Calculus

Applications of Integrals

Applications of Integrals Logo

Average Value of a Function

Solved Problems

Example 5.

Find the average value of the sine squared function \[f\left( x \right) = {\sin ^2}x\] on the interval \(\left[ {0,\pi } \right].\)

Solution.

The average value \(\bar f\) is given by

\[\bar f = \frac{1}{\pi }\int\limits_0^\pi {{{\sin }^2}xdx} .\]

Using the trigonometric identity

\[{\sin ^2}x = \frac{1}{2}\left( {1 - \cos 2x} \right),\]

we obtain

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

Example 6.

The daily temperature of the outside air is given by the equation \[T\left( t \right) = 20 - 5\cos \left( {\frac{{\pi t}}{{12}}} \right),\] where \(t\) is measured in hours \(\left( {0 \le t \le 24} \right)\) and \(T\) is measured in degrees \(C.\) Find the average temperature between \({t_1} = 6\) and \({t_2} = 12\) hours.

Solution.

We calculate the average temperature in the given interval through integration using the definition of the average value of a function:

\[{{\bar T}_{\left[ {6,12} \right]}} = \frac{1}{{12 - 6}}\int\limits_6^{12} {T\left( t \right)dt} = \frac{1}{6}\int\limits_6^{12} {\left[ {20 - 5\cos \left( {\frac{{\pi t}}{{12}}} \right)} \right]dt} = \frac{1}{6}\left. {\left[ {20t - 5 \cdot \frac{{12}}{\pi }\sin \left( {\frac{{\pi t}}{{12}}} \right)} \right]} \right|_6^{12} = \frac{1}{6}\left[ {\left( {240 - \frac{{60}}{\pi }\sin \pi } \right) - \left( {120 - \frac{{60}}{\pi }\sin \frac{\pi }{2}} \right)} \right] = \frac{1}{6}\left( {120 + \frac{{60}}{\pi }} \right) = 20 + \frac{{10}}{\pi } \approx 23.2^\text{ o}C.\]

Example 7.

Given the rational function \[f\left( x \right) = \frac{2}{{{{\left( {x + 1} \right)}^2}}}.\] Find the values of \(c\) that satisfy the Mean Value Theorem for Integrals for the function on the interval \(\left[ {0,3} \right].\)

Solution.

First we calculate the average value of the function \(f\left( x \right)\) on the interval \(\left[ {0,3} \right]:\)

\[\bar f = \frac{1}{{b - a}}\int\limits_a^b {f\left( x \right)dx} = \frac{1}{3}\int\limits_0^3 {\frac{{2dx}}{{{{\left( {x + 1} \right)}^2}}}} = \frac{2}{3}\left. {\left( { - \frac{1}{{x + 1}}} \right)} \right|_0^3 = \frac{2}{3}\left( {1 - \frac{1}{4}} \right) = \frac{1}{2}.\]

To determine the values of \(c,\) we solve the equation

\[ f\left( c \right) = \bar f .\]

Hence,

\[\frac{2}{{{{\left( {c + 1} \right)}^2}}} = \frac{1}{2},\;\; \Rightarrow {\left( {c + 1} \right)^2} = 4,\;\; \Rightarrow c + 1 = \pm 2,\;\; \Rightarrow {c_1} = 1,\;{c_2} = - 1.\]

We see that only the positive root \({c_1} = 1\) lies in the interval \(\left[ {0,3} \right]\), so the answer is \(c = 1.\)

Example 8.

Given the quadratic function \[f\left( x \right) = {\left( {x + 2} \right)^2}.\] Find the values of \(c\) that satisfy the Mean Value Theorem for Integrals for the function on the interval \(\left[ {0,9} \right].\)

Solution.

The average value of the function \(f\left( x \right)\) on the interval \(\left[ {0,9} \right]\) is given by

\[\bar f = \frac{1}{{b - a}}\int\limits_a^b {f\left( x \right)dx} = \frac{1}{9}\int\limits_0^9 {{{\left( {x + 2} \right)}^2}dx} = \frac{1}{9}\int\limits_0^9 {\left( {{x^2} + 4x + 4} \right)dx} = \frac{1}{9}\left. {\left[ {\frac{{{x^3}}}{3} + 2{x^2} + 4x} \right]} \right|_0^9 = \frac{{441}}{9} = 49.\]

To find the values of \(c,\) we solve the equation \(\bar f = f\left( c \right).\) This yields

\[{\left( {c + 2} \right)^2} = 49,\;\; \Rightarrow c + 2 = \pm 7,\;\; \Rightarrow {c_1} = - 9,\;{c_2} = 5.\]

We see that the solution is \(c = 5.\)

Example 9.

A sawtooth signal has the period \(T = 1\) and is given by the equation \[f\left( t \right) = At\left({\text{mod} A}\right).\] Find the RMS value of the sawtooth waveform.

Solution.

Root Mean Square value of the sawtooth waveform y=At (mod A) over the period T=1.
Figure 2.

The sawtooth signal is periodic, so we integrate over one cycle from \(t = 0\) to \(t = 1.\) The \(RMS\) value is expressed by the formula

\[RMS = \sqrt {\frac{1}{{1 - 0}}\int\limits_0^1 {{{\left( {At} \right)}^2}dt} } = \sqrt {{A^2}\int\limits_0^1 {{t^2}dt} } = A\sqrt {\left. {\frac{{{t^3}}}{3}} \right|_0^1} = \frac{A}{{\sqrt 3 }}.\]

Example 10.

Find the RMS value of the sine function \[f\left( t \right) = A\sin t\] over the interval \(\left[ {0,2\pi } \right].\)

Solution.

Root Mean Square value of the sine function over the interval [0,2pi]
Figure 3.

The \(RMS\) value is given by the integration formula

\[RMS = \sqrt {\frac{1}{{2\pi }}\int\limits_0^{2\pi } {{{\sin }^2}tdt} } .\]

Using the trig identity

\[{\sin ^2}t = \frac{1}{2}\left( {1 - \cos 2t} \right),\]

we have

\[RMS = \sqrt {\frac{1}{{2\pi }}\int\limits_0^{2\pi } {{A^2}{{\sin }^2}tdt} } = \sqrt {\frac{1}{{2\pi }}\int\limits_0^{2\pi } {\frac{{{A^2}}}{2}\left( {1 - \cos 2t} \right)dt} } = \sqrt {\frac{{{A^2}}}{{4\pi }}\int\limits_0^{2\pi } {\left( {1 - \cos 2t} \right)dt} } = \sqrt {\frac{{{A^2}}}{{4\pi }}\left. {\left[ {t - \frac{{\sin 2t}}{2}} \right]} \right|_0^{2\pi }} = \sqrt {\frac{{{A^2}}}{{4\pi }} \cdot 2\pi } = \frac{A}{{\sqrt 2 }}.\]
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