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Substitution Rule for Definite Integrals
We now need to go back and revisit the substitution rule as it applies to definite
integrals. At some level there really isn’t a lot to do in this section. Recall that the
first step in doing a definite integral is to compute the indefinite integral and that
hasn’t changed. We will still compute the indefinite integral first. This means that we
already know how to do these. We use the substitution rule to find the indefinite
integral and then do the evaluation.
There are however, two ways to deal with the evaluation step. One of the ways of
doing the evaluation is the probably the most obvious at this point, but also has a point
in the process where we can get in trouble if we aren’t paying attention.
Let’s work an example illustrating both ways of doing the evaluation step.
Example 1
Evaluate the following definite integral.
Solution
Let’s start off looking at the first way of dealing with the evaluation step. We’ll need to be careful
with this method as there is a point in the process where if we aren’t paying attention we’ll get the
wrong answer.
Solution 1 :
We’ll first need to compute the indefinite integral using the substitution rule. Note however, that
we will constantly remind ourselves that this is a definite integral by putting the limits on the
integral at each step. Without the limits it’s easy to forget that we had a definite integral when
we’ve gotten the indefinite integral computed.
In this case the substitution is,
Plugging this into the integral gives,

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solution method. The limits given here are from the original integral and hence are values of
t
.
We have
u
’s in our solution. We can’t plug values of
t
in for
u
.
Therefore, we will have to go back to
t
’s before we do the substitution. This is the standard step
in the substitution process, but it is often forgotten when doing definite integrals. Note as well
that in this case, if we don’t go back to
t
’s we will have a small problem in that one of the
evaluations will end up giving us a complex number.
So, finishing this problem gives,

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