Page 80 - DMTH202_BASIC_MATHEMATICS_II

Unit 5: Definite Integrals by Substitution

5 4t Notes
(b) 3  2 8t 2 dt

Solution:

5 4t
(a) 5  2 dt

 2 8t
Be cautious with this integral. The denominator is zero at and both of these are in the gap
of integration. Thus, this integrand is not constant in the interval and so the integral can’t
be completed.
Be cautious with definite integrals and be on the lookout for division by zero problems.
In the preceding section they were simple to spot as all the division by zero problems that
we had there were at zero. Once we go into substitution problems though they will not
always be so easy to spot so confirm that you first take a rapid look at the integrand and
observe if there are any continuity problems with the integrand and if they take place in
the interval of integration.
5 4t
(b) 3  2 8t 2 dt

Now, here the integral can be completed since the two points of discontinuity, x   1 2 , are
both outside of the interval of integration. The substitution and converted limits here are,
1
2
u  2 8t du   16t dt  t dt   dt

16
t  3  u   70 t  5  u  198
The integral is then,

5 4t 4  198 1
3  2 8t 2 dt   16  70  u du

1  198
  In u
4  70
1
  In 198  In 70  
4

Example: Evaluate each of the following:
In(1+ ) x x

(a) 0  e cos 1 e dx 
6  4
e In t
(b) 2  dt
e t
 sec 3P tan 3P 
(c)   9 3 dP

12 2 sec 3P 

(d)   2 cos  cos sinx    dx
x
2
2 e w
(e) 1  2 dw
50 w
Solution:
In 1+ 
(a) 0  e x cos 1 e dx x 

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