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:

f_avg = \frac{1} {b - a} \int ^{b}_{a} f(x) dx

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:

\int^{b}_{a} f(x) dx = f(c) (b - a)

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].

f(x) = 2x^2

Solution

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

f_avg = \frac{1} {b - a} \int ^{b}_{a} f(x) dx

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

f_avg = \frac{1} {2 - 0} \int ^{2}_{0} 2x^2 dx

f_avg = \frac{1} {2} \int ^{2}_{0} 2x^2 dx

To integrate the function f(x) = 2x^2, first, we will find the antiderivative of the function. The antiderivative of this function is \frac{2x^3}{3} + C. Suppose C = 0. The fundamental theorem of calculus says that:

\int _{a} ^ {b} f(x) = [F(x)]_{a} ^{b} = F (b) - F(a)

\int _{0} ^ {2} (2x^2) dx = F (3) - F(0)

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

=\frac{2 (2)^3}{3} - \frac{2 (0)^3}{3}

= \frac{16}{3}

Hence, our definite integral is \frac{16}{3}. Now, we will substitute this value in the equation below like this:

f_avg = \frac{1} {2} \int ^{2}_{0} 2x^2 dx

f_avg = \frac{1} {2} \cdot  \frac {16}{3}

f_avg = \frac{16}{6} = \frac{8}{3}

Hence, the average value of the function at the interval [0,2] is \frac{8}{3}.

 

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Example 2

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

f(x) = x^3 + 3

Solution

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

f_avg = \frac{1} {b - a} \int ^{b}_{a} f(x) dx

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

f_avg = \frac{1} {3 - 0} \int ^{3}_{0} x^3 + 3 dx

f_avg = \frac{1} {3} \int ^{3}_{0} x^3 + 3 dx

To integrate the function f(x) = x^3 + 3 dx, first, we will find the antiderivative of the function. The antiderivative of this function is \frac{x^4}{4} + 3x + C. Suppose C = 0. The fundamental theorem of calculus says that:

\int _{a} ^ {b} f(x) = [F(x)]_{a} ^{b} = F (b) - F(a)

\int _{0} ^ {3} (x^3 + 3) dx = F (3) - F(0)

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

=( \frac{(3)^4}{4} + 3(3)) - ( \frac{(0)^4}{4} + 3(0))

= (\frac{81}{4} + 9) - (0)

= \frac{129}{4}

Hence, our definite integral is \frac{129}{4}. Now, we will substitute this value in the equation below like this:

f_avg = \frac{1} {3} \int ^{3}_{0} x^3 + 3 dx

f_avg = \frac{1} {3} \cdot \frac{129}{4}

f_avg = \frac{39}{4}

Hence, the average value of the function at the interval [0, 3] is \frac{39}{4}.

 

Example 3

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

f(x) = \sqrt{x} + 1

Solution

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

f_avg = \frac{1} {b - a} \int ^{b}_{a} f(x) dx

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

f_avg = \frac{1} {3 - 2} \int ^{3}_{2} \sqrt{x} + 1 dx

f_avg = \frac{1} {1} \int ^{2}_{0} \sqrt{x} + 1 dx

To integrate the function f(x) = \sqrt{x} + 1, first, we will find the antiderivative of the function. The antiderivative of this function is \frac{2x^{\frac{3}{2}}} {3} + x + C. Suppose C = 0. The fundamental theorem of calculus says that:

\int _{a} ^ {b} f(x) = [F(x)]_{a} ^{b} = F (b) - F(a)

\int _{2} ^ {3} (\sqrt {x} + 1) dx = F (3) - F(2)

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

=(\frac{2 (3)^ {\frac{3}{2}}} {3} + 3) - (\frac{2 (2)^ {\frac{3}{2}}} {3} + 2)

= 2 \sqrt{3} - \frac{4 \sqrt{2}}{3} + 1

Hence, our definite integral is   2 \sqrt{3} - \frac{4 \sqrt{2}}{3} + 1. Now, we will substitute this value in the equation below like this:

f_avg = \frac{1} {1} \int ^{3}_{2} \sqrt{x} + 1 dx

f_avg = 1 \cdot 2 \sqrt{3} - \frac{4 \sqrt{2}}{3} + 1

f_avg = 2 \sqrt{3} - \frac{4 \sqrt{2}}{3} + 1

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].

f(x) = sin x

Solution

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

f_avg = \frac{1} {b - a} \int ^{b}_{a} f(x) dx

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

f_avg = \frac{1} {4 - 2} \int ^{4}_{2} sin x dx

f_avg = \frac{1} {2} \int ^{4}_{2} sin x dx

To integrate the function f(x) = sin x, 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:

\int _{a} ^ {b} f(x) = [F(x)]_{a} ^{b} = F (b) - F(a)

\int _{2} ^ {4} (- cos x) dx = F (4) - F(2)

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

=- cos (4) - (- cos (2))

- cos 4 + cos 2

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

f_avg = \frac{1} {2} \int ^{4}_{2} sin(x) dx

f_avg = \frac{1} {2} \cdot  (- cos (4) + cos (2))

Hence, the average value of the function at the interval [2,4] is \frac{1} {2} \cdot  (- cos (4) + cos (2)).

Example 5

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

f(x) = x^2 + 2x + 3

Solution

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

f_avg = \frac{1} {b - a} \int ^{b}_{a} f(x) dx

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

f_avg = \frac{1} {5 - 0} \int ^{5}_{0} x^2 + 2x + 3 dx

f_avg = \frac{1} {5} \int ^{5}_{0} x^3 + 2x + 3 dx

To integrate the function f(x) = x^2 + 2x + 3, first, we will find the antiderivative of the function. The antiderivative of this function is \frac{x^3}{3} + x^2 + 3x + C. Suppose C = 0. The fundamental theorem of calculus says that:

\int _{a} ^ {b} f(x) = [F(x)]_{a} ^{b} = F (b) - F(a)

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

= \frac{5^3}{3} + 5^2 + 3(5) - 0

= \frac{125}{3} + 25 + 15

= \frac{125}{3} +40

= \frac{245}{3}

Hence, our definite integral is \frac{245}{3} . Now, we will substitute this value in the equation below like this:

f_avg = \frac{1} {5} \int ^{5}_{0} x^2 + 2x + 3 dx

f_avg = \frac{1} {5} \cdot  \frac {245}{3}

f_avg = \frac{245}{15} = \frac{49}{3}

Hence, the average value of the function at the interval [0,5] is \frac{49}{3}.

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Rafia Shabbir