# Calculus 2 : Definite Integrals

## Example Questions

### Example Question #21 : Definite Integrals

Evaluate the indefinite integral .

Explanation:

The integral itself is not too difficult to take, simply use the Power Rule on the  and  terms. The trick is to be careful when integrating  is a constant value (about ) not a variable, so it must be integrated accordingly.

### Example Question #22 : Definite Integrals

Evaluate  .

Explanation:

This integral requires integration by parts followed by u-substitution. Here are the details

. Start

. Factor out the

Set up integration by parts with . We then have  and . Afterward, we use the integration by parts formula .

.

Now at this point we use u-substitution to evaluate the 2nd integral. Let , then  and therefore . Substituting into the integral we have

. (Don't forget to change the bounds of integration by plugging them into  for our equation for .)

### Example Question #23 : Definite Integrals

Evaluate .

Not possible without a calculator

Explanation:

This integral isn't possible to integrate directly using antiderivatives, but we can still find its value by noticing that is an odd function , and that our limits of integration are negatives of each other.

Hence

. (Since is an odd function)

### Example Question #24 : Definite Integrals

Evaluate

Explanation:

We can use u-substitution for this integral

. Start

Let , then , and our integral becomes

. (Don't forget to change the bounds of integration by plugging them into our equation for )

### Example Question #25 : Definite Integrals

Evaluate .

Explanation:

We proceed by using integration by parts.

. Start

Let , then we get
. Then using the integration by parts formula , we get

### Example Question #26 : Definite Integrals

Evaluate the following integral:

Explanation:

To evaluate this integral, we must integrate by parts, according to the following formula:

So, we must assign our u and dv, and differentiate and integrate to find du and v, respectively:

The derivative and integral were found using the following rules:

Note that we ignore the constant of integration.

Now, use the above formula:

Note that both the product of u and v and the integral are being evaluated from zero to

The integral was performed using the following rule:

Simplifying the above results, we get .

### Example Question #27 : Definite Integrals

Find the area between  and  between

Explanation:

We can write this problem as:

Integrating:

By the fundamental theorem of calculus:

### Example Question #21 : Definite Integrals

Explanation:

Compute the Indefinite Integral

Evaluate the integral

### Example Question #29 : Definite Integrals

Suppose , where  is a constant

Find  such that

Explanation:

By the fundamental theorem of calculus:

### Example Question #30 : Definite Integrals

Evaluate this integral.