SM 025

Lecture Note: 1 of 5 Mathematics 2 SM025
Session 2020/2021
Ch. 1 Integration

Topic: 1. INTEGRATION

Sub-Topic: 1.1 Integration of Functions

Learning Outcomes: At the end of this lesson, students should be able to
(a) relate integration and differentiation.
(b) define the basic rules of integration.

(c) find the integral of ex, 1 , ax and abxc.
x

(d) find the integral of the forms:

(i)  f x dx,
f x

(ii) f xefxdx,

(iii)  f  x f xn dx,

(iv) abxcdx.

Integration as Anti-Differentiation

 The process of finding anti-derivatives is called integration.
 Integration is the reverse process of differentiation.

 If d F x  f x, then  f xdx  F x  C.
dx

 The symbol  is called an integral sign and C is called the constant of integration.

 A single function has many anti-derivatives F x, but the functions have only one

derivative f x.

 There are two types of integral:
o Indefinite integral in which we aren't given the upper and lower limits of integration.

For example,  2x dx  x2  C.

o Definite integral in which we have upper and lower limits on the integral.

2

For example, 1 2xdx  3.

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Lecture Note: 1 of 5 Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

EXAMPLE 1

Given y  2x  3 , find dy . Hence, find   10 22 dx.
3x  2 dx
3x 

dy  (3x  2)(2)  (2x  3)(3) u  2x  3, v  3x  2
dx (3x  2)2
u  2 v  3
5
  (3x  2)2

d  2x  3    (3x 5 2)2
dx 3x  2 

2 d 32xx  32 dx  2  (3x 5 2)2 dx
dx  

  10 2)2 dx  2  2x  3   C
(3x  3x  2

 4x  6  C
3x  2

Basic Rules of Integration

(a) dx  x  C, where C is a constant.
(b)  kdx  k dx, where k is a constant.

(c) xndx  xn1  C, where C is a constant.
n 1

(d)  kf xdx  k f xdx, where k is a constant.
(e)  f x  gx dx   f xdx   g xdx.

EXAMPLE 2 (b) 3x6dx
Find the following integrals.
 3x6dx  3 x6dx
(a)  4dx
 4dx  4x  C

 3  x7   C
7

 3 x7  C
7

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Lecture Note: 1 of 5 Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

(c)  3x  2x2dx (d)  1  y dy
y3

 3x  2x2dx   3x3  2x2dx  1  y  1  dy
y3
 3  x4   2 x3   C dy  1  y2
4 3 y3 y3

 3 x4  2 x3  C  y3  5
43
y 2dy

 y2  3  C
2
y2
3
2

  1  2 C
2y2
3

3y 2

Integration of ax +bn dx

 ax  bn dx  ax  bn1  C, n  1
n  1a

EXAMPLE 3
Find the following integrals.

(a)  3x  56 dx (b)  2 dx

3x  57 1  x2
7 3
 3x  56 dx   C  2 dx   21  x2dx

 3x  57  C 1  x2

21 21  x1

 C

11

 2 C
1x

(c)  4x  2dx (d)  4 dx

1 4  x5

 4x  2dx   4x  22 dx  4 dx  44  x25dx

3 4  x5

4x  22  4 4  x32 C
 324  C
 3 1
3 2

 4x  22  C  8 3 C

6 34  x2

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Lecture Note: 1 of 5 Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

Integration of  1 dx,  1 dx and  f xdx.
x ax + b f x

(a)  1 dx  ln x  C, where C is a constant.
x

(b)   1 b  dx  ln ax  b  C, where C is a constant.
ax  a

(c)  f x dx  ln f x  C, where C is a constant.
f x

EXAMPLE 4
Integrate the following functions with respect to x.

(a)  2 dx (b)  1 dx
x 5  3x

 2 dx  2 ln x C  1 dx  ln 5  3x C
x 5  3x 3

  1 ln 5  3x  C
3

(c)  2 3 1 dx

x 

 3 1 dx  3  1 dx
2 x1
2x 

 3  ln x1   C
2 1

 3 ln x  1  C
2

(d)   x 2 4  3 1 dx
 2x 

  2  3 1 dx  2 ln x4  3 ln 2x  1 C
 2x  1 2
x 4

 2 ln x  4  3 ln 2x  1  C
2

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Lecture Note: 1 of 5 Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021
EXAMPLE 5

Integrate the following functions with respect to x.

(a)  2x  1 1 dx (b) 3x2  x dx
x2  x  2x3  x2

Let f x  x2  x  1, Let f x  2x3  x2,

f x  2x  1 f x  6x2  2x

 2x  1 1 dx   f x dx  3x2  x dx  1 6x2  2x dx
x2  x  f x 2 2x3  x2
2x3  x2

 ln f x  C  1  f x dx
2 f x
 ln x2  x  1  C
 1 ln f x  C

2
 1 ln 2x3  x2  C

2

Integration of Exponential Functions

(a) eaxbdx  eaxb  C, where C is a constant.
a

(b) abxcdx  abxc  C, where C is a constant.

b ln a

(c) f xefxdx  efx  C, where a  0 and C is a constant.

EXAMPLE 6
Find the following integrals.

(a) 5e5xdx
5 e5xdx  5e5x  C

5
 e5x  C

(b) e2x1dx (c)  3x dx

 e2x1dx  e2x1  C  3x dx  3x 3  C
2
 1 e2x1  C 1 ln
2
 3x  C
ln 3

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Lecture Note: 1 of 5 Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

(d) 52x1dx   (e) 3x2  1 ex3x2dx

 52x1 dx  52x1  C Let f x  x3  x  2,

2 ln 5 f x  3x2  1
 52x1  C
   3x2  1 ex3x2dx  f  xefxdx
ln 25
 efx  C
 ex3 x2  C

Integration of  f x f xn dx

 f x f xn dx  f xn1  C, where n  1 and C is a constant.

n 1

EXAMPLE 7 (b)  ln x dx
Find the following integrals. x

(a)  2xx2  34 dx Let f x  ln x,

Let f x  x2  3, f x  1

f x  2x x

 2xx2  34 dx   f x f x4 dx  ln x dx   1 ln xdx
x
 f xn1  C x

n 1   f x f x dx

x2  35  f xn1  C

 C n 1
5

 ln x2  C

2

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Lecture Note: 2 of 5 Mathematics 2 SM025
Session 2020/2021
Ch. 1 Integration

Topic: 1. INTEGRATION

Sub-Topic: 1.2 Integration of Trigonometric Functions

Learning Outcomes: At the end of this lesson, students should be able to
(a) solve the integral of trigonometric functions sin ax, cos ax and sec2 ax.

*Include: Constant multiples, sums and differences involving trigonometric functions sin ax, cos ax

and sec2 ax.
**Include: Integration of sin2 ax and cos2 ax.
***Include: Use of double angle and compound angle formulae.

Integration of Trigonometric Functions

(a)  sin ax  b dx  1 cosax  b  C, where C is a constant.
a

(b)  cos ax  b dx  1 sin ax  b  C, where C is a constant.
a

(c)  sec2 ax  bdx  1 tan ax  b  C, where a, b and C are constants.
a

EXAMPLE 8
Find the following integrals.

(a)  sin 3xdx (b)  cos 54 x dx

 sin 3xdx  1 cos 3x  C  cos  4 x dx  1 sin  4 x  C
3 5 5
 4 
5

 5 sin  4 x  C
4 5

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Lecture Note: 2 of 5 Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

(c)  sec2 3x  2dx

 sec2 3x  2 dx  1 tan 3x  2  C
3

(d)  sin x  cos 3xdx

 sin x  cos 3x dx   cos x  1 sin 3x  C
3

(e)  cos 4x  4 sin 3x  5 dx

 cos 4x  4 sin 3x  5 dx  1 sin 4x  4 cos 3x  5  C
4 3

(f)   1 2x  5 cos 4x dx
cos2

  1 2x  5 cos 4x dx   sec2 2x  5 cos 4xdx
cos2

 1 tan 2x  5 sin 4x  C
24

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Lecture Note: 2 of 5 Mathematics 2 SM025
Session 2020/2021
Ch. 1 Integration

Topic: 1. INTEGRATION

Sub-Topic: 1.3 Techniques of Integration

Learning Outcomes: At the end of this lesson, students should be able to
(a) find the integral by using substitution method.

*Only one particular technique is used to solve the problem at one time.
(b) perform integration by parts.

*Erroneous to integrate when the integrand is not of the same variable with respect to variable of
integration.
**The choice of u is such that its differential at some stage is in dependent of x.
***Exclude: Integration involving product of exponential and trigonometric function.
****Use ‘LOPET’.
*****Only one particular technique is used to solve the problem at one time.
(c) solve the integral of a rational function by means of decomposition into partial fractions.
*Exclude: Improper partial fraction.
** Only one particular technique is used to solve the problem at one time.

Integration by Substitution

 The substitution method turns an unfamiliar integral into one we can evaluate.
In other words, substitution gives us a simpler integral involving the variable u.

 Let's now review the five steps for Integration by Substitution:
 Choose a new variable u.
 Determine the differential dx in terms of u and du.
 Substitute.
*When a new variable is substituted, the whole integral must be in terms of the new variable.
 Integrate the resulting integral.
 Return to the initial variable x.

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Lecture Note: 2 of 5 Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021
EXAMPLE 9
Use suitable substitutions to evaluate the following integrals.

(a)  x3  2x3x2  2dx

Let u  x3  2x,

du  3x2  2
dx

3x2  2dx  du

 x3  2x3x2  2dx   udu

 u2  C
2

x3  2x2

 C
2

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Lecture Note: 2 of 5 Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

(b) 3x2 sin x3dx (c) e3x 1 dx
e3x 
Let u  x3,
Let u  e3x  1,
du  3x2
dx du  3e3x
3x2dx  du dx
e3xdx  du
 3x2 sin x3dx   sin udu
3
 cos u  C
 cos x3  C  e3x 1 dx  1  d3u 
e3x  u

  1 du
3u

 1 ln u  C
3

 1 ln e3x  1  C
3

(d)  ln 2x3 dx (e)  tan 3xdx

x sin 3x
cos 3x
Let u  ln 2x,  tan 3xdx   dx

du  1 Let u  cos 3x,
dx x
1 dx  du du  3 sin 3x
x dx
sin 3xdx   du
 ln 2x3 x1 dx   u3du
3
 u4  C
4  tan 3xdx   1 sin 3xdx
cos 3x
 ln 2x4  C
  1  du 
4 u 3

 1  1 du
3 u

  1 ln u  C
3

  1 ln cos 3x  C
3

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Lecture Note: 3 of 5 Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

Integration by Parts

 Integration by Parts used for some of the integrands that cannot be integrated by
substitution or any common method.

 Let u and v are differentiable functions of x.
By integrating the product rule of differentiation, then Integration by Parts states that

 udv  uv   vdu

 Choose u based on LOPET,
i.e. LO = Logarithm, P = Polynomial, E = Exponent, T = Trigonometry.

*u is a function of x which can be easily differentiated, while dv can be integrated.
**LOPET is just an approach in selecting u, NOT a technique of integration.

 The second integral  vdu must be simpler than  udv.

 In some cases, we need to repeat integration by parts more than once.

EXAMPLE 10 (b) 2xe3x1dx P ET
Find the following integrals. 2x e3x1
LO
(a)  xexdx

LO P E T
x ex

u  x,  dv   exdx u  2x,  dv  e3x1dx

du  dx v  ex du  2 v  e3x1
dx 3
 xexdx  uv   vdu du  2dx

 xex  exdx  2xe3x1dx  uv  vdu

 xex  ex  C

 e3x1  e3x1 2dx
3
2x  3

  2 xe3x1  2e3x1 dx
3
3

  2 xe3x1  2e3x1  C

3 33

  2 xe3x1  2 e3x1  C
39

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Lecture Note: 3 of 5 (d)  ln xdx Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021
LO
(c)  x cos 3xdx ln x P ET
1
LO P E T
x cos 3x

u  x,  dv   cos 3xdx u  ln x,  dv   1dx

du  dx v  sin 3x du  1 vx
3 dx x
du  dx
 x cos 3xdx  uv   vdu
x

 x  sin 3x    sin 3x dx  ln xdx  uv   vdu
3 3
x  dx 
 1 x sin 3x   cos 3x  C  ln xx   x
3 9

= 1 x sin 3x  1 cos 3x  C  x ln x   dx
39
 x ln x  x  C

Integration by Partial Fraction

 It is useful method to integrate rational functions which cannot be integrated by any of the
previously discussed method.

 There are two types of fraction which is proper fraction and improper fraction.

Some functions and their corresponding partial fractions are given below:

Function Partial fraction

x AB
x1 x1
x  1x  1

1 x A 1  (x B
  1)2
x  12

x1 A  Bx  C
x x2  1
xx2  1

2 A B  C
x 2x 2x
x4  x2 

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Lecture Note: 3 of 5 Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021
EXAMPLE 11

Find the following integrals.

(a)  x 5 x dx

1 

5 x  A  B
x 1x
x1 

5  A 1  x  Bx

When x  1, When x  0,

5  B 1 5  A 1

B5 A5

 x 5 x dx    5  1 5 x  dx
x 
1 

 5 ln x  5 ln 1  x  C

(b)  x2 2x  1 3 dx
 2x 

x2 2x  1 3  x 2x  1
 2x 
 3x  1

 A B
x3 x1

2x  1  A x  1  B x  3

When x  1, When x  3,

3  B 4 5  A 4

B3 A5
4 4

 x2 2x  1 dx    4 5 3  3 1 dx
 2x  3
x  4x 

 5 ln x  3  3 ln x  1  C
44

14 of 44

Lecture Note: 3 of 5 Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

(c) 2x  4 dx
x3  2x2

2x  4  2x  4
x3  2x2
x2 x  2

 A  B  C
x x2 x2

2x  4  Axx  2  B x  2  Cx2

When x  0, When x  2, When x  1,

4  B 2 8  C 4 6  A 11  2 1  2 1

B  2 C 2 A  2

 2x  4 dx    2  2  x 2 2 dx
x x2 
x2 x  2

  2  2x2  x 2 2  dx
x 

 2 ln x  2x1  2 ln x  2  C
1

 2 ln x  2  2 ln x  2  C
x

(d)  5x2  4x  12 dx

x  2x2  4

5x2  4x  12  A 2  Bx  C
x x2  4
x  2x2  4

5x2  4x  12  A x2  4  Bx  C x  2

When x  2 When x  0, When x  1,

24  A 8 12  A 4  C 2 21  3 5  B 3

A3 C  6  2A B2

 6  2 3

C 0

 5x2  4x  12 dx    x 3 2  2x 4  dx
 x2 
x  2x2  4

  x 3 dx   f x dx
2 f x

 3 ln x  2  ln x2  4  C

15 of 44

Lecture Note: 3 of 5 Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

(e)  r3 1 r dr


r3 1 r  1

rr2  1

 r r  1  1)

1(r

r r  1  1  A  r B 1  r C
r  1
1r

1  Ar  1r  1  Brr  1  Crr  1

W hen r  0, W hen r  1, W hen r  1,

1  A 11 1  B 12 1  C 12

A  1 B1 C1
2 2

 r3 1 dr    1  1  1  dr
r  r
2r  1 2r  1

 ln r  1 ln r  1  1 ln r  1  C
22

16 of 44

Lecture Note: 4 of 5 Mathematics 2 SM025
Session 2020/2021
Ch. 1 Integration

Topic: 1. INTEGRATION

Sub-Topic: 1.4 Definite Integrals

Learning Outcomes: At the end of this lesson, students should be able to
(a) use the properties of definite integral.
(b) solve definite integrals.
(c) find the area of a region bounded by

(i) a curve and the x or y axis,
(ii) two curves,
(iii) a line and a curve.
(d) determine the volume of a solid of revolution bounded by
(i) a curve and the x or y axis,
(ii) two curves,
(iii) a line and a curve.

*Rotate about the coordinate axes.
**Exclude: Rotation about x  a or y  b.

Properties of Definite Integral

b

(a) a cdx  c b  a, where c is a constant.

(b) b f x  g x dx  b f x  d x  b g xdx.

a a a

bb

(c) a cf xdx  ca f xdx, where c is a constant.

a

(d) a f xdx  0.

ba

(e) a f xdx  b f xdx.

(f) c  b c where ab  c.

a f xdx a f xdx  b f xdx,

b

(g) If f x  0 for a  b  c, then f xdx  0.

a

17 of 44

Lecture Note: 4 of 5 Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

EXAMPLE 12

58 5
  Given f xdx  5, f xdx  10 and gxdx  6. Evaluate each of the following.
25 2

5 2

(a) 2 3f xdx (b) 5 gxdx

55 25

2 3f xdx  32 f xdx 5 g xdx  2 g xdx

 35  6

 15

(c) 5 2f x  3g x dx 8

2 (d) 2 f xdx

5 2f x  3g x dx  5 f xdx 85

2 22 2 f xdx  2 f xdx

5 8

32 g xdx 5 f xdx

 25  36  5  10
 15
 28

Determine the Value of the Definite Integral

EXAMPLE 13
Evaluate each of the following.

(a) 2 e4x2dx (b) 1 2x3 dx
0 0 1  x4

 2 e4x2dx   e4x2 2  1 2x3 dx  1  du 
0   0 1  x4 2
 4 0 1 u  1  x4

 e422  e402 u2 du  4x3
44 dx
 12 1
2x3dx  du
u 2du 2

 e6  1 1
4 4e2
 u2  C

  1  x4 1
0

 1  14  1  0

 2 1

18 of 44

Lecture Note: 4 of 5 Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

Area between the curve and the x -axis

If f is continuous throughout a,b, then the area of the region between the curve y  f x and

b

the x-axis from x  a to x  b is given by A  a ydx.

y  f x

y  f x

b b

A   ydx A   ydx

a a

Area between the curve and the y -axis

If f is continuous throughout c, d, then the area of the region between the curve x  g y and

d

the y -axis from y  c to y  d is given by A  c xdy.

x  g y x  g y

d d

A   xdy A   xdy

c c

19 of 44

Lecture Note: 4 of 5 Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

EXAMPLE 14

(a) Sketch the graph of y  xx  2 (b) What is the area between the curve y  xx  2

for 2  x  4. and the x-axis, bounded by x  2 and x  2 ?

02
A   xx  2dx  0 x x  2dx
2

 20   4
33

 8 units2

EXAMPLE 15
Find the area of the region bounded by the curve y  x3  3, the y -axis and the lines y  2
and y  4.

y4
y2

When x  0,

y  03  3

3

A   xdy

3 1 41

y  33dy y  33dy
  
23

 4  3  4 4

 3 y  33  3 y  33 

 41 2  41 3

33
44

 3 unit2
2

20 of 44

Lecture Note: 4 of 5 Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

Area Bounded by : Two Curves OR A Line and A Curve

Take a look at the following sketch to get an idea of what we are initially going to look at.

y  f x

x  g y

x  f y

y  g x

b d

A   f x  gx dx A   f y  g y dy

a c

However, it is sometimes easy to forget that these always require the first function to be the
larger of the two functions. So, instead of these formulas we will instead use the following
“word” formulas to make sure that we remember that the area is always the “larger” function
minus the “smaller” function.

A  b fuunpcpteiorn  fulnowcteiorn dx, a  x  b A  d furnigcthiton  funlecfttion dy, c  y  d
a c

21 of 44

Lecture Note: 4 of 5 Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

EXAMPLE 16

Calculate the area of the region bounded by the curve y  x2  3, 3y  2x  14 and y  3.

x2  3   2 x  14
33

3x2  9  2x  14
3x2  2x  5  0

3x  5x  1  0

x 5,x 1
3

 2 x  14  3
33
2x  14  9

x5
2

1 x2 3  3 5  2 14  3
 A  0  dx  2 3 x  3  dx

1

1 x2dx  5  2 5 
2 3 3
0
  1 x  dx

  x3 1   x2  5 5
 3 3
 3 0 2
x1

13
34

 13 unit2
12

22 of 44

Lecture Note: 4 of 5 Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

EXAMPLE 17

Find the area of the region bounded by the curves y  x2  5x and y  25  x2.

x2  5x  25  x2
2x2  5x  25  0

2x  5x  5  0

x 5, x  5
2

A  5 25  x2  x2  5x dx
5
2

   5x2  2x3 5
25x 2 3 
5
2

  5 52  253   25 25  5  5 2 2  5 3 
 2 2
255  2 3 
2 3

 140 5 units2
8

23 of 44

Lecture Note: 5 of 5 Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

Volume of the Solids Generated by Revolving the Region by the Curve & the x -axis

The volume of the solid generated by revolving the region about the x-axis between the graph

of a continuous function y  f x and the x-axis from xa to xb is V   b y2dx.
a

y  f x

360

Volume of the Solids Generated by Revolving the Region by the Curve & the y -axis

The volume of the solid generated by revolving the region about the y -axis between the graph

of a continuous function x  g y and the y -axis from yc to yd is V  d x2dy.
c

x  g y
360

24 of 44

Lecture Note: 5 of 5 Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

EXAMPLE 18

Find the volume of the solid formed when the area bounded by the curve y  1 , y-axis, y  2
x

and y  5 is rotated about y-axis.

V   x2dy

  5  1 2 dy
2 y

  5 y2dy
2

   1 5
y 
2

   1   12
5

 3  unit3
10

25 of 44

Lecture Note: 5 of 5 Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

EXAMPLE 19

Sketch the curve y2  123  x in the first quadrant. Given that R is the area bounded by

the axes in the first quadrant.

6

Find

(a) V1, the volume of the solid of revolution obtained by revolving the region R about the

x-axis.

3
V1   123  xdx

0

 12 3x  x2 3
2 
 0

 12 33  32  0

 2

 54 units3

(b) V2, the volume of the solid of revolution obtained by revolving the region R about the
y -axis.

V2   6 3  y2 2 dy
0 12

  6 9  6y 2  y4  dy
0 12 144

  9y  y3  y5 6
6 
720 0

= 1454  0

 144  units3
5

(c) Hence, show that V1 : V2  15 : 8.

V1 : V2
54 : 144 

5

15 : 8 shown

26 of 44

Lecture Note: 5 of 5 Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

Volume of Solids Generated by Revolving the Region between :
Two Curves OR A Line and A Curve

Revolve about x -axis

 Suppose that y1  f x and y2  g x are non-negative continuous function such that,
f x  g x for a  x  b.

 Let R be the region enclosed between the graphs of these functions and the lines x  a and
x  b. When this region is revolved about the x-axis, it generates a solid.

y1  f x

y2  g x 360

 The volume of the solid generated by revolving R about x-axis is V   b



a
y12  y22 dx.

Revolve about y -axis

 Suppose that x1  f y and x2  g y are non-negative continuous function such that,
f y  g y for c  y  d.

 Let R be the region enclosed between the graphs of these functions and the lines y  c and
y  d. When this region is revolved about the y-axis, it generates a solid.

360

x2  g y

x1  f y

   d



c
The volume of the solid generated by revolving R about y -axis is V  x12  x22 dy.

27 of 44

Lecture Note: 5 of 5 Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

EXAMPLE 20

A region R is enclosed by the curve y  x2  4 and the line y  2x  4.
(a) Sketch the graphs and shade the region R.

(b) Find the volume of the solid generated when the region R is rotated through 2 radian

about the x-axis.

V   2 x2  42  2x  42  dx
0

  2 x4  8x2  16  4x2  16x  16 dx
0

   2 x4  12x2  16x dx
0

   x5  12x3  16x2 2
5 3 2 0

 32  unit3
5

28 of 44

Lecture Note: 5 of 5 Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

EXAMPLE 21

The diagram shows a region R in the first quadrant bounded by the curves y  1 4  x2 ,

4

y  1 4  x2  and the y-axis. Calculate the volume of the solid formed when R is rotated

2

through 360o about the y-axis.

y  1 4  x2 

4
4  x2  4y

x2  4  4y

y  1 4  x2

2
4  x2  2y

x2  4  2y

21

V  0 4  2ydy  0 4  4ydy

  4y  y2  2   4y  2y 2 10
0

  4  0  2  0

 2 unit3

29 of 44

Tutorial 1: 6 Hours Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

1. Find the following integrals.

(a) 6p 2  2p  4 dp (b)  x x3  x2 dx
3

 6p 2  2p  4dp  3 6 p 5   2p2  4p C  x x3  x2dx  x 7  1 x 3  dx
3 5 3 2 2 2 2

 18 5  p2  4p  C  2 9  1 5 C

p3 x2 x2
5 95

1 ln x

(c)  2e2 dx

 1 lnx 1

2e2 dx  2eln x2 dx

1

  2x2dx

 4 3  C

3 x2

Given that F x  x  5 x2 d F x 2x2  5x  4 dx.
2.  4, find dx . Hence, find 4 x2  4

d F x  d x  5x2  1  u x5 1
dx dx u  1
42 v  x2  42

  x  5 x  1 1
x2   x2  4 1 v  x2  2x
4 2 42

 x2  5x  x2  4 x
x2  4 x2  4

 2x2  5x  4
x2  4

d   5x2  1   2x2  5x  4
dx  x2  4
x 42

1  d x  5x2  1 dx  1  2x2  5x  4 dx
4 dx 4 x2  4
42

 2x2  5x  4 dx  1 x  5x2  1  C

4 x2  4 4 42

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Tutorial 1: 6 Hours Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021
3. Find the following integrals:
(b)  3  ey 2  ey dy
(a) 53x dx  3  ey 2  ey dy   7  3ey  2ey dy

 53x dx  53x  C  7y  3ey  2ey  C
3 ln5

(c) e2x  4 dx
ex

   e2x 4

ex
dx  ex  4ex dx

 ex  4ex  C

4. Find the following integral by using substitution method.

(a)  x dx (b)  ln x2 dx
x1
2x

 x dx   u  1 du u  x1  ln x2 dx  1  u2du u  ln x
x 1 u 2
du  1 2x du  1
dx dx x
 1 1  1  u3   C 1 dx  du
2 3 x
 u2  u 2du
3 dx  du

 2u 2 1 C  1 ln x3  C

3  2u2 6

 2 x 3  2x 1  C
3
 12  12

(c) x2  1  e x3 xdx
3

 x21 ex3 x dx  eu  du  u  x3  x
3 3
du  3x2  1
 1  eudu dx
3 du  3x2  31dx

 1 eu  C x2 
3

 1 ex3 x  C  1 dx  du
3 3 3

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Tutorial 1: 6 Hours Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021
5. Find the following integrals:
(b) 1 1 dx
(a)  x cos2  4x2dx  ex

 x cos2  4x2 dx   cos u  du  1 ex u  1  ex
8  ex  ex
 1 dx  1 dx du  ex
dx
u  2  4x2  1  cos udu   1 du exdx  du
8 u
du  8x
dx   1 sin u  C  ln u  C
xdx   du 8
 ln 1  ex  C
8   1 sin 2  4x2   C

8

6. By using integration by parts, find

(a)  ln 3xdx

 ln 3xdx  ln 3xx   x  1 dx u  ln 3x,  dv   dx
x
du  1 vx
 x ln 3x   dx dx x
du  1 dx
 x ln 3x  x  C
x

(b) xexdx u  x,  dv   exdx

 xexdx  xex  exdx du  dx v  ex

 xex  ex  C

(c)  x ln x2dx u  ln x,  dv   2xdx

 2x ln xdx  ln xx2   x2 dxx du  1 v  x2
 x2 ln x   xdx dx x
du  dx
 x2 ln x  x2  C
2 x

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Tutorial 1: 6 Hours Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

(d)  x3x dx

 x3x dx  x3x  3x dx u  x,  dv   3x dx
du  dx
ln 3 ln 3 v  3x
ln 3
 x3x  1  3x   C
ln 3 ln 3 ln 3

 x3x  3x C
ln 3
ln 32

7. Show that e x ln xdx  1 1  e2 .
14

 e ln x x2 1e e  x2  dx 
1 2 1 2 x
x ln xdx   u  ln x,  dv   xdx

  x2 ln x  x2 e du  1 v  x2
 2 4  dx x 2
1 du  dx

 e2  1 x
4

 1 1  e2 

4

8. By expressing as partial fraction, find the following integrals.

(a)  x3 1 dx
x

x3 1 x  1  x3 1 x dx    x1  x2 x 1 dx u  x2  1
  
xx2  1
du  2x
 A  Bx C  ln x   1  d2u  dx
x x2 1 u xdx  du

1  A x2  1  Bx  C x  ln x  1  1 du 2
2 u
1  Ax2  A  Bx2  Cx
 ln x  1 ln u  C
1  A  Bx2  Cx  A 2

By comparing the coefficients:  ln x  1 ln x2  1  C
 A  1, B  1, C  0 2

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Tutorial 1: 6 Hours Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

(b)  1 3x dx
x  2x2

3x  3x
1  x  2x2
x  12x  1

A B
x  1 2x  1

3x  A 2x  1  B x  1

When x  1, When x   1 ,
2
3  A 3
3  B  3 
A  1 2 2

B  1

 1 3x dx    x 1  1 1 dx
x  2x2 1 2x 

  ln x  1  1 ln 2x  1  C
2

9. 2 3x2  7x  6 dx. the answer in the of a  ln b.
Evaluate 1 Give form
x  32 x  1

3x2  7x  6  A  B  x C
x1 x3
x  32 x  1  32

3x2  7x  6  A x  32  B x  3x  1  C x  1

When x  1, When x  3, When x  0,
16  16A 12  4C 6  9  3B  3
A1 C 3 B2

 3x2  7x  6  x 1  x 2  x 3
1 3
x  32 x  1  32

3x2  7x  6  1 2 3 
    
x  32 x  1
 2 dx  2 x 1  x 3  x 32 dx
1 1

 ln x  1  2 ln x  3  3 2

 x  31

 3  ln 3
28

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Tutorial 1: 6 Hours Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

x2  3 B x2 3
x2  4 x2  x2 4
10.
Show that can be written in the form of A  4 . Hence, find dx.

x2  4 1
x2  3

x2  4

7

x2  3  1  x2 7 4
x2  4 

x  7  2  x A 2  x B 2
 
2x

7  Ax  2  B x  2

When x  2, When x  2,
7  4A 7  4B
A7 B  7

4 4

x  7  2  7 2)  7 2)
4(x  4(x 
2x

 x2  3dx   1 7  7 dx
x2 4 4(x  2) 4(x  2)

 x  7 ln x  2  7 ln x  2  C
44

 x  7 ln x  2  C
4 x2

11. Find the following integrals.

(a)  sin 2x cos 2xdx (b)  cot ydy

 sin 2x cos 2xdx  1  sin 4xdx  cot ydy   cos y dy u  sin y
2 sin y du  cos y
dy
 1  cos44x  C   1 du cos ydy  du
2 u

  1 cos 4x  C  ln u  C
8
 ln sin y  C

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Tutorial 1: 6 Hours Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

(c)  x sin 2x    dx
4

 x sin 2x    dx  1 x cos2x     1  cos 2x    dx ux
4 2 4 2 4 du  dx

 1 x cos 2x     1 sin 2x     C sin 2x  
2 4 4 4 4
 dv    dx

v   1 cos 2x   
2 4

12. Evaluate the following integrals.



(a) 2 sin 5x cos xdx
0

   1 sin 5x  x  sin 5x  x dx
2 2
2 sin 5x cos xdx 
0
0

  1 sin 6x  sin 4xdx
2

02


  1 cos 6x  1 cos 4x 2
 12 8 0

  1 cos 6     1 cos 4      1 cos 6 0  1 cos 4 0
12 2 8 2  12 8

1
6

(b) cos3  d
sin2 

 cos3  d  cos  cos2  d
sin2  sin2 
u  sin 
 cos  1  sin2  d
du  cos 
sin2  d
cos d  du
 1  u2 du
u2

  u2  1du

1 uC
u

  1  sin   C
sin 

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Tutorial 1: 6 Hours Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

(c) 3 tan x  45 sec2 xdx u  3 tan x  4
du  3 sec2 x
 3 tan x  45 sec2 xdx  1  u5du dx
3 sec2 xdx  du

 1  u6   C 3
3 6

 1 3 tan x  46  C

18

(d) cot2 2 sin3 2d

 cot2 2 sin3 2d  cos2 2 sin3 2d
sin2 2

  sin 2 cos2 2d u  cos 2

   u2 du du  2 sin 2
2 d
sin 2d   du
  u3  C
6 2

  cos3 2  C
6

(e)  sin2 2xdx

 sin2 2xdx  1  1  cos 4xdx
2

 1 x  sin44x  C
2

 1 x  1 sin 4x  C
28

(f)  cos3 2xdx

 cos3 2xdx   cos 2x cos2 2xdx

  cos 2x1  sin2 2xdx u  sin 2x

  1 1  u2 du du  2 cos 2x
dx
2 cos 2xdx  du

  12  u2  du 2
2

 1 u  1 u3  C
26

 1 sin 2x  1 sin3 2x  C
26

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Tutorial 1: 6 Hours Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

13. Sketch and find the area of the region

(a) between the curve y  x3 and the lines y  x and y  1.

y  x y  x3

y 1

A  1 11  1 1  x3  dx
2
0

 1  x  x4 1
2 4 
 0

 5 units2
4

(b) enclosed by curve y  7  x  x2 and the line y  2x  3.

7  x  x2  2x  3
x2  3x  10  0

x  2x  5  0

x  2, x  5

A  5 7  x  x2   2x  3 dx
2

 5 x2  3x  10dx
2

  1 x3  3 x2  5
10x
 3 2 2

 343 units2
6

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Tutorial 1: 6 Hours Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

14. Find the volume of the solid formed by revolving the given region through 360.

(a) y  x3, y  1, y-axis; about the y-axis.

V   1 x2dy
0

12

  y3dy
0

   3 y 5 1
5 3 0

 3  units3
5

(b) y  ex, x  0, x  1, x-axis; about the x-axis.

V   1 y2dx
0

  1 e2xdx
0

   e2x 1
 
 2 0

 e2  1  units3
2

(c) y  2x  3, y  x2; about the x-axis.

V   3 y2dx
1

  3 2x  32  x4  dx
1

  2x  33  x5 3
6 5 1


 1088  units3
15

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Tutorial 1: 6 Hours Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

15. Determine the coordinates of the points of intersection between the curve y2  x and the

line y  x  2. Find the area bounded by the curve and the line. Find also the volume of

the solid of revolution generated when this area is rotated through 360 about the y-axis.

y2  x

x  22  x

x2  4x  4  x
x2  5x  4  0

x  1x  4  0

x  1, x  4

W hen x  1, W hen x  4,
y  1  2 y  4  2

1  2

 the points of intersection are 1,1 and 4,2.

A  1 2  y   y 2  dy
2

 2y  y2  y3 1
2 3 
2

 9 units2
2

V  12 y2  y2 2  dy

2

  1 4  4y  y2  y4 dy
2

 4y y3 y 5 1
3 
  2y2   5 2

 72  units3
5

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Tutorial 1: 6 Hours Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

16. A region R is bounded by the curve y  xx  2 and y  x.

(a) Sketch the graphs and shade the region R.

yx

y  xx  2

(b) Find the area of R.

A  3 x  x2  2x dx

0

 3 3x  x2  dx

0

 3 1  3
 
 x2  x3
 2 3 0

 9 units2
2

(c) Find the volume of the solid obtained when the part of R above the x-axis is rotated

through 360 about the x-axis.

   V   2 x2dx   3 x2  x2  2x 2 dx
02

     2 x2dx   3 x4  4x3  3x2 dx
02

   1 x3 2    1 x5  x4  x3 3
 3 0  5  2

 97  units3
15

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Tutorial 1: 6 Hours Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

17. Sketch and shade the region R bounded by the curve y  x, line y  2  x and y-axis.
Hence, find the area of the region R.

x 2x y x
y 2x
x  2  x2

x2  5x  4  0

x  1x  4  0

x  1, x  4

 1

R  0 2  x  x dx

 2x  x2  2 3 1
2 3 
x2 0

 5 unit2
6

(a) If R1 is a region bounded by the curve y  x, line y  2  x and x-axis, deduce the
ratio of R : R1.

R1  1 22  5 5
6
2 R  6
R1 7
7
6 6

R 5
R1 7

 R : R1  5 : 7

(b) Find the volume of the solid generated when the region R is rotated through 360 about
the x-axis.

 V 1 2  x2 x 2  dx
 0  

   2  x3  x2 1
3 2 
 0

 11  unit3
6

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Tutorial 1: 6 Hours Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

18. Given that f x  1 and g x  x .
x3 4

(a) On the same axes, sketch the graphs of f and g for the values of x between x  0 and

x  2. Shade the region R bounded by f, g, x  0 and x  2.

x 1 g x  x
4 x3
x2  3x  4 4
x2  3x  4  0
f x  1
x  4x  1  0
x3
x  4, x  1 x2

(b) Find the area of region R.

 R  1  x 1 3  x4  dx  2  x  x 1 3 dx
0  1 4 

 ln x  3  x2 1   x2  ln x  2
8   8
 0 31

 0.315 unit2

(c) Find the volume of the solid generated when the region R is rotated through 2 radian
about the x-axis.

 V   1  x 1 3 2   x 2  dx   2  x 2   x 1 32  dx
0  4 1 4 

   1  1  x2  dx   2  x2  1 32  dx
 32 16  1 16 x  
0 x

   1  x3 1    x3  1 2
   48  
x 3 48 0 x 3 1

 19  unit3
120

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Tutorial 1: 6 Hours Mathematics 2 SM025
Ch. 1 Integration Session 2020/2021

Answers

1. (a) 18 5  p2  4p  C 29 15 43
p3 (b) x2  x2  C (c) x2  C
5
95 3

2x2  5x  4 1 x  5 1
, 2
 2. x2
4 4 C

x2  4

3. (a) 53x C (b) 7y  3ey  2ey  C (c) ex  4ex  C

3 ln 5

4. (a) 2 x  3 1 C (b) 1 ln x3  C (c) 1 ex3 x  C
3
3 12  2 x  12 6

5. (a) 1 sin 2  4x2   C (b) ln 1  ex  C

8

6. (a) x ln 3x  x  C (b) xex  ex  C (c) x2 ln x  x2  C x3x 3x
7. DIY 2 (d)  C
ln 3 ln 32

8. (a) 1 x2  1  C (b)  ln x  1  1 ln 2x  1  C

ln x  ln
22

3 3 7 x2
9.  ln 10. x  ln C
28 4 x2

11. (a) 1 cos 4x  C (b) ln sin y  C (c) 1 x cos 2x     1 sin 2x     C
  4 4 4
8
2

1 (b)  1  sin   C (c) 1 3 tan x  46  C (d)  cos3 2 C 11
12. (a) (e) x  sin 4x  C
6 sin  18 6
28

1 sin3 2x
14. (f) sin 2x  C
26

5 343
13. (a) (b)
46

3 (b)  e2  1  1088
14. (a)  2 (c) 

5 15

15. 1, 1,4, 2, 9 , 72 

25

16. (a) DIY (b) 9 (c) 97

2 15

5 11
17. DIY, (a) 5 : 7 (b) 
66

18. (a) DIY (b) 0.315 (c) 19 
120

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