Be Careful: There are two ways to use substitution to evaluate definite integrals. When evaluating a definite integral, make sure you know which way you're using them.
Way 1: First integrate the indefinite integral by substitution.
We know how to use substitution to find indefinite integrals. This means we know how to use substitution to find antiderivatives.
To use substitution to find an antiderivative, we first use substitution to find the indefinite integral. Then we drop the + C, which is the same thing as setting C = 0. The indefinite integral is an infinite family of antiderivatives, and setting C = 0 gets us one antiderivative out of that infinite family.
Way 2: Change the variables, and never go back.
The other way to evaluate definite integrals by substitution is to change the variable of integration and also bf change the limits of integration to match the new variable. With this method, we never have to put the original variable back in.
Here are the two steps to follow when integrating a definite integral via Way 2:
- Do the substitution in the definite integral, remembering to change the limits of integration.
- Use the FTC to integrate the resulting definite integral.
Going back to definite integrals, the Fundamental Theorem of Calculus says that we can evaluate the definite integral

by finding an antiderivative F of f and taking the difference
F ( b ) – F ( a )
which is abbreviated by
.
So long as we can use substitution on the integrand, we can use substitution to evaluate the definite integral.
There are two steps:
1. Use integration by substitution to find the corresponding indefinite integral. This gets us an antiderivative of the integrand.
2. Use the FTC, with the antiderivative from (1), to find the definite integral.
Remember, first you do substitution to get a new definite integral, with a new variable of integration and new limits of integration. Then you evaluate the new definite integral using the FTC.
Way 1 or Way 2? Help!
There are two obvious questions that might be bouncing around in your head right now.
- "Should I use Way 1 or Way 2?"
- "How do I know when to use Way 1 and when to use Way 2?"
There's good news and bad news...no, actually there's just good news. Either way will work. As long as you know which way you're using and do it correctly, you'll get the right answer. To test this, work out these integrals using Way 2 and work out these integrals using Way 1. You should get the same answers you got before. So the easiest answer to both questions is "Use whichever way you like better."
It's true that some definite integrals may be shorter with one way than with the other. The best way to figure out when one way is shorter than the other is to do a bunch of practice problems.
Practice:
Find an antiderivative of 3(3x + 1)5. | |
First we find the indefinite integral of 3(3x + 1)5. Let u = 3x + 1 and u' = 3. Then 
This is the family of all antiderivatives of 3(3x + 1)5. To get one antiderivative, we pick a value of C. The simplest choice is C = 0. This gives us the antiderivative 
| |
Evaluate . | |
First we use integration by substitution to find the corresponding indefinite integral. We did the work for this in a previous example: 
This means is an antiderivative of 3(3x + 1)5. | |
Evaluate . | |
First we use substitution to evaluate the indefinite integral .
Take u = 4x2 + 1 u' = 8x. We need to introduce a factor of 8 to the integrand, so we multiply the integrand by 8 and the integral by . 
Letting C = 0, the simplest antiderivative of the integrand is .
We use this antiderivative in the FTC. 
If we're going to give an exact answer (which is advisable), it doesn't get any nicer than this. | |
Evaluate . | |
Since the variable of integration is x, the limits of integration are values of x. The integral 
is read "the integral of 3(3x + 1)5 as x goes from 0 to 1." To be explicit, we could write the integral as 
If we take u = 3x + 1 du = 3dx and start the substitution, we run into a problem: 
This could be read " the integral of u5 as x goes from 0 to 1," which doesn't make sense! The problem is that the limits of integration are values of x, but now the variable of integration is u. We can fix this by changing the limits of integration to values of u. We know that u = 3x + 1. So when x = 0, the corresponding value of u is u = 3(0) + 1 = 1. This lets us change the lower limit of integration from a value of x to a value of u: 
When x = 1 we have u = 3(1) + 1 = 4. This lets us change the upper limit of integration: 
Now that we have everything in terms of u, there's no reason to go back to x. We can evaluate the integral we have, with no x in sight: 
It's reassuring that we got the same answer as when we did this problem the other way! | |
Use Way 2 to evaluate 
| |
We take u = 4 – x du = (-1)dx Let's label the limits of integration as x-values so we don't mess up. 
We're not done with the substitution yet. We still have to change the limits of integration so we have u-values instead of x-values. When x = 2, u = 4 – (2) = 2. When x = -1, u = 4 – (-1) = 5. Here's what the integral looks like after the limits of integration are changed: 
Now that we've changed the limits of integration, we're done with the substitution. From here, we can use the FTC to integrate. 
We conclude that 
Be Careful: When using substitution to evaluate a definite integral, you aren't done with the substitution part until you've changed the limits of integration. | |
A test contained the following question:

Laurie wrote down the following answer:

What did Laurie do wrong? What is the correct value of the integral?
Answer
When finding the indefinite integral, Laurie forgot to put back the original variable. She should have written

When we use the FTC with the correct antiderivative, -ln|4 – x|, we get

The moral of Laurie's story is to keep the two steps separate when evaluating a definite integral using Way 1. First, evaluate the indefinite integral. Remember to put the original variable back in. Then, after you're all done with the indefinite integral, use the FTC. Keep your calculations for the two steps separate so you don't get mixed up!
Evaluate the definite integral using Way 1(first integrate the indefinite integral, then use the FTC).

Answer
First we work out the indefinite integral. We have

So

Now we use the FTC with the antiderivative

Here goes:

Evaluate the definite integral using Way 1(first integrate the indefinite integral, then use the FTC).

Hint
Rewrite the integrand.
Answer
Following the hint, we can rewrite the integrand:

This is a little easier to work with. First we evaluate the indefinite integral using substitution. Take
u = 3x + 4, so u' = 3. Then

Now we use the FTC with the antiderivative
.

Evaluate the definite integral using Way 1(first integrate the indefinite integral, then use the FTC).

Answer
Using substitution, we find that

Now we use the FTC.

Evaluate the definite integral using Way 1(first integrate the indefinite integral, then use the FTC).

Answer
Using substitution, we find that

We use the FTC:

Evaluate the definite integral using Way 1(first integrate the indefinite integral, then use the FTC).

Answer
First we figure out the indefinite integral. Take u = ln x. Then
. So

Using the FTC, we get

A test contained the following question:

KT wrote down the following answer:

What did KT do wrong? What is the correct value of the integral?
Answer
KT took u = x + 1, so du = dx.
She neglected to change the limits of integration when doing the substitution. The limits 1 and 2 are values of x, not values of u. She wrote this:

Which means this:

However, KT continued as though it meant this, which is not true:

To find the correct value of the integral, we need to finish the substitution. Since u = x + 1, we have
x = 1 arrow u = 2
x = 2 arrow u = 3
Here's the integral with the limits of integration fixed:

Now we can use the FTC to evaluate the integral.

We conclude

Evaluate the definite integral by substitution, using Way 2.

Answer
First we need to do the substitution. Take

Then

So

Now we can use the FTC on this simpler integral.

That's an exact answer, so we leave it like that. We conclude

Evaluate the definite integral by substitution, using Way 2.

Answer
Take
u = -x
du = (-1)dx
Then

We still need to change the limits of integration. Since u = -x, this part is pretty simple.

So

Now we can use the FTC. The function sec2 u is the derivative of tan u, so

We conclude

Evaluate the definite integral by substitution, using Way 2.

Answer
First we do the substitution. Take
u = (x3 – 3x + 2)
du = (3x2 – 3)dx
= 3(x2 – 1)dx
Then

To change the limits of integration we have to do a little work this time. When x = -2, we have
u = (-2)3 – 3(-2) + 2 = 0.
When x = 0 we have
u = (0)3 – 3(0) + 2 = 2.
We fix the limits of integration:

Now we use the FTC:

Evaluate the definite integral by substitution, using Way 2.

Answer
We do the substitution first, remembering that this includes changing the limits of integration. Take

We introduce a factor of 4 to the integrand in order to start the substitution.

We still need to change the limits of integration. When
we have

and when x = π we have

After we change the limits of integration, the substitution is complete:

Now we can use the FTC to evaluate the integral.

We conclude

Evaluate the definite integral by substitution, using Way 2.

Answer
We take

When x = e2 we have
u = ln e2 = 2
and when x = e3 we have
u = ln e3 = 3.
Let's do the substitution all at once:

Now we use the FTC:

We conclude
