# Calculus 2 : Definite Integrals

## Example Questions

### Example Question #11 : Definite Integrals

Find the integral of this equation:

Explanation:

Distribute the polynomial first:

Then integrate:

And plug and solve:

### Example Question #12 : Definite Integrals

Evaluate the integral:

Explanation:

Factor the polynomial:

Then integrate (To integrate, add 1 power of each term. So . Then divide the term by the value of the exponent. .

Then evaulate the integral:

Explanation:

Simplify:

Integrate:

Evaluate:

### Example Question #14 : Definite Integrals

Find the integral:

Explanation:

Integrate new equation:

Use natural log rules:

### Example Question #15 : Definite Integrals

Evaulate the integral:

Explanation:

Simplify:

Seperate the terms:

Now integrate the simplified form:

### Example Question #11 : Definite Integrals

Evaluate the definite integral

Explanation:

Because integration is a linear operation, we are able to anti-differentiate the function term by term.

We use the properties that

• The anti-derivative of    is
• The anti-derivative of    is

to solve the definite integral

And by the corollary of the Fundamental Theorem of Calculus

### Example Question #372 : Integrals

Evaluate the definite integral

DNE

Explanation:

We use the property that

• The antii-derivative of    is

to solve the definite integral

And by the corollary of the Fundamental Theorem of Calculus

And since

the definite integral becomes

### Example Question #381 : Integrals

Evaluate the following definite integral:

Explanation:

is an integral of a rational function, so we need to make the denominator into one of the basic rational functions we know the antiderivatives of and then split it up into multiple parts if necessary.

The most obvious thing to do then, is to substitute :

So our new integral is:

which can then be split up into two pieces:

both of these are simple:

### Example Question #12 : Definite Integrals

Evaluate .

Explanation:

Instead of evaluating the integral directly using antiderivatives, it is much eaiser to notice that the bounds of integration are negatives of each other, and that  is an odd function

(in other words,)

Hence

. Since  is odd.

### Example Question #383 : Integrals

Evaluate

Explanation:

We use integration by parts for this integral.

Let , and , then taking the derivative of  and the antiderivative of , we get , and .

Using the integration by parts formula, , we have

. Start

. Use the integration formula. (do not evaluate the left part yet).

. Evaluate the last integral, and begin to evaluate the entire expression for the antiderivative.

. Evaluate

. Use rules of logs

. Simplify