### All Calculus 2 Resources

## Example Questions

### Example Question #21 : Finding Integrals

Find the integral of this equation:

**Possible Answers:**

**Correct answer:**

Distribute the polynomial first:

Then integrate:

And plug and solve:

### Example Question #22 : Finding Integrals

Evaluate the integral:

**Possible Answers:**

**Correct answer:**

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:

=

### Example Question #23 : Finding Integrals

**Possible Answers:**

**Correct answer:**

Simplify:

Integrate:

Evaluate:

### Example Question #24 : Finding Integrals

Find the integral:

**Possible Answers:**

**Correct answer:**

Integrate new equation:

Use natural log rules:

### Example Question #25 : Finding Integrals

Evaulate the integral:

**Possible Answers:**

**Correct answer:**

Simplify:

Seperate the terms:

Now integrate the simplified form:

### Example Question #26 : Finding Integrals

Evaluate the definite integral

**Possible Answers:**

**Correct answer:**

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 #11 : Definite Integrals

Evaluate the definite integral

**Possible Answers:**

DNE

**Correct answer:**

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 #12 : Definite Integrals

Evaluate the following definite integral:

**Possible Answers:**

**Correct answer:**

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:

And that's the final answer.

### Example Question #13 : Definite Integrals

Evaluate .

**Possible Answers:**

None of the other answers

**Correct answer:**

None of the other answers

The correct answer is .

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 #14 : Definite Integrals

Evaluate

**Possible Answers:**

None of the other answers

**Correct answer:**

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

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