Integration

Learning Objectives:

In this chapter, you will learn about

• the concept of indefinite integral

Learning Outcomes:

•Determine integrals by reversing differentiation

2y x

2dy

dx

2 3y x

2dy

dx

2 4y x

2dy

dx

y mx c

2 dx 2x c2 dx

2x c

2 dx

2x c

2y x c

2y x c 2y x c

( ), ( )dy

if f x then y f x dxdx

3.1.2 Integration of algebraic expressions

m dx mx c

8dx 8x c

3.5 dx 3.5 x c

Integrate (a) 8 (b) 3.5 (c) 3

4

3

4dx

3

4x c

3.1.2 (a) Integral of Constant

3.1.2 Integration of algebraic expressions

3.1.2 (b) Integral of nax

Differentiation Integration

2dy

xdx

22dy

xdx

32dy

xdx

ndynx

dx

2x dx 22

2

xc 2x c

22x dx 32

3

xc

32x dx 42

4

xc 41

2x c

42x dx 52

5

xc

During differentiation, we carry out two operations on each term in x: multiply the term with the index, and reduce the index by 1.

ndyax

dx

1nnax

42dy

xdx

nax dx 1

1

naxc

n

Integrate each of the following with respect of x:

(a) 8x8x dx

8 1

8 1

xc

9

9

xc

(b) 4

6

x

4

6dx

x

4 16

4 1

xc

46x dx

36

3

xc

32x c

3

2c

x

If the derivative of a function is given as find the function y. 3

1,

9

dy

dx x

3

1,

9

dy

dx x

31,

9

dyx

dx

31

9x

3 11

9 3 1

xc

2

18

xc

2

1

18c

x

2y x

2dy

dx

2 3y x

2dy

dx

2 4y x

2dy

dx

y mx c

2 dx2x c

2 dx

2x c

2 dx

2x c

2y cx 2y cx 2y cx

3.1.3 Determine the constant of Integration

If and y=5 when x=3, find the value of y when x=5

23 6 4dy

x xdx

23 6 4y x x dx 3 23 6

43 2

x xy x c

3 23 3 4 (1)y x x x c

Subsitute x=3 and y=5 into (1)

3 25 3(3) 3(3) 4(3) c

5 27 27 12 c 5 12 c

7 c

3 23 3 4 7y x x x

,therefore

5when x 3 23(5) 3(5) 4(5) 7y

63y

3.1.4 Equations of curve from functions of gradients

Find the equation of curve.

The gradient of a curve passing through the point (-1, 2) is given by 4 5dy

xdx

sin 4 5,dy

ce xdx

by integration,

4 5y x dx 245

2

xy x c

22 5y x x c

The curve passing through the point (-1, 2)x=-1 when y=2

22 2( 1) 5( 1) c

2 2 5 c 5 c

22 5 5y x x

The equation of the curve is 22 5 5y x x

The gradient function of a curve passing through the point (-1, 2) and (0,k) is 23 10x x . Find the value of k.

y

x

2

2

4

4

6

6

8

8

10

10

– 2

– 2

– 4

– 4

– 6

– 6

– 8

– 8

– 10

– 10

2

2

4

4

6

6

8

8

10

10

– 2

– 2

– 4

– 4

– 6

– 6

– 8

– 8

– 10

– 10

(-1, 2)

(0, k)

23 10 ,dy

x xdx

23 10y x x dx 3 23 10

3 2

x xy c

3 25y x x c

The curve pass through (-1, 2)

,Therefore3 22 ( 1) 5( 1) c

8 c

Therefore, the equation of the curve is

3 25 8y x x

At point (0, k),

3 20 5(0) 8k

8k

Exercise 3-1-09 t0 6-1-09

1. Given that and that y=5 when x= -1, find the value of y

when x=2

2 7dy

xdx

2. Given that and that v=2 when t = 1, find the value of v

when t = 2.

3 2dv

tdt

3.1.5 Integrate by substitution

Find the integration by substitution

4(2 5)x dx2 5let u x

, 2du

Hencedx

1

2dx du

4 1.2

u du

4 1.2

u du 51

2 5

uc

51

10u c

51(2 5)

10x c

3.1.5 Integrate by substitution

Find the integration by substitution

2

1

(3 2)dx

x 3 2let u x

, 3du

Hencedx

1

3dx du

2

3 2x dx

2

1

(3 2)dx

x

2 1.3

u du

11

3 1

uc

1

3c

u

1

3(3 2)c

x

3.1.5 (a) Integral of ( )nax b

( )nax b dx 1( )

( 1)( )

nax b

n a

4(2 5)x dx4 1(2 5)

(4 1)(2)

x

5(2 5)

10

xc

51(2 5)

10x c

The slope of a curve at any point P(x, y) is given by . Find the equation of the curve given that its passes through the point ( 4, -3)

5(3 )x

The gradient of the curve, 5(3 )

dyx

dx

Integrate with respect to x, we have

5

3y x dx 63

6( 1)

xy c

Since the curve passes through (4, -3)

61(3 )6

y x c

613 (3 4)

6c

17

6c

,Therefore

The equation of the curve is

61 17(3 )6 6

y x or

66 (3 ) 17y x

Area under a curve

3.2.2(b) Area under a curve bounded by x= a and x=b

A

The area A under a curve by y = f(x) bounded by the x-axis from x=a to x=b is given by

b

aA y dx

Integration as Summation of Area

3.2.2(b) Area under a curve bounded by x= a and x=b

A

2 2y x

2

1A y dx

1 2

2 2

12A x dx

23

1

23

xA x

23 2

1

2 12(2) 2(1)

3 3A

8 14 2

3 3A

213

3A unit

Step (1) Find x-intercept

a b

On the x-axis, y =0

2 2x x y

2 2x x 02 2 0x x

( 2) 0x x 0x 2and x

0 2

2 2

02A x x dx 23 2

0

2

3 2

x xA

23

2

03

xA x

23 2

2 2

0

2 02 0

3 3A

24

3A unit

24

3A unit

Area under a curve bounded by 2 4 4y x x

2( 1)y x Area under a curve bounded by

The area under a curve which is enclosed by y = a and y = b is

b

aA x dy

The area under a curve is 2 2

12A y ydy

1

2

2 2x y y

23 2

1

2

3 2

y yA

3 32 23 13 1

3 3A

19 9 1

3A

20

3A

22

3A unit

A

Area under a curve bounded by curve

Area under a curve bounded

Exercise 19-2-09 23-2-09

1

2

Exercise Text Book Page 71 23-2-09

10- ( a) ( b ) ( c )

11- ( a) ( b ) ( c )

12- ( a) ( b )

Exercise Text Book Page 72

13- ( a) ( b ) ( c )

14

17 ( a ) (b ) ( c )

Volume of Revolutions

The resulting solid is a cone

To find this volume, we could take slices (the yellow disk shown above), each dx wide and radius y:

The volume of a cylinder is given by

V = πr2h

Because radius = r = y and each disk is dx high, we notice that the volume of each slice is:

V = πy2dx

Adding the volumes of the disks (with infinitely small dx), we obtain the formula:

y = f(x) is the equation of the curve whose area is being rotated

a and b are the limits of the area being rotated

dx show that the area is being rotated abount the x-axis.

Example 2

Find the volume if the area bounded by the curve y = x3 + 1, the x-axis and the limits of x = 0 and x = 3 is rotated around the x-axis..

When the shaded area is rotated 360° about the x-axis, we again observe that a volume is generated:

2y x

2 2

1V y dx

2 2

1(2 )V x dx

2 2

14V x dx

23

1

4

3

xV

3 34(2) 4(1)

3 3V

32 4

3 3V

28

3V

3193

V unit