### All AP Calculus AB Resources

## Example Questions

### Example Question #41 : Antiderivatives By Substitution Of Variables

Evaluate the following indefinite integral:

**Possible Answers:**

**Correct answer:**

First we distribute the which will allow us to break the integral up into two parts.

.

For the first integral, we use substitution of variables. Letting , we find that or that . Substituting these two equations in, we have

At this point, you can use the formula for integrating exponentials, or just recall the simple derivation for it

The second integral is simple power rule.

Combining these two expressions and substituting back in for u, we get a final answer of

Note, if you didn't distribute the at the beginning, this is fine. Substitution of variables will still work correctly. Just be careful that when you integrate you remember that this becomes u, and not x. The extra -8 you get when doing this can just be combined with the C term.

### Example Question #41 : Antiderivatives By Substitution Of Variables

Integrate:

**Possible Answers:**

**Correct answer:**

To integrate, the following substitution was made:

Now, rewrite the integral in terms of u and integrate:

The following rule was used to integrate:

Finally, rewrite the answer in terms of x:

### Example Question #86 : Techniques Of Antidifferentiation

Integrate:

**Possible Answers:**

**Correct answer:**

To integrate, we must make the following substitution:

Rewriting the integral in terms of u and integrating, we get

The integral was performed using the identical rule.

Finally, replace u with the original x term:

### Example Question #41 : Antiderivatives By Substitution Of Variables

Calculate the following integral:

**Possible Answers:**

**Correct answer:**

Use substitution to solve the integral

Rewrite the integrand as a negative exponent:

Therefore:

Make the following substitution:

Apply the substitution to the integrand:

Solve the integral:

Re-substitute the value of u:

Solution:

### Example Question #41 : Antiderivatives By Substitution Of Variables

Calculate the following integral:

**Possible Answers:**

**Correct answer:**

Solve by u substitution

Make the following substitution:

Apply the substitution the integral:

Solve the integral:

Re-substitute the value for u:

### Example Question #42 : Antiderivatives By Substitution Of Variables

Integrate:

**Possible Answers:**

**Correct answer:**

To integrate, we can split the integral into the sum of two separate integrals:

We can solve them independently and add the results together.

To solve the first integral, we make the following substitution:

Rewriting the integral in terms of u and integrating, we get

The integral was found using the identical rule.

Rewriting the integral in terms of x, we get

We solve the second integral using the following substitution:

Rewriting the integral in terms of u, we get

The integral was performed using the identical rule.

Now, rewrite the answer in terms of x:

Combining the results from each integral, we get

Note that the two constants of integration combine to make a single constant.

### Example Question #43 : Antiderivatives By Substitution Of Variables

Integrate:

**Possible Answers:**

**Correct answer:**

To integrate, the following substitution must be made:

Rewriting the integral in terms of u and integrating, we get

The integration was performed using the following rule:

Finally, we rewrite our answer in terms of x:

### Example Question #91 : Techniques Of Antidifferentiation

Evaluate the integral

**Possible Answers:**

**Correct answer:**

Evaluating integrals requires knowledge of the basic integral forms. In this problem, there is a group raised to an exponent, which is the .

This arrangement typically follows the basic integral form , where is a variable expression, is some constant number and is the constant of integration, which stays unknown. In this case, we can use u-substitution to match this form.

The exponent is 5, so the in the basic integral form will be5.

Since the inside of the exponent group is , set .

Now we differentiate to find . Recall that the derivative of .

Notice that the parts of our match up with the integral we are evaluating, and there are no variables that aren't accounted for. This is clearer if we write out and simplify what our substitution says.

This is exactly what we are asked to integrate.

This means we can evaluate the integral by using basic integral form directly. Plugging in our and into the right side of , we get

This is equivalent to the answer , which is the correct answer.

### Example Question #49 : Antiderivatives By Substitution Of Variables

Evaluate the integral

**Possible Answers:**

**Correct answer:**

The first thing to recognize about this integral is that the derivative of is . is in the denominator and is in the numerator. This arrangement usually results in the natural log, , of the bottom. The basic integral form is as follows

This means we need to set the denominator as in our substitution.

Now we find by differentiating . Remember that the derivative of is . This derivative involves chain rule and power rule.

This is similar to the numerator. We have the variables but the is not present in the given integral. Fortunately, is a constant and can be corrected by dividing our equation by on both sides. Doing so yields

This adjustment means that when we plug our information into the basic integral form, we will write instead of just . Doing so, we get

The is a constant, and thus can pass to the left of the integral sign.

From the basic intergral form, the is equal to . So we replace just the integral with the equivalent answer, but we leave the .

Plugging in 's substitution, we get

The last thing we do is distribute the to the natural log and the constant, .

The constant, , is unknown, so when we multiply it by , it is still unknown, so we leave it as .This yields

, which is the correct answer.

### Example Question #50 : Antiderivatives By Substitution Of Variables

Evaluate the integral, .

**Possible Answers:**

**Correct answer:**

To evaluate this integral requires the basic integral form, .

In this form, the is the exponent of . For our problem, this exponent is . Thus, we will set for our u-substitution.

Now we need to find and verify that it contains all the other variables in our integral. Finding the derivative of will require the Chain rule. Differentiating , we get

This matches all the variables not accounted for by the , but our has an extra factor of 4. We address this by dividing both sides by 4 to match what we have in our integral.

Now we can rewrite the original integral using variables.

Since the is a constant, we can pass it to the left of the integral sign. Then we will apply the basic integral form to the remaining integral.

When we distribute the to the inside of the group, remember the the will still be unknown, so it stays as is.

Finally, we "un-substitute" our for what it equals originally, .

This is the correct answer.

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