# Calculus 2 : Average Values and Lengths of Functions

## Example Questions

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### Example Question #1 : Average Values And Lengths Of Functions

What is the arc length if  from ?

Explanation:

Write the formula for arc length.

Calculate the derivatives.

Substitute the derivatives and the bounds into the integral.

### Example Question #1 : Average Values And Lengths Of Functions

What is the average value of the function

from  to

Explanation:

The average value of a function p(t) from t=a to t=b is found with the integral

In this case, we must compute the value of the integral

.

A substitution makes this integral clearer. Let . Then . We should also rewrite the limits of integration in terms of u. When t = 0, u=1, and when t = 2, u = 5. Making these substitutions results in the integral

Evaluating this integral using the fact that

yields

### Example Question #1 : Average Values And Lengths Of Functions

What is the average value of the function  over the interval ?

Explanation:

In general, the average value of a function  over the interval  is

This means that the average value of  over  is

.

Since the antiderivative of  is , the integral evaluates to

.

### Example Question #4 : Average Values And Lengths Of Functions

What is the length of the curve  over the interval ?

Explanation:

The general formula for finding the length of a curve  over an interval  is

In this example, the arc length can be found by computing the integral

.

The derivative of  can be found using the power rule, , which leads to

At this point, a substitution is useful.

Let

.

We can also express the limits of integration in terms of  to simplify computation. When , and when .

.

Now use the power rule, which in general is , to evaluate the integral.

### Given the interval, find the average value of the following function:

Explanation:

When is integrable on [a,b], the average value of f(x) on [a,b] is defined to be:

For the problem statement, we are given f(x) and the intervals [a,b]. All that needs to be done is solving the integral over this interval and dividing the result by the difference between the two intervals.

So:

To solve this integral, we have two options. We can FOIL the terms out for  and solve the integral of the resulting polynomial, or we can use a simple u-substitution. Either way, the result will always be the same. We will try both ways, to prove that this holds true:

FOIL Method:

This is one of the answer choices!

U-Substition Method:

Next, we must adjust the bounds of the new integral which will be in terms of u.

So the new integral becomes:

As you can see both methods results in the same answer!

### Example Question #6 : Average Values And Lengths Of Functions

What is the average of value of  between the intervals  and  ?

Explanation:

When asked for the average value between intervals of form  and  then integral becomes

so in this case we can rewrite our average value problem as

.

The integral comes  evaluated at  and  which simplifies to

(note that  is 1).

However, since we are looking for the average value, we must divide the whole thing by b-a, which in this case is 2, so the final answer is

.

### Example Question #7 : Average Values And Lengths Of Functions

What is the average value of the function  on the interval ?

Explanation:

The average value of a function over an interval is defined to be the integral of the function divided by the length of the interval.

The general rule for this type of integration is as follows.

So we get:

### Example Question #8 : Average Values And Lengths Of Functions

What is the average value of the function

on the interval, ?

Explanation:

The average value of a function is found by taking the integral of the function over the interval and dividing by the length of the interval.

The general rule for this type of integration is as follows.

So we have:

### Example Question #9 : Average Values And Lengths Of Functions

What is the average value of the function  on the interval ?

Explanation:

The average value of a function is found by taking the integral of the function over the interval and dividing by the length of the interval.

The general rule for this type of integration is as follows.

So we have:

### Example Question #10 : Average Values And Lengths Of Functions

What is the average value of the function  on the interval ?