Page 76 - DMTH202_BASIC_MATHEMATICS_II

Unit 5: Definite Integrals by Substitution

Solution 1: Notes
Firstly, we calculate the indefinite integral by means of the substitution rule.
Here, the substitution is,
1
2
2
u  1 4t 3 du   12t dt  t dt   du

12
Plugging this into the integral provides,
0 2 3 1 0 1 2
2  2t 1 4t dt   2  u du

 6 
0
3
1
  u 2
9
 2
Observe that we didn’t perform the evaluation yet. This is where the latent problem occurs with
this solution method. The limits specified here are from the original integral and thus are
values of t. We have u’s in our solution. We can’t plug values of t in for u.
Thus, we will have to get back to t’s before we perform the substitution.

Did u know? Getting back to t’s before performing the substitution is the standard step in
the substitution procedure, but it is frequently forgotten when performing definite
integrals.
Note also that here, if we don’t move back to t’s we will have a little problem in that one of the
evaluations will finish up providing us a complex number.

So, concluding this problem provides,

0 2 3 1 3 2 3 0
2  2t 1 4t dt   1 4t 

 9
 2
1  1 3 
33
      2 

9  9 
1
  33 33 1 
9
Thus, that was the first solution method. Now let us observe the second method.
Solution 2:

Notes This solution method isn’t actually all that dissimilar from the first method. In this
method we are going to keep in mind that when performing a substitution we want to
remove all the t’s in the integral and write all in terms of u.
When we say all here we actually mean all.

!

Caution Keep in mind that the limits on the integral are also values of t and we’re going to
translate the limits into u values.
Converting the limits is quite simple as our substitution will tell us how to relate t and u therefore
all we require to do is plug in the original t limits into the substitution and we’ll obtain the
new u limits.

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