Page 78 - DMTH202_BASIC_MATHEMATICS_II

Unit 5: Definite Integrals by Substitution

 6 4 5 Notes
(b) 2  3  dx


1 2x  1 2x
1
y
(c) 0  2 e  2cos   y dy
0  z 
(d)   3sin    5cos    z dz
3  2 
Solution:

Because we’ve done performed a few substitution rule integrals, here we aren’t going to put a lot
of exertion into elucidating the substitution part of things here.
5 2 5
(a) 1  1 w  2w w  dw

The substitution and converted limits are,

1
u  2w w 2 du  2 2w dw   1 w dw   dw

2
w   1  u   1 w 5  u 35
At times a limit will stay similar after the substitution. Don’t get thrilled when it takes
place and don’t imagine it to occur all the time.
Here is the integral,

5 2 5 1 35 5
1  1 w 2w w  dw  1  u du
 2 
1 35
 u 6  153188802
2  1
 6 4 5
(b)  2  1 2x  3  1 2x dx

Here is the substitution and converted limits for this problem,

1
u  1 2x du  2dx  dx  du

2
x   2  u   3 x   6  u  11
The integral is then,
 6 4 5 1  11  3 5
2  3  dx  3  4u  du

  1 2x 2  u
1 2x
1  11
  2u  2  5 In u 
2 3 
1  2  1  2 
    5 In 11     5 In3


2  121  2  9 
112 5 5
  In 11  In 3
1089 2 2
1
y
(c) 2 e  2cos   y dy
0 
This integral requires to be divided into two integrals as the first term doesn’t need a
substitution and the second does.

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