# How to Change Limits of Definite Integrals Video

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• 0:04 Expressing the Same Idea
• 0:38 U-Substitution
• 1:38 Example One
• 3:48 Example Two
• 5:07 Trigonometric Substitution
• 8:16 Lesson Summary
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Lesson Transcript
Instructor: Carmen Andert

Carmen has two master's degrees in mathematics has has taught mathematics classes at the college level for the past 9 years.

In this lesson, we will discuss changing the limits of definite integrals. This is useful when using u-substitution or trigonometric substitution to solve definite integrals and saves time as there is no need to revert back to the original variable.

## Expressing the Same Idea

You are new to town and ask a couple friends for directions to the park. One tells you to drive north for 2 miles, then drive west for 1 mile. The other tells you to drive north for about 4 minutes, then turn left and drive for about 2 minutes. Both friends get you to the park. Is one way better than the other? No! Some people prefer to measure precise distances, and others are better off with time. In a similar manner, the idea of changing the limits of definite integrals when substituting is just expressing the same problem in a different way.

## U-Substitution

When evaluating an integral using u-substitution, an expression involving the original variable is replaced by a new variable. The end result is a simpler integral.

When we use u-substitution to evaluate a definite integral, we must change each part of the integral to use the new variable (usually u) instead of the old variable (often x). There are three pieces that must be changed:

1. The function itself (change an expression involving x into u)
2. The differential (change dx into an expression involving du)
3. The bounds (change the bounds of integration from values for x into values for u)

Hopefully, the first two are familiar to you. We will focus on the third step. To change the bounds, use the expression that relates x and u. Plug in the original lower bound for x and solve for u. This gives the new lower bound. Then plug in the original upper bound for x and solve for u to find the new upper bound.

## Example One

Evaluate the definite integral âˆ« âˆš(4x+1)dx from 0 to 2.

What do you do first? As with most definite integrals, you should ignore the bounds (0 and 2) at first and focus on how to find an antiderivative of the function inside the integral. We need an antiderivative of âˆš(4x+1).

This one is simple enough that you might be able to do it without using u-substitution, but especially for beginning students, it might be easier to substitute u = 4x + 1. Then, take the derivative to see that du = 4dx, and solve for dx to get dx = 1/4 * du.

Here is the new part. We have changed all the expressions involving x to expressions involving u. Remember that the bounds 0 and 2 on the integral were bounds for x, not u! So now we can change those also.

Remember that u = 4x + 1. That means that when x = 0, u = 4(0) + 1 = 1, and when x = 2, u = 4(2) + 1 = 9. So, while the old bounds for x were 0 and 2, the new bounds for u are 1 and 9. Putting all these pieces together, the original integral is rewritten as follows:

âˆ« âˆš(4x+1)dx from 0 to 2 = âˆ« âˆšu * 1/4 * du from 1 to 9.

Do you see how everything that talked about x (the x itself, the dx, and the bounds) has been changed into talking about u instead? That means that we can now forget about x completely and just evaluate the integral using the variable u.

Finishing up, an antiderivative for âˆšu is 2/3*u3/2, so we can finish:

• âˆ« âˆšu * 1/4 * du from 1 to 9
• = 2/3 * u3/2 * 1/4 evaluated from 1 to 9
• = 1/6(93/2 - 13/2)
• = 1/6(27 - 1)
• = 13/3

## Example Two

Evaluate the definite integral âˆ« sin3x cosx dx from Ï€/3 to Ï€/2.

You should notice that the function sin3x cosx is pretty complicated, so it will be helpful to try a u-substitution to simplify the expression.

Let's use u = sinx. Then du = cosx dx, so the integral now has u3 du in it, and all that remains is to change the bounds.

Since u = sinx, when x = Ï€/3, u = sin(Ï€/3) = âˆš(3)/2. Also, when x = Ï€/2, u = sin(Ï€/2) = 1. Our integral is rewritten as follows:

• âˆ« sin3x cosx dx from Ï€/3 to Ï€/2
• = âˆ« u3 du from âˆš(3)/2 to 1

We now solve it as usual, using the variable u and forgetting that the variable x ever existed.

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