INTEGRATION IN MATHEMATICS ENGINEERING 2(DBM20023)

DBM2013 Engineering Mathematics 2 Chapter 3 Integration

Solve the following integral.

1.  (x 4x − 2) dx
+1)(2x

2.  1 dx
x2 (1 − 4x)

Solution:

1.  (x 4x − 2) dx
+1)(2x

4x = A + B Proper fraction :
(x +1)(2x − 2) (x +1) (2x − 2) Linear Factor

4x = A(2x − 2) + B(x +1)
When x = −1,
4(−1) = A(2(−1) − 2) + B(−1+1), so A = 1

When x = 1,
4(1) = A(2(1) − 2) + B(1+1), so B = 2.

Substitute the value of A and B. Integral of reciprocal
 4x = 1 + 2 function

(x +1)(2x − 2) (x +1) (2x − 2)

 4x dx =  1+ 2 dx
 (x +1)(2x − 2) (x +1) (2x − 2)

=  (x +1) −1 dx + 2 (2x − 2) −1 dx

= ln (x +1) + 2 ln (2x − 2)  + c
1 2 

= ln (x +1) + ln (2x − 2) + c

2.  1 dx
x2 (1 − 4x)

Proper fraction:
Repeated Linear Factor

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

1 = A+ B + C
x 2 (1 − 4x) x x 2 1 − 4x

1 = A(x)(1 − 4x) + B(1 − 4x) + C(x 2 )

When x = 0 , B = 1

When x = 1 , C = 16
4

Expand :1 = Ax - 4Ax2 + B - 4Bx + Cx 2

Compare x : 0 = A - 4B, B = 1

Then, A = 4

 1 = 4 + 1 + 16
x 2 (1 − 4x) x x 2 1 − 4x

 1 dx =  4 dx +  1 dx +  16 dx
x2 (1− 4x) x x2 1− 4x

  = 4 x−1 dx + x−2dx +16 (1− 4x)−1 dx

= 4 ln x − x −1 + 16 ln (1− 4 x)  + c
−4 

 1 dx = 4 ln x − 1 − 4 ln (1 − 4x) + c
x2 (1− 4x) x

EXERCISE 9: Page 180

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

Solve the following integral.

a)  x x +1 dx b)  (3x x+3 + 1) dx
(x − 2) + 2)(x

c)  6 dx d)  3x − 5 dx
x2 −1 x2 − x − 2

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

e)  x (x 10 + 3) dx f) 3x − 2 dx
+1) (x x3 − x2

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

g)  (x x+3 + 7) dx  x2 + x −1
+ 1)2 (x
h) x (x2 −1) dx

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

3.10 Techniques of integration

3.10.1 Area of the graph by using integration along x-axis or y-axis.

Area between a curve and the x-axis Area between a curve and the y-axis

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration
y
y
b

ab x a x
Formula: = ∫ ( ) Formula: = ∫ ( )

Example 14: Page 185
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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

Find the area bounded by the graph and x-axis b) y
Solution:

y
a)

x 13 x

12 Fig 12.2b

Fig 12.2a 2

a) Shaded area = 12 x2 dx =  x3 1 = 8−1 = 7 unit2
2  6 66 6


1  3 3
2  
( )b) 3  2x 2  =4
Shaded area = 1  3  3 27 −1 unit2 = 5.59 unit2
2x dx =

 2 1

Example 15:

Find the area bounded by the graph of y = x2 and the x-axis on [0, 2]

Solution:

b

We have A= f (x)dx , then
a

2 2 x 2 dx

f (x)dx =
 A =
00

= x3 2 = 8 unit2

 3  0 3

Example 16:

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

Find the area bounded by the graph of y = x2 + 2x and the x-axis on [-2, 2]

Solution:

Area of region A Area of region B

AA = 0 (x2 + 2x) dx AB =2 (x2 + 2x) dx
−2 0

=  x3 + 2x2 0 =  x3 + 2x2 2
 2   2 
 3  −2  3 0

 (0) 3   (−2) 3    (2) 3   (0) 3 
  3     3 
=  + (0) 2 −  + (−2) 2 =  + (2) 2 −  + (0) 2
 
3   3  

= −4 = 20 unit 2
3 3

= 4 unit 2
3

Therefore, total area bounded by the graph is,
AT = AA + AB
= 4 + 20
33
= 8 unit 2

Example 17: Page 187

Find the area of the shaded region.
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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

1. 2
2.
Area , A = 1 f ( y) dy

= 2 y 2 dy
1

=  y3 2
 3 
 1

 (2) 3   (1) 3 
 3   3 
=   −  

=8−1
33

= 7 unit 2
3

2

Area , A = f ( y) dy
1

= 2 y2 − 4 y + 3) dy

(
1

 y3 4y2  2
 3 2 
=  − + 3 y

1

=  (2) 3 − 2(2)2 +  −  (1) 3 − 2(1) 2 + 
 3(2)  3(1)
 3   3


=  2  −  4 
3   3

= −2
3

= 2 unit 2
3

Find the area of the shaded region. Page 188
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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

Solution:

To find the area of shaded region, we have to sum of the two regions under different
functions.

 Area , A = 1 x2 dx + 2 (−x + 2) dx
01

=  x3 1 + − x2 + 2
   2 2x
 3 0
1

=  (1)3  −  (0)3  +  − (2)2 + 2(2)  − − (1)2 + 2(1) 
3 3 2 2

=1+1
32

= 5 unit 2
6

EXERCISE 10 Page 189
3441:

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

Find the area of the shaded region in each graph below.
1. 2.

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

3. 4.

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

5. 6. Find the area bounded by the graph of
y = sin x and the x-axis on [0,2π]

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

EXERCISE 11
121213123441:

Find the area of the shaded region.
1.

2. Page 193
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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

3.

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

4. Find the area bounded by 2 = − 2 from = 0 and = 3.

EXERCISE 12
3441:

Find the area of the shaded region.

1.

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration
2.

3.10.2 Volume of Revolution
3.10.2.1 Volume generated when the bounded region is revolved along x-axis

Generated Volume, V = b y2 dx
a

Example 19:
Find the volume of the object below revolved along x-axis.

a) b)

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

Solution:

a) V = b y2 dx b)V = 2  x2 2 dx
a 2
1
V = 2 2x dx
0 V = 1  2 x4 dx
41
 2x 2  2
 2  0
=    x5 2
5 
 =  (2)2 − (0)2 = 1   1
4 


= 4 unit 3  = 1  (2)5 − (1)5
20

= 31  unit 3
20

3.10.2.2 Volume generated when the bounded region is revolved along y-axis

Generated Volume, V = b x2 dy
a

Example 20:

Find the volume of the object below revolved along y-axis.
a) b)

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

Solution:

a) V = b x2 dy b) V = 3 (3 − y) dy
a 0
V = 2 y + 1dy
0 V =  − y2 3
3 y 2 
 y2 2  0
 2 y
=   + 0  3(3) (3) 2   3(0) (0) 2 
2 2
=  − − −

 (2) 2   (0) 2 + (0)
2 2
=  + (2) − = 9  unit 3
2
= 4 unit 3

EXERCISE 13
3441:

1. Find the generated volume when these shaded regions are rotated 3600 around the x-axis
a) b)

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

2. A region bounded by a curve y2 = x + 1, y-axis and a line x = 4 is rotated 360o around the
x-axis. Find the generated volume in .

3. Find the generated volume when these shaded regions are rotated 3600 around the y-axis.
a) b)

0
-2

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

4. A sphere is formed when the area bounded by a curve y = 4 − x 2 and the x-axis as shown
below is rotated around the x-axis. Find the volume of the sphere if the radius is 2.

5. Find the generated volume when this shaded region are rotated through 360o around:

a) x-axis b) y-axis

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

6. Find the generated volume when the shaded region bounded by a straight line y = x , a curve
y2 = x − 2 , a straight line x = 4 and x-axis is rotated 360o around the x-axis.

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