Average Value of a Function

Definition of Average Value

One of the main applications of definite integrals is to find the average value of a function y = f (x) over a specific interval [a, b].

In order to find this average value, one must integrate the function by using the Fundamental Theorem of Calculus and divide the answer by the length of the interval.

So, the average (or the mean) value of f (x) on [a, b] is defined by

\[\bar f = \frac{1}{{b - a}}\int\limits_a^b {f\left( x \right)dx} .\]

The Mean Value Theorem for Definite Integrals

Let \(y = f\left( x \right)\) be a continuous function on the closed interval \(\left[ {a,b} \right].\) The mean value theorem for integrals states that there exists a point \(c\) in that interval such that

\[f\left( c \right) = \frac{1}{{b - a}}\int\limits_a^b {f\left( x \right)dx} .\]

In other words, the mean value theorem for integrals states that there is at least one point \(c\) in the interval \(\left[ {a,b} \right]\) where \(f\left( x \right)\) attains its average value \(\bar f:\)

\[f\left( c \right) = \bar f = \frac{1}{{b - a}}\int\limits_a^b {f\left( x \right)dx} .\]

Geometrically, this means that there is a rectangle whose area exactly represents the area of the region under the curve \(y = f\left( x \right).\) The value of \(f\left( c \right)\) represents the height of the rectangle and the difference \(\left( {b - a} \right)\) represents the width.

Geometric interpretation of the average value of a function and the mean value theorem for integrals. — Figure 1.

Root Mean Square Value of a Function

The root mean square value \(\left( {RMS} \right)\) is defined as the square root of the average (mean) value of the squared function \({{{\left[ {f\left( x \right)} \right]}^2}}\) over an interval \(\left[ {a,b} \right].\) The corresponding integration formula is written in the form

\[RMS = \sqrt {\frac{1}{{b - a}}\int\limits_a^b {{{\left[ {f\left( x \right)} \right]}^2}dx} } .\]

The \({RMS}\) value has many applications in mathematics, physics and engineering. For example, in physics, the RMS value of an alternating current \(\left( {AC} \right)\) is equal to the value of the direct current \(\left( {DC} \right)\) that dissipates the same power in a resistor.

Solved Problems

Example 1.

The average value of a function \(y = f\left( x \right)\) over the interval \(x \in \left[ {1,5} \right]\) is \(2.\) What is the value of \[\int\limits_1^5 {f\left( x \right)dx}?\]

Solution.

By definition,

\[\bar f = \frac{1}{{b - a}}\int\limits_a^b {f\left( x \right)dx} ,\;\; \Rightarrow \int\limits_a^b {f\left( x \right)dx} = \bar f\left( {b - a} \right).\]

Substituting the given values, we obtain

\[\int\limits_1^5 {f\left( x \right)dx} = 2\left( {5 - 1} \right) = 8.\]

Example 2.

Find the average value of the cubic function \[f\left( x \right) = {x^3}\] on the interval \(\left[ {0, 1} \right].\)

Solution.

We use the integration formula

\[\bar f = \frac{1}{{b - a}}\int\limits_a^b {f\left( x \right)dx} .\]

Hence,

\[\bar f = \frac{1}{{1 - 0}}\int\limits_0^1 {{x^3}dx} = \int\limits_0^1 {{x^3}dx} = \left. {\frac{{{x^4}}}{4}} \right|_0^1 = \frac{1}{4}.\]

Example 3.

Find the average value of the square root function \[f\left( x \right) = \sqrt{x}\] on the interval \(\left[ {0, 25} \right].\)

Solution.

Using the definition of the average value, we can write

\[\bar f = \frac{1}{{b - a}}\int\limits_a^b {f\left( x \right)dx} = \frac{1}{{25}}\int\limits_0^{25} {\sqrt x dx} = \frac{1}{{25}}\int\limits_0^{25} {{x^{\frac{1}{2}}}dx} = \frac{1}{{25}} \cdot \left. {\frac{{2{x^{\frac{3}{2}}}}}{3}} \right|_0^{25} = \frac{1}{{25}} \cdot \frac{{2{{\left( {\sqrt {25} } \right)}^3}}}{3} = \frac{{250}}{{75}} = \frac{{10}}{3}.\]

Example 4.

Find the average value of the cosine function \[f\left( x \right) = \cos{x}\] on the interval \(\left[ {0, \frac{\pi }{2}} \right].\)

Solution.

We use the formula

\[\bar f = \frac{1}{{b - a}}\int\limits_a^b {f\left( x \right)dx} .\]

So we set up and evaluate the following integral

\[\bar f = \frac{1}{{\frac{\pi }{2} - 0}}\int\limits_0^{\frac{\pi }{2}} {\cos xdx} = \frac{2}{\pi }\int\limits_0^{\frac{\pi }{2}} {\cos xdx} = \frac{2}{\pi }\left. {\sin x} \right|_0^{\frac{\pi }{2}} = \frac{2}{\pi }.\]

See more problems on Page 2.