Maths Methods 2011

MATHEMATICAL METHODS UNITS 3 AND 4

01: Anti-differentiation

ANTIDIFFERENTIATION Anti-differentiation, or integration, is the reverse of differentiation. f’(x) f(x)

a cax +

nx c1n

x 1n+

+

+

1n −≠

nax c1n

ax 1n+

+

+

1n −≠

Where ‘c’ is any real number, called the constant of integration. If the function f(x) is a derivative, then the set of all anti-derivatives of f is called the indefinite integral of f, and is denoted by ∫ dx)x(f . The symbol ∫ is called the integral sign, the function f is the

integrand of the integral and x is the variable of integration. Properties:

∫ ∫ ∫±=± g(x)dxf(x)dxg(x))dx(f(x)

∫ ∫= dx)x(fkdx)x(kf

Example 1

If f’(x) = 2x1 , then f(x) could be

A x1 B 3x

2− C x1− D loge(x2) E logex

Solution.

2x1 can be rewritten as x-2 ∴ ∫ −2x dx = c

12x 12

++−

+−

= 1

x 1

−

−

= x1− ∴ C

Maths Methods 2011

By definition 02: Integrating (ax + b)n

∫ + n)bax( = c)1n(a

)bax( 1n+

++ +

for n ≠ -1.

Example 1

∫ + 4)5x3( dx is equal to

A c15

)5x3( 5+

+ B c5

)5x3( 5+

+ C c3

)5x3( 5+

+

D 15

)cx5( 52x3 2

++ E 12(3x + 5)3 + c

Solution Using the formula, where a = 3, b = 5, and n = 4.

∫ + 4)5x3( dx = c)14(3)5x3( 5

+++ becomes c

15)5x3( 5

++ ∴ A

Example 2

Find ∫ + 3)1x3(1 dx.

Solution

∫ + 3)1x3(1 dx = ∫ −+ 3)1x3( dx Using a = 3, b = 1, and n = -3

∫ −+ 3)1x3( dx = c)13(3

)1x3( 13+

+−+ +−

= c)2(3)1x3( 2

+−+ −

= c)1x3(6

12 +

+−

Example 3

An anti-derivative of 2)1x2(1−

is equal to

A ½loge(2x – 1)2 B 1x2

1−

− C )1x2(2

1−

− D 3)1x2(4−

E 12(3x + 5)3

Solution Note that none of the options has a ‘c’ term. This is acceptable because the question is saying that an anti-derivative of . . . . . is equal to, not the anti-derivative is equal to. The ‘c’ may be zero, therefore these are possible answers.

∫ − 2)1x2(1 dx = ∫ −− 2)1x2( dx Using a = 2, b = -1, and n = -2

∫ −− 2)1x2( dx = )12(2

)1x2( 12

+−− +−

= )1(2)1x2( 1

−− −

= )1x2(2

1−

− ∴ C

Maths Methods 2011 Example 4

An anti-derivative of 4)5x2(1+

is equal to

A 4loge(2x + 5) B 3)5x2(61+

− C )5x2(

6+

− D 3)5x2(31+

− E 5)5x2(10+

Solution

∫ + 4)5x2(1 dx = ∫ −+ 4)5x2( dx Using a = 2, b = 5, and n = -4

∫ −+ 4)5x2( dx = )14(2

)5x2( 14

+−+ +−

= )3(2)5x2( 3

−+ −

= 3)5x2(61+

− ∴ B

Example 5

If c is an arbitrary constant and f’(x) = 1x3

6−

then f(x) is equal to

A 12 c1x3 +− B 4 c1x3 +− C c1x3 +− D c1x33

4+

− E c

1x34

+−

Solution

∫ − 1x36 dx = ∫ −− 2

1)1x3(6 dx = ∫ −− 2

1)1x3(6 Using a = 3, b = -1, and n = -½

∫ −− 21

)1x3(6 dx = c)1(3

)1x3(6

21

121

++−

− +−

= c)(3

)1x3(6

21

21

++ = 4 1x3 + + c ∴ B

Example 6

An anti-derivative of 25

)4x3(

1

− is equal to

A 23

)4x3(

1

− B

23

)4x3(2

3

−

− C 23

)4x3(2

9

−

− D 23

)4x3(

3

− E

23

)4x3(9

2

−

−

Solution

∫− 2

5)4x3(

1 dx = ∫ −− 25

)4x3( dx Using a = 3, b = -4, and n = - 25

∫ −− 25

)4x3( dx = )1(3

)4x3(

25

125

+−− +−

= )(3)4x3(

23

23

−− −

= 23

)4x3(9

2

−

−− ∴ E

Maths Methods 2011

03: The Integration of trigonometric functions

From the concept that integration is the reverse of differentiation, we can see the following:

dxd cos(kx) = -ksin(kx)

and dxd sin(kx) = kcos(kx)

Therefore: sin(kx) dx∫ = -k1 cos(kx) + c

and : cos(kx) dx∫ = k1 sin(kx) + c

Example 1 An anti-derivative 3x2 + sin(x) is A x3 – cos(x) B 6x – cos(x) C x3 + cos(x) D 6x + cos(x) E x3 + sin(x) Solution Divide the problem into 2 parts.

∫ x3 2 dx = x3 the sin(x)∫ dx = -cos(x) ∴ A

note that since it asks for an anti-derivative, you don’t need the ‘c’. Example 2 Find the derivative of xsin(x) and hence

find an anti-derivative of xcos(x)

If y = xsin(x) Then y’ = sin(x) + xcos(x) ∴ ∫ sin(x) + xcos(x) dx = xsin(x)

∴ ∫ xcos(x) dx = xsin(x) - ∫ sin(x) dx

∴ ∫ xcos(x) dx = xsin(x) + cos(x)

Hence

means that you must use the information provided to find your answer.

You can always check your answers by finding the derivative of your answer.

Eg. dxd xsin(x) + cos(x) = sin(x) + xcos(x) – sin(x)

= xcos(x)

Maths Methods 2011

04: The Integration of exponential functions

From the concept that integration is the reverse of differentiation, we can see the following:

dxd ex = ex ∴ ∫ ex dx = ex + c

Further dxd ekx = kekx ∴ ∫ ekx dx =

k1 ekx + c

Example 1 If g’(x) = 6e2x, then g(x) is equal to A 3e2x + c B 6e2x + c C 12e2x + c D 3xe2x + c E

2xe6 + c Solution

From ∫ ekx dx = k1 ekx + c we get ∫ 6e2x =

26 e2x + c ∴ A

Example 2 If f’(x) = x3

1e - 2x

1 , then, for any real number c, f(x) is equal to

A x31 3

1e -

x1 + c B x

31 3

1e +

x1 + c C x3

1e3 -

x1 + c

D x31

e3 + x1 + c E x

31 3

1e - 3x

2 + c

Solution Do this problem in parts.

∫ x31

e dx = 3 x31

e and - ∫ 2x1 dx = - -

x1 =

x1 ∴ D is the answer

Example 3

∫ (ex – e-x)2 dx is equal to

A 2x+ c B e2x – 2x + c C ½e2x - ½e-2x + c D ½e2x + ½e-2x – 2x + c E ½e2x – ½e-2x – 2x + c Solution Since the bracket is squared, it is probably easiest to expand the bracket first. ∴ (ex – e-x)2 = e2x –2e0 + e-2x = e2x + e-2x - 2 ∴ ∫ (ex – e-x)2 dx becomes ∫ (e2x + e-2x - 2)dx= ½e2x – ½e-2x - 2x + c ∴ E

Maths Methods 2011 Example 4 An anti-derivative of e3x + sin(3x) is equal to

A 3

e x3-

3)x3cos( B 3e3x + 3cos(3x) C e3x -

3)x3cos(

D 3

e x3 - cos(3x) E e3x – cos(3x)

Solution Do this in parts

∫ e3x dx = 31 e3x and ∫ sin3x = -

31 cos3x ∴ A

Maths Methods 2011

05: Definite Integrals

When we find an indefinite integral, the result is an expression, but when we find the definite integral we end up with a number. To find the integral, the function needs to be continuous over the domain. If we let F(x) be the integral

of f(x), then ∫b

a

dx)x(f is called the definite integral, and is equal to F(b) – F(a).

We usually write this like

∫b

a

dx)x(f = [ ]ba)x(F = F(b) – F(a)

So the process is quite simple Step 1 Find an anti-derivative F(x) of f(x) Step 2 Write your result as [ ]ba)x(F Step 3 Substitute a and b into F(x) Step 4 Subtract: F(b) – F(a) to obtain your value. The constant of integration can be omitted because it disappears in the subtraction.

In ∫b

a

dx)x(f , x is called the variable of integration, a and b are called the lower and upper limits

respectively. There is no reason why b should be greater than a. Properties of the definite Integral

1. ∫a

a

dx)x(f = 0

2. ∫b

a

dx)x(f = ∫−a

b

dx)x(f

3. ∫b

a

dx)x(f = ∫c

a

dx)x(f + ∫b

c

dx)x(f where a ≤ c ≤ b.

4. ∫b

a

dx)x(kf = ∫b

a

dx)x(fk , k ∈ R

5. ∫ ±b

a

dx))x(g)x(f( = ∫b

a

dx)x(f ± ∫b

a

dx)x(g

Maths Methods 2011 Integration by recognition This type of integration is based on observational skills, some algebraic manipulation and the result that if f’(x) = g(x) then ∫ dx)x(g = f(x) + c

Example 1

Given that dxd ( )3x2

e + = 2x 3x2e + , find the exact value of ∫ +

1

0

3x dxxe52

Solution From the given information we have:

∫ +1

0

3x dxxe52

= 25∫ +1

0

3x dxxe22

= 25 ( )∫ +

1

0

3x dxedxd 2

= 25 [ ]103x2

e + = 25 (e4 – e3)

Example 2

Given that dxd xsin(x) = sin(x) + xcos(x), find ∫ dx)xcos(x

Solution

From dxd xsin(x) = sin(x) + xcos(x) we can rewrite it as xcos(x) =

dxd xsin(x) – sin(x).

This gives us the term we need to integrate isolated.

∴ ∫ dx)xcos(x = ∫

− dxxsin)xsinx(

dxd = ∫ dx)xsinx(

dxd - ∫ dx)x(sin

= xsinx + cosx + c Example 3

Find dxd (xlogex) and hence evaluate ∫ dx)x(loge

Solution Using the product rule,

dxd (xlogex) =

dxd (x) × loge(x) + x ×

dxd (logex)

= loge(x) + 1

∴ loge(x) = dxd (xlogex) – 1

∴ ∫ dx)x(loge = ∫ dx)xlogx(dxd

e - ∫ dx1

= xlogex – x + c

Maths Methods 2011 Example 4

Given that dxd ( ))3x(log 2

e + = 3x

x22 +

, find ∫ +dx

3xx6

2 .

Solution

Since ∫ +dx

3xx6

2 = ∫ +dx

3xx23 2 . ∫ +

dx3x

x62 = ∫ +

dx3x

x23 2 = 3loge(x2 + 3) + c

Example 5

If ∫ −a

0

dx)6x3( = 0, a ≠ 0, then a is equal to

A -4 B -2 C 2 D 4 E 6 Solution

∫ −a

0

dx)6x3( = a

0

2x6

2x3

− = 0 ∴ (

2a3 2

- 6a) - (2)0(3 2

- 6(0)) = 0

∴ (2a3 2

- 6a) = 0

∴ 2a3 2

= 6a

∴ 3a2 - 12a = 0 ∴ a(3a – 12) = 0 ∴ a = 0 or a = 4 ∴ D since a ≠ 0

Example 6

∫−

−−1

1

x dx)ex2( is equal to

A 0 B -e – e-1 C e-1 - e D 2(1 + e-1) E 2(1 – e-1) Solution

∫−

−−1

1

x dx)ex2( = ∫−

1

1

dx)x2( - ∫−

−1

1

x dx)e( = 1

1

2

2x2

−

- [ ]1 1

xe −−−

= [ ]1 12x − + [ ]1 1

xe −−

= (1 – 1) + (e1 - e)

= e1 - e ∴ C

Maths Methods 2011 Example 7

∫ +2π

0

dx)xcosx(sin is equal to

A 0 B 1 C -1 D 2 E -2 Solution

∫ +2π

0

dx)xcosx(sin = ∫2π

0

dx)x(sin + ∫2π

0

dx)x(cos = [ ]2π

0xcos− + [ ]2π

0xsin

= -(cos( 2π )- cos(0)) + sin( 2

π ) – sin(0) = -(0 – 1) + (1 – 0) = 1 + 1 = 2 ∴ D Example 8

∫ +4

2

dx)10)x(f5( is equal to

A ∫4

2

dx)x(f5 + 2 B ∫4

2

dx)x(f5 + 10 C ∫4

2

dx)x(f5 + 10x

D ∫4

2

dx)x(f5 + 50 E ∫4

2

dx)x(f5 + 20

Solution

∫ +4

2

dx)10)x(f5( = ∫4

2

dx)x(f5 + ∫4

2

dx10 = ∫4

2

dx)x(f5 + [ ]42x10 = ∫4

2

dx)x(f5 + (40 – 20)

= ∫4

2

dx)x(f5 + 20 ∴ E

Example 9

∫ +

1

03 dx

)1x2(1 =

A 92 B

98 C

185

− D 910

− E 23

−

Solution

∫ +

1

03 dx

)1x2(1 = ∫ −+

1

0

3dx)1x2( = 1

0

13)1x2()13(2

1

+

+−+− =

1

0

2)1x2(41

+− − =

1

02)1x2(4

1

+−

Maths Methods 2011

= (-361 ) – (-

41 ) = -

361 +

369 =

368 =

92 ∴A

Example 10

∫ +−3

0

2 dx)3x4x( is exactly equal to

A 0 B 1 31 C 1 3

2 D 2 E 2 32

Solution

∫ +−3

0

2 dx)3x4x( = ∫3

0

2dxx - ∫3

0

xdx4 + ∫3

0

dx3

= 3

0

3

3x

-

3

0

2

2x4

+ [ ]30x3 = 9 – 18 + 9 = 0 ∴ A

Example 11

∫ +1

0

3dx)1x2( is exactly equal to

A 10 B 10 81 C 13 3

1 D 20 E 80 Solution

∫ +1

0

3dx)1x2( = 1

0

4)1x2()13(2

1

+

+ =

1

0

4)1x2(81

+ =

881 -

81 = 10 ∴ A

Example 12

∫3

0

dx)x(f = 4 and g(x) = 2f(x) – 1, then ∫3

0

dx)x(g is equal to

A -5 B 5 C 7 D 11 E -11 Solution 3

0

g(x)dx∫ = 3

0

(2f(x) 1)dx−∫ = 3

0

2f(x)dx∫ - 3

0

1dx∫ = 3

0

2 f(x)dx∫ - 3

0

1dx∫

= 2 × -4 - [ ]03x = -8 + 3 = -5 ∴ A

Maths Methods 2011 Example 13

Evaluate ∫2π

0

dx)x2sin(4

A -4 B -2 C 0 D 2 E 4 Solution

∫2π

0

dx)x2sin(4 = ∫2π

0

dx)x2sin(4 = 4 [ ]2π

021 x2cos− = [ ]2

π

0x2cos2− = -2cos π + 2cos0

= 2 + 2 = 4 ∴ E Example 14

∫ +4

1

dx)6)x(f2( can be written as

A ∫ +4

1

6dx)x(f2 B ∫ +4

1

dx)6)x(f(2 C ∫ +4

1

18dx)x(f2

D ∫4

1

dx)x(f2 + ∫4

1

dx3 E ∫4

1

dx)x(f2 + 6x

Solution

∫ +4

1

dx)6)x(f2( = ∫4

1

dx)x(f2 + ∫4

1

dx6 = ∫4

1

dx)x(f2 + [ ]41x6 = ∫4

1

dx)x(f2 + (24 – 6) = ∫ +4

1

18dx)x(f2 ∴ C

Example 15

Evaluate ∫4

12 dx

x1

A 19261 B 2

1 C 43 D 16

15 E 1 Solution

∫4

12 dx

x1 = ∫ −

4

1

2dxx = [ ]411x−− = 4

1x1

− = -

41 - -1 =

43 ∴ C

Example 16

The value of ∫ +2

1

2 dx)1x3( is

A 18 B 22 C 26 D 31 E 38

Maths Methods 2011 Solution

∫ +2

1

2 dx)1x3( = 2

1

3)1x3()12(3

1

+

+ =

2

1

3)1x3(91

+ =

9343 -

964 =

9279 = 31 ∴ D

Example 17

Given that k > 1 and x > 1, ∫ −

x

1k1 dt

t1 is equal to

A )1x(k1 k − B )1x(

k21 k2 −−

− C )1x(k2

1 k2 −+

+

D )1x

1(k2

1k2 −

− − E (k – 1)(xk – 2 – 1)

Solution

∫ −

x

1k1 dt

t1 = ∫ −

x

1

1k dtt = x

1

ktk1

=

k1 (xk – x) ∴ A

Maths Methods 2011

06: The calculation of area

We define the area bounded by the function and the x-axis to be the area under the curve. This is found by finding ∫ dx)x(f .

If we want to find the area under the curve between x = a and x = b, then the area is ∫b

a

dx)x(f .

We can approximate the area under a curve by constructing rectangles, and then finding the sum of the areas. Consider the problem of finding the shaded area in the diagram below.

We can use strips of width one unit. If the equation is f(x) = 10 – x2, then the heights of the rectangles are 6, 9, 9, 6, 1. ∴ the area is 6 × 1 + 9 × 1 + 9 × 1 + 6 × 1 + 1 × 1 = 31 square units. Our approximation would be much better if the width of the rectangles was reduced to ½.

The smaller the width of the rectangles the better the approximation. Because all the rectangles lie below the curve, we say that we have the sum of the lower triangles. However, we could just have easily chosen the have rectangles that lie above the curve. We would then have the sum of the upper triangles.

Maths Methods 2011 Lower Sum

This always gives an estimate lower than the actual value, if the function is increasing Exact area

Upper sum

This always gives an estimate higher than the actual value, if the function is increasing Lower Sum = SL < Exact area < SU = Upper Sum The rectangles can give either the upper value or the lower value for the area. To improve this approximation, we can make the width of the rectangles as small as possible. If f(x) is positive and continuous on the interval [a, b], the area, A sq units, bounded by y = f(x) and the lines x = a and x = b is given by

Area (A) = ∫b

a

dx)x(f = ∫b

a

ydx .

Maths Methods 2011 If f(x) is negative and continuous on the interval [a, b], (i.e. f(x) ≤ 0 for a ≤ x≤ b)then the integral

∫b

a

dx)x(f is a negative number. We write the area as - ∫b

a

dx)x(f or use the absolute value of the integral,

∫b

a

dx)x(f .

If the function crosses the x-axis then the total area is the sum of the areas between the points of intersection with the x-axis. The signed area is found by integrating across the domain as a whole.

A1

A2 The signed area is A1 - A2. But the total area is A1 + 2A Example 1 To find an approximation to the area between the graph with equation y = x2 and the x-axis between the lines with equations x = 1 and x = 4, the partitioning shown, using rectangles, can be used. The area of the shaded rectangles is equal to A 14 B 15 C 21 D 30 E 54

1 2 3 4

Maths Methods 2011 Solution When x = 1, x2 = 1, ∴ the area of the left rectangle is 1 × 1. When x = 2. x2 = 4, ∴ area of middle rectangle = 1 × 4 = 4, and the right hand side area is 1 × 9/ ∴ total area = 1 + 4 + 9 = 14 square units. A Example 2 The area between the curve with equation y = cos

2x and the x-axis between the lines with equations x =

0 and x = π, is equal to A 0 B ½ C 1 D 2 E 4

Y = cos2x

Solution

∫π

02x dxcos = [ ]π02

xsin2 = 2sin 2π - 2sin0 = 2 × 1 – 2 × 0 = 2 ∴ D

Example 3 Let A denote the area in the first quadrant bounded by the lines with equation x = 0 and y = 0 and the parabola y = 16 – x2.

A

(i) Express the area A as an integral. Solution

Maths Methods 2011

The area A is ∫ −4

0

2dxx16

(ii) Evaluate the area A

Solution

A = ∫ −4

0

2dxx16 = 4

0

3

3xx16

− = (16 × 4 – 43 ÷3) – 0 = 64 -

364 = 42 3

2

Example 4 The area, in square units, enclosed by the curve y = 4 -

x1 , the x-axis and the lines x = 1 and x = 4 is

A 16 – loge4 B 12 + loge4 C 12 D 12 – loge4 E 11 161

Solution It is always best to draw a sketch (on your calc.) first.

The function is positive so the area is ∫ −4

1x14 dx = ∫

4

1

4 dx - ∫4

1x1 dx

= [ ]41x4 - [ ]41e xlog = 12 – (loge4 – loge1) = 12 – loge4 ∴ D Example 5 The graph with the equation y = ax(x – 2) is shown below. Given that the shaded region has an area of 6 square units, the value of a must be

A 9 B -9 C 4 D -4 E 29

−

Solution

Maths Methods 2011

∫ −1

0

2 )ax2ax( dx = 6 ∴1

0

3

3xa

-

1

0

2

2ax2

= 6

∴ 3a - a = 6

∴ a – 3a = 18 ∴ -2a = 18 ∴ a = -9 ∴ B Example 6 The graph of a particular cubic function, shown below, has three intercepts: a, b and c.

a b c The shaded region is given by:

A ∫c

a

dx)x(f B ∫b

a

dx)x(f + ∫c

b

dx)x(f C ∫b

a

dx)x(f - ∫c

b

dx)x(f

D ∫c

a

dx)x(f - ∫c

b

dx)x(f E ∫c

b

dx)x(f - ∫b

a

dx)x(f

Solution

You need to integrate separately across both regions, so it is ∫b

a

dx)x(f + ∫c

b

dx)x(f . Since the second

area will give a negative value, the total is ∫b

a

dx)x(f - ∫c

b

dx)x(f . ∴ C

Maths Methods 2011 Example 7 The graph below represents part of the function defined by f(x) = 1 - ex.

y = 1 The shaded region (in square units) is given by: A (e – 2) B 1 C (e – 1) D 2 E e Solution

The area is given by ∫ −1

0

x dx)e1( = [ ]10xex − = (1 – 0) – (e – 1) = 1 – e + 1 = 2 - e

e2 − = (e – 2) ∴ A Example 8 The total area of the shaded rectangles can be used as an approximation for the area between the curve

with equation y = x1 , the x-axis, and the lines with equation x = 1 and x = 5. This approximation is equal

to

A 1 B 1.61 C 2 D 2.08 E 1212

Solution

Each rectangle is 1 unit wide, ∴ the area of each is 1 × 11 + 1 ×

21 + 1 ×

31 + 1 ×

41 = 1 +

126 +

124 +

123

= 1212 ∴ E

Maths Methods 2011 Example 9 The area between the curve y = ex + 1, the x-axis and the lines x = 0 and x = 1 is equal to A e – 1 B 1 C e D e + 1 E e + 2 Solution

The area is ∫ +1

0

x dx)1e( = [ ]10x xe + = (e1 – e0) + (1 – 0) = e – 1 + 1 = e ∴ C

Example 10 The graph shown below has equation y = 4cos(2x).

The area shaded (in square units) is equal to

A 2 B 6 C 4 D 8 E ∫4π3

0

dx)x2cos(4

Solution

The area is given by ∫4π

0

dx)x2cos(4 + ∫4π3

4π

dx)x2cos(4 = 4π

02)x2sin(4

+ 4π3

4π2

)x2sin(4

= (2sin(2π ) - 2sin(0)) + )

2πsin(2)

2π3sin(2 −

= (2 – 0) + )12()12( ×−−× = 2 + 4 = 6 ∴ B

Maths Methods 2011 Example 11 For the graph with the equation y = 2x3, 0 ≤ x ≤ 3.

The approximate area enclosed (in square units) by the curve, the x-axis and the lines x = 1 and x = 3, using the trapezia shown is A 88 B 70 C 43 D 40 E 35 Solution The area of the first triangle is given by 2

1 × b × h = 21 × 1 × (2 × 23) = 8

The area of the trapezium is given by 21 (a + b) × h = 2

1 (16 + 54) × 1 = 35 ∴ total area is equal to 35 + 8 = 43 square units. Example 12 The graph of the function y = a + bsin(

2x ) is shown below.

a) Find the values of a and b? Solution When x = 0, sin(2x) = 0. ∴ a = 3, since the graph has been translated up 3. The amplitude of the sine function is 1. Since the maximum value is 5, b = 2. b) Find the area of the shaded function

Maths Methods 2011

The area is given by ∫ +π3

02x dx)sin(23 = π3

02x)cos(4x3 − = 9π - (4cos )( 2

π3 - 4cos(0))

= 9π - (4 × (0) – 4 × 1) = 9π + 4 Example 13 The area of the shaded region (bounded by the graph of f and x and y-axes) may be approximated by the sum of the areas of the triangles OAB and BCD.

A B D C (2, -1) Using this approximation, the area (in square units) is A 2

1 B 1 C 1½ D 2 E 2½ Solution The area of triangle A0C = 2

1 × 1 × 3 = 1 21

The area of triangle BCD = 21 × 2 × 1 = 1 ∴total area = 2 2

1 ∴ E Example 14 The area bounded by the curve f(x) =

x273−

and the x-axis from x = ½ to x = 2 is equal to

A 23 loge2 B 3

2 loge0.5 C 23 loge0.5 D 3

2 loge2 E 23 loge3

Solution

The are is given by dxx27

32

21∫ −

= 2

e21

)x27((log2

3−

− =

2

e21

)x27((log23

−− = ( )6log3(log23

ee −− )

= 23− loge½ =

23− × -loge2 = 2

3 loge2 ∴ A

Maths Methods 2011 Example 15

An approximate value of ∫ +3

1

2 dx)1x( is to be calculated using the area of two strips each in the shape of

a trapezium as shown in the diagram. y= x2 + 1 The value of the approximation is equal to A 10 3

2 B 11 C 11½ D 12 E 15 Solution The area of the first trapezium is given by 2

1 (a + b) × h = 21 (2 + 5) × 1 = 3½

The area of the second trapezium is given by 21 (a + b) × h = 2

1 (5 + 10) × 1 = 7½ ∴ total area = 11 ∴ B Example 16

In the diagram shown, the total area of the two regions shaded is equal to

A ∫−2

1

dx)x(f B ∫−

1

2

dx)x(f C ∫−1

1

dx)x(f + ∫−

−

2

1

dx)x(f

D ∫−

1

1

dx)x(f - ∫−

−

1

2

dx)x(f E ∫−

1

1

dx)x(f + ∫−

−

1

2

dx)x(f

Maths Methods 2011 Solution

The area is given by ∫−

1

1

dx)x(f + ∫−

−

1

2

dx)x(f which becomes ∫−

1

1

dx)x(f - ∫−

−

1

2

dx)x(f ∴ D

Example 17 The graph of f:

−

2π,

2π

→R, f(x) = kcos(x) is shown. The area of the shaded region is equal to 1.

The value of k is A 4

1 B ½ C 43 D 1 E 2

Solution

The area is given by ∫−

2π

2π

dx)xcos(k = 1 ∴ 2π

2π)xsin(k

− = k(sin( 2

π ) – sin( 2π− ))

= k(1 - -1) ∴ Area = 1 = 2k. ∴ k = ½ ∴ B Example 18 The curved defined by the equation y =

2x1−

is sketched at right for the domain x > 2. The shaded

region, bounded by the curve, the x-axis, and the ordinates x = 4 and x = 5, has an area of logek square units.

The value of k is A 1 B 1½ C 2 D 3 E 6 Solution

The area is given by ∫ −

5

4

dx2x

1 = 54e )2x(log − = loge3 – loge2 = loge1.5 ∴ k = 1½ ∴ B

Maths Methods 2011

07: Area between two curves

Consider two functions whose graphs do not intersect in the interval a ≤ x ≤ b. The area enclosed by these two curves and the lines x = a and x = b is shown below.

y = f(x) y = g(x) a b This area can be calculated by y = f(x) y = g(x) a b a b

Area between the curves = ∫b

a

dx)x(f - ∫b

a

dx)x(g

That is, if f(x) > g(x) on the interval [a, b], then the area, A square units, enclosed by the two curves and the lines x = a and x = b is given by

A = ∫b

a

dx)x(f - ∫b

a

dx)x(g = ∫ −b

a

dx))x(g)x(f(

Maths Methods 2011 Example 1 Find the area of the region enclosed by the curves f(x) = x + 2, g(x) = x2 + x – 2 and the lines x = -1 and x = 1. Solution The first step is to sketch both graphs so that you can clearly see which one of the two curves lies above the other. y = g(x) y = f(x) Because f lies above g in the interval –1 ≤ x ≤ 1, we can write

Area of shaded region = ∫−

−+−+1

1

2 dx))2xx()2x(( = ∫−

−1

1

2 dx)x4(

= 1

1

3

3xx4

−

−

= (4 - 31 ) – (-4 + 3

1 ) = 322 square units.

NOTE If the question was given as: Find the area of the region enclosed by the curves f(x) = x + 2, g(x) = x2 + x –2. This would indicate that we want the total area enclosed by the two curves. To find this area we would need to find the points of intersection of the two curves, that is, solve for f(x) = g(x) and then integrate over this region. In this case we find that x + 2 = x2 + x – 2 x2 – 4 = 0 ∴ x = ± 2

Therefore the area = ∫−

−2

2

2 dx)x4( .

Maths Methods 2011

08: Applications of anti-derivatives

Example 1 A railway bridge over a river in countryside Victoria has a central arch. In the diagram below, the straight line AB indicates the water level under the bridge, and the curved line ATB represents the underside of the arch, and T is the top of the curve ATB The arch has shape given by the relation y= -0.01x2 + 25 where y metres is the distance from a point on the curve AB to the line AB, and x metres is the distance of the point P from the line OT. Railway T x P y Water level A 50 m 50 m B a) What is the height of T above the water level? Solution: When x = 0, y = ) + 25. ∴ Height is 25 metres. b) The cross-sectional area enclosed by the underside of the arch and the water lies between the area of the triangle ATB and the area of the rectangle with base AB and with height OT. Evaluate these lower and upper approximations for the cross-sectional area. ATB = ½ × 100 × 25 = 1250 m2 (lower approximation) ATB = 100 × 25 = 2500 m2 (upper approximation) c) A more accurate approximation to the cross-sectional area is made be constructing ten equally spaced ‘upper rectangles’, as shown in the diagram below.

Maths Methods 2011 50 40 30 20 10 0 10 20 30 40 50 Find the value of this approximation. Solution Area ~ 2 × 10 (y(0) + y(10) + y(20) + y(30) + y(40))

2 × 10(25 + 24 + 21 + 16 + 9) = 1900 m2 d) Find the value of the approximation to the area if the areas of the ‘lower rectangles’ are summed. 50 40 30 20 10 0 10 20 30 40 50 Solution Solution Area ~ 2 × 10 (y(10) + y(20) + y(30) + y(40))

2 × 10(24 + 21 + 16 + 9) = 1400 m2 e) Using the information from parts ‘c’ and ‘d’ between what two values must the actual cross-sectional area lie? Solution 1400 m2 < Area < 1900 m2 f) Use calculus to find the actual area bounded by the underside of the arch and the water level, to the nearest square metre. Solution

Area = 2 ∫ +−50

0

2 dx)25x01.0( = 250

0

3x25

3x01.0

+− = 1666 3

2 = 1667 m2 to nearest square metre.