# Calculus 2 : Average Values and Lengths of Functions

## Example Questions

### Example Question #21 : Average Values And Lengths Of Functions

You will need a calculator for this question.

Find the surface area of revolution formed by the function, , from x=0 to x=2, when revolved around the y axis. Round to the nearest hundredth.

Explanation:

To find the surface area of a revolution, we must use the following formula.

, where "r" is the variable radius of the curve from the axis of revolution, and "ds" is the differential of the arc length of the curve at that radius. For this problem, "r" is simply the variable "x", because every point's radius from the axis of revolution (the y axis), is its x coordinate.

You can think of "ds" as the arc length formula after it is differentiated. Arc length is an integral, so the derivative of an integral is just what's inside that integral. Here that means .

"a" and "b" will be the x=0 and x=2 part of the question.

From this information we can rewrite the surface area formula as

Now we will find f'(x).

Plug  in to the formula.

Now simplify .

Notice that the is part of the derivative of what is under the radical. This almost follows the basic integral form ,with , and . Let's force it to match the form exactly. First I will rewrite the square root as an exponent.

Now multiply by a form of 1 (highlighted in red) to force the "du" part of the integration form.

It may help to rewrite the expression in terms of to make it easier to apply integration rules. We just need to remember to convert back to x after integrating, before we plug in the bounds.

Now integrate.

Now convert u back to x. Recall that

Now plug in the upper and lower bounds.

Now use a calculator to find the approximate answer of

### Example Question #21 : Average Values And Lengths Of Functions

Evaluate the average value of the function on the interval .

Explanation:

To solve for the average value on the interval  we follow the following formula

For this problem we evaluate

### Example Question #23 : Average Values And Lengths Of Functions

Evaluate the average value of the function on the interval .

Explanation:

To solve for the average value on the interval  we follow the following formula

For this problem we evaluate

### Example Question #24 : Average Values And Lengths Of Functions

Find the average value of  on

Explanation:

Recall the formula for average value of a function, f(x), on a closed interval, [a,b], is:

For this we first need to integrate our function on the defined interval:

Plugging this into our formula we attain

### Example Question #25 : Average Values And Lengths Of Functions

Find the average value of the function  over the interval

Explanation:

To find the average value, we need to evaluate .

This integral can be evaluated using -substitution.

Then we can proceed as follows

(Start)

(Factor out a , leaving a  in the numerator)

(Substitute the equations for )

(Integrate, recall that )

(Substitute  back in)

Now we have

.

### Example Question #26 : Average Values And Lengths Of Functions

Find the average value of the function

on the interval

Explanation:

To solve for the average value on the interval  we follow the following formula

For this problem we evaluate

### Example Question #22 : Average Values And Lengths Of Functions

Find the average value of the function

on the interval

Explanation:

To solve for the average value on the interval  we follow the following formula

For this problem we evaluate

### Example Question #163 : Integral Applications

Find the average value of  on .

Explanation:

In order to find the average value we must solve

evaluated between 1 and e.

### Example Question #29 : Average Values And Lengths Of Functions

Find the average value of the function

on the interval

Explanation:

To solve for the average value on the interval  we follow the following formula

For this problem we evaluate

### Example Question #171 : Integral Applications

Find the average value of the function

on the interval