrating by Parts.

This concept is guaranteed to appear on a C4 paper and is not too tricky.

Integration by parts is used to integrate functions that are made up of products

of functions. Not surprisingly the process is the reverse of differentiating a

product. The idea is best explained through an example.

Example 5

Use integration by parts to find the exact value of 3 5

1x lnxdx

The rule for integrating by parts is:

dv du

u dx uv v dxdx dx

The thing to remember is that one part of the question becomes u and the other

part becomesdv

dx. Since we can’t integrate lnx, it must be u and therefore x5 must

equaldv

dx. On the right of the rule we need v and

du

dx this can be achieved by

integrating and differentiating respectively.

Therefore:

3 5

1x lnxdx

Let u = lnx and 5dvx

dx

du 1

dx x

6xv

6

So by substituting into the rule:

dv du

u dx uv v dxdx dx


6 63 35

1 1

356 6 63

11

36

1

x 1 xx lnxdx dx

6 x 6

x x x xdx

6 6 6 36

5x 405 5

36 4 36

910

9


Volume of Revolution

The basic principle here is that we take very thin strips, of width δx (delta x),

underneath a curve in the xy plane and rotate them around the x axis to generate a

volume.

The diagram on the left shows the graph of y = x2 in 3D. The pink section is the

area under the graph between the x values of 0 and 1. This pink section has been

rotated through 360 degrees about the x axis to form the green solid in the

diagram on the right.

If you look at the far end of the green solid it has a circular appearance. The area

of the end is πy2 and therefore the volume of a very thin disc of width δx will be

πy2δx. We use integration to add up the volumes of all of the very thin discs to

give the volume of revolution.

∑ πy2δx = 2y dx

Curves can be rotated about the x or y axis and therefore there are two formulae

to calculate volumes.

The volume generated by a curve as it rotates around the x axis is given by:

2y dx


Example 3

The finite region bounded by the curve with the equation y = 8x – x2 and the x axis

is rotated through 360º about the x axis. Using integration find, in terms of π,

the volume of its solid form.

The volume generated by a curve as it rotates around the x axis is given by:

2y dx

The mistake a number of candidates make is to state that y2 in this case is 64x2 +

x4 WRONG

Y2 = (8x – x2)( 8x – x2) = 64x2 – 16x3 + x4

Therefore volume is given by :

2 3 4

53 4

(64x - 16x + x )dx

64 xx 4x

3 5

The quadratic, y = 8x – x2, has solutions of x = 0 and x = 8 and therefore these

become the limits on the integral.

853 4

0

64 xx 4x

3 5

16380

15


Example 4

Find the volume generated when the region bounded by the curve with equation 5

y 2x

, the x axis and the lines x = 1

3 and x = 3 is rotated through 360 degrees

about the x axis.

Volume = 2y dx

23 3

1 1 23 3

3

1

3

5 20 252 dx 4 dx

x x x

254x 20lnx

x

25 4 112 20ln3 20ln 75

3 3 3

23220ln9

3

Take care with the negative powers when integrating and the rules of logs. In

most cases the question will specifically ask for answers in terms of a + bln9 for

example. A calculator answer will loose you the final answer mark.


More complex integrations by substitution.

Example 9

Use the substitution u2 = 2x – 1, or otherwise, find the exact value of

25

13

6xdx

2x 1

From an earlier example you should remember that there are three parts that

need dealing with, 6x, the dx and the square root in the denominator

If u2 = 2x – 1 then 2udu = 2dx ( by implicit differentiation). So dx = udu.

If u2 = 2x – 1 then x = 0.5u2 + 0.5. The integral becomes:

225

13

2 3

3u 36xdx udu

u2x 1

(3u 3)du u 3u

And now for the limits. If x = 13 then u = 5 and if x = 25 then u = 7.

25 73

513

6xdx u 3u

2x 1

224


Integration of Parametrics to Find Areas under curves.

Example 10

The diagram shows a sketch of the curve C with parametric equations

x = 7tsint, y = 5sect, 0 ≤ t < 2

The point P(a, 10) lies on C.

a) Find the exact value of a.

The region R is enclosed by the curve C, the axes and the line x = a.

b) Show that the area of R is given by

3

035 tant t dt

c) Find the exact value of the area R.

a) Find the exact value of a.

The point P has y coordinate 10. Therefore:

5sect = 10 If t = 3

sect = 2 x = 7

sin3 3

cost = 0.5 x = 7 3

6


t = 3

a =

7 3

6

b) When integrating to find the area under a parametric curve you need to

integrate the following:

dx

y dtdt

The x part of the parametric is a product and therefore we need to differentiate

the first (7t) and multiply it second (sint) then add the first multiplied by the

differential of the second. This gives:

dx

7sint 7tcostdt

So to find the area R:

3

0

dxy dt 5sect 7 sint 7tcost

dt1 sint

Remembering that sect = and that tantcost cost

35 tant t dt

The limits are defined from the work in part (a).

c) The integral from part (b) becomes:

3

0

sint35 t dt

cost

The integral of tant should be in your notes. The denominator nearly

differentiates to give the numerator. Therefore the integral is given by the

negative ln of cost.


3 3

0 0

2 30

sint35 t dt 35 tant t dt

cost

35 lncost 0.5t

8.45

= 43.5

If you are asked to find the volume generated by a parametric as it rotates about

the x axis then the integral to evaluate is:

2 dxy dt

dt


Finally an integration question that has an exponential substitution!

Example 12

Use the substitution u = 3x to find the exact value of

2

0 x

x

13

dx3

The integral very quickly becomes

1u

udx but we haven’t dealt with the dx.

From example 9 of the C4 differentiation notes the differential of u is:

33dx

du x ln

Therefore dx3u

du

ln and the integral becomes:

3

5

1u3

1

1u3

du9

1

9

1

ln

ln

)ln(ln)(ln

Remember that the question said exact answer so leave it in terms of logs.


Past paper questions on Integration

1. (a) Express )2)(32(

35

xx

x in partial fractions.

(3)

(b) Hence find the exact value of xxx

xd

)2)(32(

356

2

, giving your answer as a single logarithm.

(5)

(C4, June 2005 Q3)

2. Use the substitution x = sin to find the exact value of

xx

d)1(

12

1

2

3

02

.

(7)

(C4, June 2005 Q4)

3. Using the substitution u2 = 2x – 1, or otherwise, find the exact value of

5

1

d)12(

3x

x

x. (8)

(C4, Jan 2006 Q3)

4. Figure 1

Figure 1 shows the finite region R, which is bounded by the curve y = xex, the line x = 1, the line x

= 3 and the x-axis.

y

R

O

y = xex

3 1 x


The region R is rotated through 360 degrees about the x-axis.

Use integration by parts to find an exact value for the volume of the solid generated. (8)

(C4, Jan 2006 Q4)

5. Figure 2

The curve shown in Figure 2 has parametric equations

x = t – 2 sin t, y = 1 – 2 cos t, 0 t 2.

(a) Show that the curve crosses the x-axis where t = 3

and t =

3

5.

(2)

The finite region R is enclosed by the curve and the x-axis, as shown shaded in Figure 2.

(b) Show that the area R is given by the integral

3

5

3

d)cos21( 2

tt .

(3)

(c) Use this integral to find the exact value of the shaded area.

(7)

(C4, Jan 2006 Q8)

6. Figure 3

O

R

x

y


The curve with equation y = 3 sin 2

x, 0 x 2, is shown in Figure 1. The finite region enclosed

by the curve and the x-axis is shaded.

(a) Find, by integration, the area of the shaded region.

(3)

This region is rotated through 2 radians about the x-axis.

(b) Find the volume of the solid generated.

(6)

(C4, June 2006 Q3)

O 2 x

y


7. Figure 1

The curve with equation y = )21(3

1

x, x > –

21 , is shown in Figure 1.

The region bounded by the lines x = –41 , x =

21 , the x-axis and the curve is shown shaded in Figure

1.

This region is rotated through 360 degrees about the x-axis.

(a) Use calculus to find the exact value of the volume of the solid generated.

(5)

Figure 2

Figure 2 shows a paperweight with axis of symmetry AB where AB = 3 cm. A is a point on the top

surface of the paperweight, and B is a point on the base of the paperweight. The paperweight is

geometrically similar to the solid in part (a).

(b) Find the volume of this paperweight.

(2)

(C4, Jan 2007 Q2)

A

B

y

– 0 x


8. Use the substitution u = 2x to find the exact value of

1

0

2)12(

2x

x

dx.

(6)

(C4, June 2007 Q2)

9. (a) Find

xxx d2cos .

(4)

(b) Hence, using the identity cos 2x = 2 cos2 x – 1, deduce

xxx dcos 2 .

(3)

(C4, June 2007 Q3)

10. )12)(12(

)14(2 2

xx

x A +

)12( x

B +

)12( x

C.

(a) Find the values of the constants A, B and C.

(4)

(b) Hence show that the exact value of

2

1

2

)12)(12(

)14(2

xx

x dx is 2 + ln k, giving the value of the

constant k.

(6)

(C4, June 2007 Q4)


11.

The curve shown in Figure 2 has equation y = )12(

1

x. The finite region bounded by the curve,

the x-axis and the lines x = a and x = b is shown shaded in Figure 2. This region is rotated through

360° about the x-axis to generate a solid of revolution.

Find the volume of the solid generated. Express your answer as a single simplified fraction, in

terms of a and b.

(5)

(C4, Jan 2008 Q3)

12. (i) Find

x

xd

2ln .

(4)

(ii) Find the exact value of

2

4

2 dsin

xx .

(5)

(C4, Jan 2008 Q4)


13.

Figure 3

The curve C has parametric equations

x = ln (t + 2), y = )1(

1

t, t > 1.

The finite region R between the curve C and the x-axis, bounded by the lines with equations x = ln

2 and x = ln 4, is shown shaded in Figure 3.

(a) Show that the area of R is given by the integral

2

0

d)2)(1(

1t

tt.

(4)

(b) Hence find an exact value for this area.

(6)

(c) Find a cartesian equation of the curve C, in the form y = f(x).

(4)

(d) State the domain of values for x for this curve.

(1)

(C4, Jan 2008 Q7)

14. (a) Use integration by parts to find

xx x de .

(3)

(b) Hence find

xx x de2 .

(3)

(C4, June 2008 Q2)


15.

Figure 3

Figure 3 shows the curve C with parametric equations

x = 8 cos t, y = 4 sin 2t, 0 t 2

.

The point P lies on C and has coordinates (4, 2√3).

(a) Find the value of t at the point P.

(2)

The line l is a normal to C at P.

(b) Show that an equation for l is y = –x√3 + 6√3.

(6)

The finite region R is enclosed by the curve C, the x-axis and the line x = 4, as shown shaded in

Figure 3.

(c) Show that the area of R is given by the integral

2

3

2 dcossin64

ttt .

(4)

(d) Use this integral to find the area of R, giving your answer in the form a + b√3, where a and b

are constants to be determined.

(4)

(C4, June 2008 Q8)


16.

Figure 1

Figure 1 shows part of the curve y = )41(

3

x. The region R is bounded by the curve, the x-axis,

and the lines x = 0 and x = 2, as shown shaded in Figure 1.

(a) Use integration to find the area of R.

(4)

The region R is rotated 360° about the x-axis.

(b) Use integration to find the exact value of the volume of the solid formed.

(5)

(C4, Jan 2009 Q2)

17. (a) Find

xx dtan2 .

(2)

(b) Use integration by parts to find

xxx

dln1

3.

(4)

(c) Use the substitution u = 1 + ex to show that

x

x

e1

e3

dx = 2

1e2x – ex + ln (1 + ex) + k,

where k is a constant.

(7)

(C4, Jan2009 Q6)


18. f(x) = )3)(1)(12(

24

xxx

x =

)12( x

A +

)1( x

B +

)3( x

C .


(4)

(b) (i) Hence find

xx d)(f .

(3)

(ii) Find

2

0

d)(f xx in the form ln k, where k is a constant.

(3)

(C4, June 2009 Q3)

19. (a) Find

xx d)5( .

(2)

Figure 3

Figure 3 shows a sketch of the curve with equation

y = (x – 1)√(5 – x), 1 x 5

(b) (i) Using integration by parts, or otherwise, find

xxx d)5()1( .

(4)

(ii) Hence find

5

1

.d)5( )1( xxx .

(2)

(C4, June 2009 Q6)


20. (a) Using the identity cos 2θ = 1 – 2 sin2 θ , find

.sin 2 d

(2)

Figure 4

Figure 4 shows part of the curve C with parametric equations

x = tan θ, y = 2 sin 2θ, 0 θ < 2

.

The finite shaded region S shown in Figure 4 is bounded by C, the line x = 3

1

and the x-axis. This

shaded region is rotated through 2 radians about the x-axis to form a solid of revolution.

(b) Show that the volume of the solid of revolution formed is given by the integral

k

6

0

2sin

d ,

where k is a constant.

(5)

(c) Hence find the exact value for this volume, giving your answer in the form p 2 + q√3, where

p and q are constants.

(3)

(C4, June 2009 Q8)


21.

Figure 2

Figure 2 shows a sketch of the curve C with parametric equations

x = 5t 2 − 4, y = t(9 − t 2)

The curve C cuts the x-axis at the points A and B.

(a) Find the x-coordinate at the point A and the x-coordinate at the point B.

(3)

The region R, as shown shaded in Figure 2, is enclosed by the loop of the curve.

(b) Use integration to find the area of R.

(6)

(C4, Jan2010 Q7)


22. (a) Using the substitution x = 2 cos u, or otherwise, find the exact value of

2

122

d)4(

1x

xx.

(7)

Figure 3

Figure 3 shows a sketch of part of the curve with equation y = 4

1

)4(

4

2xx , 0 < x < 2.

The shaded region S, shown in Figure 3, is bounded by the curve, the x-axis and the lines with

equations x = 1 and x = √2. The shaded region S is rotated through 2π radians about the x-axis to

form a solid of revolution.

(b) Using your answer to part (a), find the exact volume of the solid of revolution formed.

(3)

(C4, Jan2010 Q8)

23. Using the substitution u = cos x +1, or otherwise, show that

2

0

1 dsine

xxxcos = e(e – 1).

(6)

(C4, June 2010 Q2)


24. f(θ) = 4 cos2 θ – 3sin2 θ

(a) Show that f(θ) = 2

1 +

2

7 cos 2θ. (3)

(b) Hence, using calculus, find the exact value of

2

0

df

)( .

(7)

(C4, June 2010 Q6)

25. Use integration to find the exact value of

2

0

d2sin

xxx . (6)

(C4, Jan 2011 Q1)

26. The curve C has parametric equations

x = ln t, y = t2 −2, t > 0.

Find

(a) an equation of the normal to C at the point where t = 3,

(6)

(b) a cartesian equation of C.

(3)

Figure 1

The finite area R, shown in Figure 1, is bounded by C, the x-axis, the line x = ln 2 and the line

x = ln 4. The area R is rotated through 360° about the x-axis.

(c) Use calculus to find the exact volume of the solid generated. (6)

(C4, June 2011 Q6)


27.

Figure 3

Figure 3 shows part of the curve C with parametric equations

x = tan , y = sin , 0 < 2

.

The point P lies on C and has coordinates

3

2

1,3 .

(a) Find the value of at the point P.

(2)

The line l is a normal to C at P. The normal cuts the x-axis at the point Q.

(b) Show that Q has coordinates (k3, 0), giving the value of the constant k.

(6)

The finite shaded region S shown in Figure 3 is bounded by the curve C, the line x = 3 and the x-

axis. This shaded region is rotated through 2 radians about the x-axis to form a solid of revolution.

(c) Find the volume of the solid of revolution, giving your answer in the form p 3 + q 2, where

p and q are constants.

(7)

(C4, June 2011 Q7)

28. (a) Use integration by parts to find

.d3sin xxx

(3)

(b) Using your answer to part (a), find

.d3cos2 xxx (3)

(C4, June 2011 Q3)

29.


Figure 1

Figure 1 shows the curve with equation

y =

43

22x

x, x 0.

The finite region S, shown shaded in Figure 1, is bounded by the curve, the x-axis and the

line x = 2.

The region S is rotated 360° about the x-axis.

Use integration to find the exact value of the volume of the solid generated, giving your answer in

the form k ln a, where k and a are constants.

(5)

(C4, Jan 2012 Q4)

30. f(x) = 2)13(

1

xx =

x

A +

)13( x

B +

2)13( x

C.


(4)

(b) (i) Hence find

xx d)(f .

(ii) Find

2

1

d)(f xx , leaving your answer in the form a + ln b, where a and b are constants.

(6)

(C4, June 2012 Q1)


31. (a) Use integration to find

xxx

dln1

3.

(5)

(b) Hence calculate

xxx

dln1

2

13

.

(2)

(C4, Jan 2013 Q2)

32.

Figure 3

Figure 3 shows a sketch of part of the curve with equation y = 1 – 2 cos x, where x is measured in

radians. The curve crosses the x-axis at the point A and at the point B.

(a) Find, in terms of , the x coordinate of the point A and the x coordinate of the point B.

(3)

The finite region S enclosed by the curve and the x-axis is shown shaded in Figure 3. The region

S is rotated through 2 radians about the x-axis.

(b) Find, by integration, the exact value of the volume of the solid generated.

(6)

(C4, Jan 2013 Q4)


33. (a) Find 2e dxx x .

(5)

(b) Hence find the exact value of 1 2

0e dxx x .

(2)

(C4, June 2013 Q1)

34. (a) Use the substitution x = u2, u > 0, to show that

1

(2 1)x x dx = 2

(2 1)u u du

(3)

(b) Hence show that

9

1

1

(2 1)x x dx = 2lna

b

where a and b are integers to be determined.

(7)

(C4, June 2013 Q5)

35. Using the substitution u = 2 + √(2x + 1), or other suitable substitutions, find the exact

value of

4

0

1

2 (2 1)x dx

giving your answer in the form A + 2ln B, where A is an integer and B is a positive constant.

(8)

(C4, June 2013_R Q5)


36.

Figure 2


327secx t , 3tany t , 0 ≤ t ≤ 3

(a) Find the gradient of the curve C at the point where t = 6

. (4)

(b) Show that the cartesian equation of C may be written in the form

2 13 2( 9)y x , a ≤ x ≤ b

stating values of a and b. (3)

Figure 3

The finite region R which is bounded by the curve C, the x-axis and the line x = 125 is shown

shaded in Figure 3. This region is rotated through 2π radians about the x-axis to form a solid of

revolution.

(c) Use calculus to find the exact value of the volume of the solid of revolution.

(5)

(C4, June 2013_R Q7)


37. (a) Express 2

25

2 1x x in partial fractions.

(4)

Figure 2

Figure 2 shows a sketch of part of the curve C with equation

5

2 1y

x x

, x > 0.

The finite region R is bounded by the curve C, the x-axis, the line with equation x = 1 and the

line with equation x = 4.

This region is shown shaded in Figure 2.

The region R is rotated through 360° about the x-axis.

(b) Use calculus to find the exact volume of the solid of revolution generated, giving your answer

in the form a + bln c, where a, b and c are constants.

(6)

(C4, June 2014_R Q4)


38.

Figure 4

Figure 4 shows a sketch of part of the curve C with parametric equations

x = 3tan θ, y = 4cos2 θ, 0 ≤ θ < 2

The point P lies on C and has coordinates (3, 2).

The line l is the normal to C at P. The normal cuts the x-axis at the point Q.

(a) Find the x coordinate of the point Q.

(6)

The finite region S, shown shaded in Figure 4, is bounded by the curve C, the x-axis, the

y-axis and the line l. This shaded region is rotated 2π radians about the x-axis to form a solid of

revolution.

(b) Find the exact value of the volume of the solid of revolution, giving your answer in the form

pπ + qπ2, where p and q are rational numbers to be determined.

[You may use the formula V = 1

3πr2h for the volume of a cone.]

(9)

(C4, June 2014 Q7)


39.

Figure 1

Figure 1 shows a sketch of part of the curve with equation y = 4x − x

x 2

1

e , x 0.

The curve meets the x-axis at the origin O and cuts the x-axis at the point A.

(a) Find, in terms of ln 2, the x coordinate of the point A.

(2)

(b) Find

xx

x

de 2

1

.

(3)

The finite region R, shown shaded in Figure 1, is bounded by the x-axis and the curve with equation

y = 4x − x

x 2

1

e , x 0.

(c) Find, by integration, the exact value for the area of R.

Give your answer in terms of ln 2.

(3)

(C4, June 2015, Q3)


40. (i) Given that y > 0, find

3 4d

(3 2)

yy

y y

.

(6)

(ii) (a) Use the substitution x = 4sin2θ to show that

3

0

d4

xx

x

= λ3

0

2sin d

,

where λ is a constant to be determined.

(5)

(b) Hence use integration to find

3

0

d4

xx

x

,

giving your answer in the form aπ + b, where a and b are exact constants.

(4)

(C4, June 2016, Q6)

41. (a) Find 3

2(2 1) dx x ,

giving your answer in its simplest form.

(2)

Figure 3 shows a sketch of part of the curve C with equation


3

41

, 2 2

1y x x .

The curve C cuts the line y = 8 at the point P with coordinates (k, 8), where k is a constant.

(b) Find the value of k.

(2)

The finite region S, shown shaded in Figure 3, is bounded by the curve C, the x-axis, the y-axis and

the line y = 8. This region is rotated through 2π radians about the x-axis to form a solid of

revolution.

(c) Find the exact value of the volume of the solid generated.

(4)

(C4, June 2016, Q7)

42.

The finite region S, shown shaded in Figure 2, is bounded by the y-axis, the x-axis,

the line with equation x = ln4 and the curve with equation

y = ex + 2e–x, x ⩾ 0

The region S is rotated through 2π radians about the x-axis.

Use integration to find the exact value of the volume of the solid generated.

Give your answer in its simplest form.

(C4, June 2017, Q5)

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5.

S

O ln4

y

x

Figure 2

The finite region S, shown shaded in Figure 2, is bounded by the y-axis, the x-axis,

the line with equation x = ln4 and the curve with equation

y = e x + 2e–x, x 0

The region S is rotated through 2 radians about the x-axis.

Use integration to find the exact value of the volume of the solid generated.

Give your answer in its simplest form.

[Solutions based entirely on graphical or numerical methods are not acceptable.]

(7)

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43.


x = 3𝜃sin𝜃, y = sec3𝜃, 0 ⩽ 𝜃 < 2

π

The point P(k, 8) lies on C, where k is a constant.

(a) Find the exact value of k.

(2)

The finite region R, shown shaded in Figure 4, is bounded by the curve C, the y-axis,

the x-axis and the line with equation x = k.

(b) Show that the area of R can be expressed in the form

l (q sec2q + tanq sec2q )dq

a

b

ò

where 𝜆, α and β are constants to be determined.

(4)

(c) Hence use integration to find the exact value of the area of R.

(6)

(C4, June 2017, Q8)

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8.y

C

R

k xO

P(k, 8)

Figure 4


x = 3 sin , y = sec3 , 0 <2

π

The point P(k, 8) lies on C, where k is a constant.

(a) Find the exact value of k.

(2)

The finite region R, shown shaded in Figure 4, is bounded by the curve C, the y-axis,

the x-axis and the line with equation x = k.

(b) Show that the area of R can be expressed in the form

( )2 2sec tan sec

β

α

λ θ θ θ θ dθ+ò where , and are constants to be determined.

(4)

(c) Hence use integration to find the exact value of the area of R.

(6)

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44. Given that

4x3 + 2x2 + 17x + 8 ≡ (Ax + B)(x2 + 4) + Cx + D

(a) find the values of the constants A, B, C and D.

(4)

(b) Hence find

4 3 2

21

4 2 17 8d

4

x x xx

x

giving your answer in the form p + ln q, where p and q are integers.

(6)

(C34, Jan 2014, Q3)

45.

Figure 4

Figure 4 shows a sketch of part of the curve with equation

y = x(sin x + cos x), 04

x

The finite region R, shown shaded in Figure 4, is bounded by the curve, the x-axis and the line

4x

. This shaded region is rotated through 2π radians about the x-axis to form a solid of

revolution, with volume V.

(a) Assuming the formula for volume of revolution show that 4

2

0

(1 sin 2 )dV x x x

.

(3)

(b) Hence using calculus find the exact value of V.

You must show your working.

(9)

(C34, IAL Jan 2014, Q12)


46. Find

(a) 12

2 3 dx x

(2)

(b) 2

5d

4 1

xx

x

(2)

(C34, IAL June 2014, Q4)

47.

Figure 1

Figure 1 shows a sketch of part of the curve with equation e xy , x > 0.

The finite region R, shown shaded in Figure 1, is bounded by the curve, the x-axis and the lines x

= 4 and x = 9.

(a) Use the trapezium rule, with 5 strips of equal width, to obtain an estimate for the area

of R, giving your answer to 2 decimal places.

(4)

(b) Use the substitution u = √x to find, by integrating, the exact value for the area of R.

(7)

(C34, IAL June 2014, Q9)


48.

Figure 3


tanx t , 22siny t , 02

t

The finite region S, shown shaded in Figure 3, is bounded by the curve C, the line x = √3 and the

x-axis. This shaded region is rotated through 2π radians about the x-axis to form a solid of

revolution.

(a) Show that the volume of the solid of revolution formed is given by

3 2 2

0

4 tan sin dt t t

(6)

(b) Hence use integration to find the exact value for this volume.

(6)

(C34, IAL June 2014, Q12)

49. Use the substitution x = 2sin 𝜃 to find the exact value of

3

30 2 2

1d

4

x

x

(7)

(C34, IAL Jan 2015, Q4)


50.

The curve C has parametric equations

x = ln (t + 2), y = 2

4

t t > 0

The finite region R, shown shaded in Figure 2, is bounded by the curve C, the x-axis and the lines

with equations x = ln3 and x = ln5.

(a) Show that the area of R is given by the integral

3

21

4d

( 2)t

t t

(3)

(b) Hence find an exact value for the area of R.

Write your answer in the form (a + ln b), where a and b are rational numbers.

(7)

(c) Find a cartesian equation of the curve C in the form y = f(x).

(2)

(C34, IAL Jan 2015, Q9)

51. (i) Find the x coordinate of each point on the curve , 11

xy x

x

, at which the

gradient is 1

4. (4)

(ii) Given that 2

1d ln 7

a

a

tt

t

a > 0

find the exact value of the constant a. (4)

(C34, IAL June 2015, Q5)


52.

The curve C with equation y = 2

(4 3 )x,

4

3x is shown in Figure 1

The region bounded by the curve, the x-axis and the lines x = –1 and x =2

3, is shown shaded in

Figure 1

This region is rotated through 360 degrees about the x-axis.

(a) Use calculus to find the exact value of the volume of the solid generated.

(5)

Figure 2 shows a candle with axis of symmetry AB where AB = 15 cm. A is a point at

the centre of the top surface of the candle and B is a point at the centre of the base of the

candle. The candle is geometrically similar to the solid generated in part (a).

(b) Find the volume of this candle.

(2)

(C34, IAL Jan 2016, Q4)


53. f(θ) = 9cos2θ + sin2θ

(a) Show that f(θ) = a + bcos2θ, where a and b are integers which should be found.

(3)

(b) Using your answer to part (a) and integration by parts, find the exact value of

2 2

0

f( )d

(6)

(C34, IAL Jan 2016, Q8)

54. Use integration by parts to find the exact value of 2

0

2 dxx x

Write your answer as a single simplified fraction.

(C34, IAL June 2016, Q5)

55.

Figure 3 shows a sketch of the curve with parametric equations

x = 3sint, y = 2sin2t, 0 ≤ t ≤ 2

The finite region S, shown shaded in Figure 3, is bounded by the curve, the x-axis and the

line with equation x = 3

2.

The shaded region S is rotated through 2π radians about the x-axis to form a solid of

revolution.

(a) Show that the volume of the solid of revolution is given by

2 3

0

sin cos da

k t t t

where k and a are constants to be given in terms of π.

(5)


(b) Use the substitution u = sin t, or otherwise, to find the exact value of this volume,

giving your answer in the form p

q

where p and q are integers.

(Solutions based entirely on graphical or numerical methods are not acceptable)

(6)

(C34, IAL June 2016, Q12)

56.

(a) By using the substitution u = 2x + 3, show that

12

20 (2 3)

x

x dx =

1

2ln3-

2

9

(7)

The curve C has equation

y =9 x

(2x + 3), x > 0

The finite region R, shown shaded in Figure 3, is bounded by the curve C, the x-axis and

the line with equation x = 12. The region R is rotated through 2π radians about the x-axis

to form a solid of revolution.

(b) Use the result of part (a) to find the exact value of the volume of the solid generated.

(2)

(C34, IAL Jan 2017, Q9)

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9.

12O

y

x

R

C

Figure 3

(a) By using the substitution u = 2x + 3, show that

x

xx

( )ln

2 3

1

23

2

920

12

+= -ò d

(7)

The curve C has equation

yx

xx=

+>

9

2 30

( ),

The finite region R, shown shaded in Figure 3, is bounded by the curve C, the x-axis and

the line with equation x = 12. The region R is rotated through 2 radians about the x-axis

to form a solid of revolution.

(b) Use the result of part (a) to find the exact value of the volume of the solid generated.

(2)

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57. Use integration by parts to find the exact value of

ln x

x2dx

1

e

ò

Write your answer in the form a +

b

e, where a and b are integers.

(C34, IAL June 2017, Q2)

58.


x = 8cos3 𝜃, y = 6sin2 𝜃, 0 ⩽ 𝜃 ⩽

p

2

Given that the point P lies on C and has parameter 𝜃 =

p

3

(a) find the coordinates of P.

(2)

The line l is the normal to C at P.

(b) Show that an equation of l is y = x + 3.5

(5)

The finite region S, shown shaded in Figure 6, is bounded by the curve C, the line l, the

y-axis and the x-axis.

(c) Show that the area of S is given by

32 4

0

4 144 (sin cos sin cos )d

(6)

(d) Hence, by integration, find the exact area of S.

(3)

(C34, IAL June 2017, Q14)

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14.

O

y

P

l

x

CS

Figure 6


x = 8cos3 , y = 6sin2 , 0 2

π

Given that the point P lies on C and has parameter = 3

π

(a) find the coordinates of P.

(2)

The line l is the normal to C at P.

(b) Show that an equation of l is y = x + 3.5

(5)

The finite region S, shown shaded in Figure 6, is bounded by the curve C, the line l, the

y-axis and the x-axis.

(c) Show that the area of S is given by

4 + 144 3

0

π

ò (sin cos2 – sin cos4 ) d

(6)

(d) Hence, by integration, find the exact area of S.

(Solutions based entirely on graphical or numerical methods are not acceptable.)

(3)

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Edexcel - KumarmathsThe basic principle here is that we take very thin strips, of width δx (delta...

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Transcript of Edexcel - KumarmathsThe basic principle here is that we take very thin strips, of width δx (delta...