In this article, we will discuss what is the average value of a function and how to calculate it.

An **average value of a function** is one of the primary applications of definite integrals. For computing the average value of a function, we use the "Fundamental Theorem of Calculus" to integrate the function and then we divide the value by the length of the interval.

The average value of a function f(x) over the closed interval [a,b] can be calculated using the formula below:

The average value of the function can also be calculated using **mean value theorem**. The mean value theorem says that if f(x) is a continuous function on the closed interval [a,b], then there is number c in the closed interval [a,b] in such a way that:

Now, we will solve some examples in which we will calculate an average value of the given functions.

## Example 1

Find the average value of the following function on the interval [0, 2].

### Solution

We will use the following formula to calculate the average value of the function:

In this example, the value of b is 2 and a is 0. Now, substitute these values in the above equation like this:

To integrate the function , first, we will find the antiderivative of the function. The antiderivative of this function is . Suppose C = 0. The fundamental theorem of calculus says that:

Substitute 2 and 0 in the antiderivative of the function like this:

Hence, our definite integral is . Now, we will substitute this value in the equation below like this:

Hence, the average value of the function at the interval [0,2] is .

## Example 2

Find the average value of the following function on the interval [0, 3].

### Solution

We will use the following formula to calculate the average value of the function:

In this example, the value of b is 3 and a is 0. Now, substitute these values in the above equation like this:

To integrate the function , first, we will find the antiderivative of the function. The antiderivative of this function is . Suppose C = 0. The fundamental theorem of calculus says that:

Substitute 3 and 0 in the antiderivative of the function like this:

Hence, our definite integral is . Now, we will substitute this value in the equation below like this:

Hence, the average value of the function at the interval [0, 3] is .

## Example 3

Find the average value of the following function on the interval [2,3].

### Solution

We will use the following formula to calculate the average value of the function:

In this example, the value of b is 3 and a is 2. Now, substitute these values in the above equation like this:

To integrate the function , first, we will find the antiderivative of the function. The antiderivative of this function is . Suppose C = 0. The fundamental theorem of calculus says that:

Substitute 3 and 2 in the antiderivative of the function like this:

Hence, our definite integral is . Now, we will substitute this value in the equation below like this:

Hence, the average value of the function at the interval [2,3] is 2 \sqrt{3} - \frac{4 \sqrt{2}}{3} + 1.

## Example 4

Find the average value of the following function on the interval [2, 4].

### Solution

We will use the following formula to calculate the average value of the function:

In this example, the value of b is 4 and a is 2. Now, substitute these values in the above equation like this:

To integrate the function , first, we will find the antiderivative of the function. The antiderivative of this function is - cos x + C. Suppose C = 0. The fundamental theorem of calculus says that:

Substitute 4 and 2 in the antiderivative of the function like this:

Hence, our definite integral is - cos 4 + cos 2. Now, we will substitute this value in the equation below like this:

Hence, the average value of the function at the interval [2,4] is .

## Example 5

Find the average value of the following function on the interval [0, 5].

### Solution

We will use the following formula to calculate the average value of the function:

In this example, the value of b is 5 and a is 0. Now, substitute these values in the above equation like this:

Substitute 5 and 0 in the antiderivative of the function like this:

Hence, the average value of the function at the interval [0,5] is .

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