CHAP 4 INTEGRATION

MATHEMATICS 2 - SSM 0012

4: INTEGRATION

4.1 INTEGRAL OF A FUNCTION
 Integration is considered as the reverse operation of differentiation.

 Symbol of integration, 

Example :
y = 2x2 + 3
dy  4x  differentiation
dx

 4x = 2x2 + c  integration (to get back y)

 Notice that for integration, a constant term is included

Standard Integral
a) Integration of xn

 x ndx  x n1  c Ex: a) x 3dx  x 4  c
n1 4
b) x 4dx
1

c) x 2 dx

b) Integration of 1

 1dx  x  c

1

c) Integration of

x

 1 dx  ln x  c
x

1

d) Integration of ex MATHEMATICS 2 - SSM 0012

 exdx  ex  c c)  4x 3dx

e) Integration of kf(x) k is a constant

 kf(x)dx  k f (x)dx b) 1 x2dx
2
Ex : a) 2x 6dx

f) Integration of f(x)  g(x) k is a constant

 f (x)  g(x)dx   f (x)dx   g(x)dx

Ex: a) (2x2  x)dx b) (1 x 2e x  3)dx
4

2

Integration in the form [f(x)]nf ’(x) MATHEMATICS 2 - SSM 0012
* f ’(x) is dy of function f(x)
 [ f (x)]n f '(x)dx  [ f (x)]n1  c
n1 dx

Example: Notice that 2x is the differentiation of d) 4x dx
(x2 + 1) x2 1
a) (x2 1)4 (2x)dx
c) (e x 1)3 (e x )dx
= (x 2  1)5  c
5

b) 3x2 (x 3 2)4 dx

e) (ln x)3 dx f) e x dx
x 1ex

3

4.2 INTEGRATION TECHNIQUES MATHEMATICS 2 - SSM 0012
Based on method substitution
 Linear function
c)  2e5x2dx
1.  (ax  b)n dx   (ax  b)n1  c
a(n  1)

2.  1 b dx   1 ln ax  b  c
ax  a

 3. eaxbdx  1 eaxb  c b)  1 dx
3x 1
a

Example:

a)  (2x 1)3 dx

4. Example:

 f '(x) dx  ln f (x)  c a)  2x b)  3x2  2x c)  4x
f (x) x2 1 x3  x2 x2 3

5. Example : b)  2e2x1 c)  3x 2e2x3

 f '(x)e dxf (x)  e f (x)  c a)  2xe x2

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4.21 Integration by substitution MATHEMATICS 2 - SSM 0012

 f (x)dx   g(u) dx du 5
du

Example :

Find x 2x 5 dx
2

Solution :
Let u = x2 + 5  f ‘(x) = 2x

 x2x dx is f'(x) dx
2 5 of the form f (x)

Therefore, 2x dx  ln x 2  5  c
x2 
5

Example :

Find the following integrals.

a)  x
x4

Solution :

a) Let u2 = x + 4
x = u2 – 4

dx = 2u du

Therefore,

 x dx  u2  42udu
x4 u

 2 u3  4   c
3

 2 u(u2 12)  c
3

 2 ( x  4)(x  4) 12)  c
3

 2 ( x  4)(x  8)  c
3

MATHEMATICS 2 - SSM 0012

Exercises:

1. Find each of the following integrals.

a) (2x x 1) 3 dx b) x 1 dx
 x 2

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MATHEMATICS 2 - SSM 0012

Tips of Integrations:

1. For single function, integrate by formula (standard integral).

2. If there is product of two functions, where one of the function is the differentiation of the

other function use  [ f (x)]n f '(x)dx  [ f (x)]n1  c
n1

Intergrate [f(x)]n only

3. For rational function (fraction):

a.  a dx  ln |f(x)| + c ; f(x) is linear function, a is a constant
f (x) ; f(x)’ is the differention of f(x)

b.  f (x)' dx  ln |f(x)| + c
f (x)

c. a dx  change to [f(x)]-n and integrate by formula (standard intergral)
[ f (x)]n

d.  f (x) dx  power of function f(x) is higher than g(x), use long division and
g(x)

integrate

e.  f (x) dx  f(x) and g(x) are linear functions, adjust f(x)
g(x)

Example :

 2x  1 dx   2x  3  4 dx
2x  3 2x 3

  2x  3  4 dx
2x 3 2x 3

  1  4 dx
2x  3

 x + 4 ln|2x-3| + c

4. For other functions, use substitution

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MATHEMATICS 2 - SSM 0012

4.3 DEFINITE INTERGRALS

b

 f (x)dx  [F(x)]ab

a

= F(b) – F(a)

Example 1:

Find the following definite integrals.

2

a) (4x  5)dx
1

 = 2
2x 2  5x 1

= [2(2)2 – 5(2)] – [2(1)2 – 5(1)]

= -2 – (-3)

=1

b) 9 1 dx 3
3 x3
c) 6e3xdx
0

n

d) If  (3x 1)2 dx 14, find n.

0

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MATHEMATICS 2 - SSM 0012

4.4 AREA OF REGIONS
a) Area Bounded by a Curve and the x-axis
The definite integral of a function f(x) with respect to x over the [a, b] is defined as the area
bounded by the graph y = f(x), the x-axis and lines x = a and x = b

y
y= f(x)

ab x

b b

Area , A =  f (x)dx  When an integral ydx has a
a a
negative value, it indicates that
b the bounded area is below the
x- axis.
A   ydx
a  The value of the area is always
positive.

Example 1:
Find the area of the region bounded by the curve y = 3x2 + 4x – 4, the x-axis, x=1 and x=2.

Solution:

2

Area   ydx
1

2

  3x2  4x  4dx
1

  2
x3  2x2  4x 1

= (8 + 8 – 8) – (1 + 2 – 4)

=8+1

= 9 unit2

9

MATHEMATICS 2 - SSM 0012

Example 2:
Find the area of the region bounded by the curve y = x2 - 3x and the x-axis.

Solution:
When y = 0

0 = x2 – 3x
0 = x(x – 3)
x = 0, x = 3

3

Area   ydx
0

3

  x2  3xdx
0

 x3  3x 2 3
 2 
 3 0

  33  3(3)2   0  3(0) 
 3 2   3 2 
 

 9  27   0
2 

 9
2

 9 unit2
2

10

MATHEMATICS 2 - SSM 0012

b) Area Bounded by a Curve and the y-axis
The area bounded by the graph x = f(y), the y-axis and the lines y = a and y = b is given by

y
y= f(x)

b

a

b x

A   xdy b
a
 When an integral xdy has a
a
negative value, it indicates that
the bounded area is on the left
of the y- axis.

 The value of the area is always
positive.

Example 1:

Find the area of the region bounded by the curve y = x3 + 3, the y-axis and the lines y = 2 and y = 4.

Solution:

y = x3 + 3 Sketch the graph of y = x3 + 3

1 y y= x3 + 3

x  y  3 or x  (y  3)3

Total area = area A + area B 4B When x = 0
A y = 03 + 3
314 1 y=3
2
  (y  3)3 dy + (y  3)3 dy x

23

 3 3  2  3 3 4
 4   4 
= (y  3)  (y  3)
4 3 4 3

= 0  3    3  0 = 3   3 
4   4 4  4 

= 3 unit2
2

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MATHEMATICS 2 - SSM 0012

c) Area Bounded by Two Curve or Area Between a Curve and a Line a and b is the point
y of intersection
y= g(x) between the two
curves
y= f(x) x
ab

y
y= g(x)
a and b is the point
of intersection
between the two
curves

y= f(x)
a bx

b

A   [ f (x)  g(x)]dx
a

12

MATHEMATICS 2 - SSM 0012

Example:
Find the area of the region bounded by the curve y = x2 – 4x + 4 and the line y = 4 – x
Solution:
Sketch the curve and the line

Point of intersection , 0 3

x2 – 4x + 4 = 4 – x
x2 – 3x = 0
x(x – 3) = 0

x = 0 or x = 3

3

Area = [(4  x) (x2  4x  4)]dx
0

13

MATHEMATICS 2 - SSM 0012

4.5 VOLUME OF REVOLUTION
Revolving a Region Bounded by a Curve and Axis About the Axis

y y
a)

x
x

b) y y

x x

a) Volume of a solid, if a region is rotated completely about x-axis 14

V  π b y 2 dx



a

b) Volume of a solid, if a region is rotated completely about y-axis

V  b x 2 dy

π

a

MATHEMATICS 2 - SSM 0012

Example 1:

The region bounded by the curve y = x2 + 1, the x-axis and the lines x = 1 and x = 3 is rotated through

360o about the x-axis. Show that the volume generated is 67 11 π .
15

Solution:

Volume generated = π 3 y 2dx



1

= 3 (x 2  1)2 dx

π

1

= 3 (x 4  2x 2  1)dx

π

1

= π  x5  2x3 3
 5 3  x

1

= π 243  18  3   1  2  1
5   5 3

= π 242  2  20
 5 3 

= 67 11 π unit3
15

15

MATHEMATICS 2 - SSM 0012
Revolving a Region Bounded by a Two Curve and Axis About an Axis

y y= g(x)

R y= f(x)
a bx

b

V  π {[ f (x)2  [g(x)]2 }dx
a

Example 1:

Calculate the volume generated when the region enclosed by the curve y = 4x – x2 and the lines y=x and
x = 2 is rotated through four right angles about the x-axis.

Solution : y

2 x
024
Volume generated = π (4x  x2 )2 (x)2 dx
0 When y=0
0 = x( 4 - x)
2 x = 0, x = 4

 π (16x2  8x 3  x 4  x2 )dx
0

16x 3  2x 4 x5 x3 2
5 3 
 π  3   0


 π  3  2x 4  x5 2
5x 5 
 0

 π5(2)3  2(2)4  (2)5   0
 5 


 π 40  32 32   0
 5 

 14 2 π unit3
5

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MATHEMATICS 2 - SSM 0012
Exercises :
1. Find the volume of the solid of revolution formed when the region bounded by the curve y2 = x and

the line y = x – 2 is rotated completely about the y-axis.
2. Find the volume generated when the curve x2 + y2 = 9 is rotated 180o about the y-axis.

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