CHAPTER 3 INTEGRATION

DBM 20023 – ENGINEERING MATHEMATICS 2

When = −2
−5(−2) + 11 = (−2 − 1) + (−2 + 2)
21 = −3

= −

Step 4: Solve the indefinite integral using integration of proper rational partial fraction
(choose suitable method above, for this question )

∫ −5 +11 = ∫ −7 + ∫ 2
2+ −2 +2 −1

= − | + | + | − | +

2. ∫ 2 +1
( +1)( −2)

[ : 1 ln| + 1| + 5 ln| − 2| + ]
3 3

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DBM 20023 – ENGINEERING MATHEMATICS 2

3. ∫ 2 +5
2− −2

[ : 3 ln| − 2| − ln| + 1| + ]

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DBM 20023 – ENGINEERING MATHEMATICS 2

4. ∫ 3 +4
2−5 +6

[ : 13 ln| − 3| − 10 ln| − 2| + ]

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DBM 20023 – ENGINEERING MATHEMATICS 2

5. ∫ 3 +2
( −1)(2 +3)

[ : ln| − 1| + ln|2 + 3| + ]

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DBM 20023 – ENGINEERING MATHEMATICS 2

6. ∫ 2 −1
( −1)( +1)( −3)

[ : − 1 ln| − 1| − 3 ln| + 1| + 5 ln| − 3| ]
4 8 8

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DBM 20023 – ENGINEERING MATHEMATICS 2

7. ∫ +1
2+4 −5

[ : 1 ln| − 1| + 2 ln| + 5| + ]
3 3

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DBM 20023 – ENGINEERING MATHEMATICS 2

8. ∫ 2+ −3
( +1)( −2)( −5)

[ : − 1 ln| + 1| − 1 ln| − 2| + 3 ln| − 5| + ]
6 3 2

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DBM 20023 – ENGINEERING MATHEMATICS 2

3.7.2 REPEATED LINEAR FACTOR

When the denominator has some or all linear factors repeated n-times, like ( − ) , then there

are exists a partial fraction of the form + + + … … … … . terms
( − ) ( − )2 ( − )3

EXERCISE:

Find the indefinite integral:

1. ∫ 2 +1
( −1)2

2 +1 resolve into partial fraction
( −1)2

Partial Fraction

2 +1 = + +
( −1)2 −1 ( −1)2

Multiply both side with ( − 1)2

2 + 1 = ( − 1)2 + ( − 1) + ( )

When = 0, When = 1,

2(0) + 1 = (0 − 1)2 + (0)(0 − 1) + (0) 2(1) + 1 = (1 − 1)2 + (1)(1 − 1) + (0)

= = 3
When = 2,

2(2) + 1 = 1(2 − 1)2 + (2)(2 − 1) + 3(2)

5 = 1 + 2 + 6

5 = 7 + 2

= −

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DBM 20023 – ENGINEERING MATHEMATICS 2

Step 4: Solve the indefinite integral using integration of proper rational partial fraction (choose
suitable method above, for this question )

∫ 2 +1 = ∫ 1 + ∫ −1 + ∫ 3
( −1)2 −1 ( −1)2

= − | − | − +
−

For ∫ 3 , use substitution method:
( −1)2

= − 1 = 1 =


∫ ( 3 1)2 = ∫ 3
− 2

= 3 ∫ −2

= ∫ 3 −2

= 3 [ − −11] +

= − 3 1 +
−

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DBM 20023 – ENGINEERING MATHEMATICS 2

2. ∫ 3 −1
( −2)2

[ : 3 ln| − 2| − 5 + ]
−2

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DBM 20023 – ENGINEERING MATHEMATICS 2

3. ∫ 4
( −1)2( +1)

[ : ln| − 1| − 2 − ln| + 1| + ]
−1

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DBM 20023 – ENGINEERING MATHEMATICS 2

4. ∫ 3−2
( +3)2

[ : −2 ln| + 3| − 9 + ]
+3

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DBM 20023 – ENGINEERING MATHEMATICS 2

5. ∫ 2 2+ +4
( +1)( −4)2

[ : 1 ln| + 1| + 9 ln| − 4| − 8 + ]
5 5 −4

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DBM 20023 – ENGINEERING MATHEMATICS 2

3.8 Apply the techniques of integration

3.8.1 Calculate the graph area along x-axis or y-axis
(Under a Curve)

Area left along x-axis

X-AXIS

X-AXIS

We consider the case where the curve is below
the x-axis for the range of x values being
considered.

In this case, we find the area by simply finding the
integral:

In this case, the integral gives a negative
number. Area cannot be a negative sign,
= ∫ ( ) therefore the value becomes positive.



= ∫ ( )



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DBM 20023 – ENGINEERING MATHEMATICS 2

Example: 1 Example: 2

Find the area of shaded
region:

Find where the curve cuts the x-axis
4 − 2 = 0
(4 − ) = 0

= 0 , = 4

5 The curve cuts the x-axis at (0,0) and (4,0)

= ∫( 2 − 4 + 5) 0 and 4 are the limits of integration

0

= [ 3 3 − 4 2 + 5 4
2
5 ] = ∫(4 − 2)

0 0

= [ 3 3 − 2 2 + 5 = [42 2 − 3 3]40
= [2 2 − 3 3]04
5 ]
= [(2(4)2 − (43)3) − (2(0)2 − (03)3)]
0

= (533 − 2(5)2 + 5(5))

− (033 − 2(0)2 + 5(0))
[ ]

= 50 2 32 2
3 3
=

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Example: 3 DBM 20023 – ENGINEERING MATHEMATICS 2

Case 3: Part of the curve is below the x-axis,
part of it is above the x-axis

Find where the curve cuts the x-axis In this case, we have to sum the individual
parts, Area = ∫0 ( ) + ∫ ( )

2 − 6 = 0

( − 6) = 0 = 0 , = 6

The curve cuts the x-axis at (0,0) and (6,0)

0 and 6 are the limits of integration

6

= − ∫( 2 − 6 )

0

= − [ 3 3 − 6

3 2]

0

= − [ 3 3 − 0

3 2]

6

= (0) − (72 − 108)

= 36 2

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DBM 20023 – ENGINEERING MATHEMATICS 2

EXERCISE:16

Find the area of the shaded 2.
region for each of the
following: = 2 − 2

1.

[ : 141 2]
2
[ : 4 2]
3

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DBM 20023 – ENGINEERING MATHEMATICS 2

3 . 4. = ( − 2)2
= 3 − 4

[ : 8 2] [ : 35 2]
3
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DBM 20023 – ENGINEERING MATHEMATICS 2

Area left along y-axis

Y-AXIS Y-AXIS

We consider the case where the curve at the
left along y - axis for the range of y values being
considered.

d

In this case, we find the area by simply finding the c
integral:

In this case, the integral gives a negative
number. Area cannot be a negative sign,
therefore the value becomes positive.



= ∫ ( ) = ∫ ( )



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DBM 20023 – ENGINEERING MATHEMATICS 2

Example: 4 Example: 5

Find the area of the region bounded by the Find the area of the bounded region by the
curve = 2 − 2
curve = √ − 1 , the y-axis and the line =
1 and = 5

In this case, we express as a function of : 2

= √ − 1 = ∫( 2 − 2 )
2 = − 1
= 2 + 1 0

= [ 3 3 − 2

2]

0

So, the area is given by: = [(233 − (2)2) − (0)]

5 = 8 − 4
3
= ∫( 2 + 1)

1

= [ 3 3 + 5 = − 4
3
]
= 4 2
1 3

136 (Area always positive)
3
= 2

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DBM 20023 – ENGINEERING MATHEMATICS 2

EXERCISE:17

Find the area of the shaded 2.
region for each of the
following:
2 = 2 − 4 + 3
1.

1

[ : 8 2] [ : 2 2]
3 3

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3 . DBM 20023 – ENGINEERING MATHEMATICS 2
= 2 − 1
4.

= ( − 2)2

2

2 0
0

−2

[ : 2.797 2] [ : 64 2]
3
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DBM 20023 – ENGINEERING MATHEMATICS 2

3.8.1 Calculate the graph area along x-axis or y-axis
(Bounded by a curve and a straight line)

On x-axis

= ( )

ab

= ( ) = ∫ ( ) − ( )



d On y-axis
c = ( )



= ∫ ( ) − ( )



= ( )

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DBM 20023 – ENGINEERING MATHEMATICS 2

EXERCISE:18

Find the area of the shaded 2. =
region for each of the
following:

1. = 2 + 6 + 2 -1

-3

-2

= (3 − )

= 2 − 1

[ : 4 2] [ : 4 2]
3 3

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DBM 20023 – ENGINEERING MATHEMATICS 2

3. 4.

= 4 + 16 = 1 2 − 3
2

= 2 2 + 10 = + 1

-1 3

[ : 64 2] [ : 18 2]
3

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DBM 20023 – ENGINEERING MATHEMATICS 2

3.8.2 Calculate the volume of a region which is generated through
° about the x-axis or y-axis (Bounded by a curve)

i. Bounded by a curve and rotating 3600 with x-axis
The volume ( ) of a solid generated by revolving the region bounded by = ( ) and
the − on the interval [ , ] about the − is:



= ∫ 2



ii. Bounded by a curve and rotating 3600 with y-axis
If the region bounded by = ( ) and the − on [ , ] is revolved about the − ,
then its volume ( ) is:



= ∫ 2



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DBM 20023 – ENGINEERING MATHEMATICS 2

Example: 1

Find the volume of the solid formed when the
b o u n d e d a r e a i s r o t a t e d t h r o u g h 3600 a b o u t x - a x i s



= ∫ 2



2

= ∫ (2 )

0

= [22 2]02

= [ 2]02

= [(2)2 − (0)2]

= 4 3

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DBM 20023 – ENGINEERING MATHEMATICS 2

Example: 2

Find the volume of the solid formed when the
b o u n d e d a r e a i s r o t a t e d t h r o u g h 3600 a b o u t y - a x i s

2 = − 1



= ∫ 2



2

= ∫ ( − 1)

1

= [ 2 2 − 2

]

1

= [((22)2 − 2) − ((12)2 − 2)]

= [0 + 12]

= 1 3
2

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DBM 20023 – ENGINEERING MATHEMATICS 2

EXERCISE:18

Find the volume of the solid formed when the bounded area
is rotated through 3600 about x-axis:

a) b) = 2 + 3

= 2 − 2

2
13

[ : 16 3 ] [Ans: 592 3]
15 5

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DBM 20023 – ENGINEERING MATHEMATICS 2

c) F i n d t h e v o l u m e o f t h e d) F i n d t h e v o l u m e o f t h e

solid formed when the solid formed when the

area between the curve area between the curve
= 3 a n d t h e x - a x i s
= 2 and the x-axis from f r o m = 1 a n d = 2
2 r o t a t e d t h r o u g h 3600

= −4 a n d = −1 r o t a t e d about x-axis.

t h r o u g h 3600 a b o u t x -

axis.

[Ans: 21 3 ] [Ans: 127 3]
16 7

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DBM 20023 – ENGINEERING MATHEMATICS 2

EXERCISE:19

Find the volume of the solid formed when the bounded area
is rotated through 3600 about y-axis:

a) b)

= 2 − 2 5

2 = 5 −

[Ans: 2 3] [Ans: 125 3]
3
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DBM 20023 – ENGINEERING MATHEMATICS 2

c) Find the volume of the d) F i n d t h e v o l u m e o f t h e

solid formed when the solid formed when the

area between the curve area between the curve
2 = 4 − a n d t h e y - a x i s
f r o m = −2 a n d y = 2 2 = and the y-axis
r o t a t e d t h r o u g h 3600 4

about y-axis. from y=0 and y=2

r o t a t e d t h r o u g h 3600

about y-axis.

[Ans: 512 3] [Ans: 1 3]
15 2

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DBM 20023 – ENGINEERING MATHEMATICS 2

3.8.2 Calculate the volume of a region which is generated through
° about the x-axis or y-axis
(Bounded by a curve and a straight line with illustrated graph)

On x-axis

= ( )

ab = ∫ ( ( )) − ( ( ))

= ( )

On y-axis
= ( )

d

= ∫ ( ( )) − ( ( ))



c = ( )

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DBM 20023 – ENGINEERING MATHEMATICS 2

EXERCISE:20

Find the volume of the solid formed when the bounded area
is rotated through 3600
1)

= 2 − 4 + 3

1 4

= − 1

[Ans: 555 3]

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DBM 20023 – ENGINEERING MATHEMATICS 2

2)

= 3√

=
4

[Ans: 512 3 ]
21

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DBM 20023 – ENGINEERING MATHEMATICS 2
3)

=

= 2 − 2

[Ans: 153 3 ]
5

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DBM 20023 – ENGINEERING MATHEMATICS 2
4)

= 2√ − 1
= − 1

[Ans: 96 3 ]
5

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