Calculus

Integration of Functions

Integration of Functions Logo

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 \(y = f\left( x \right)\) over a specific interval \(\left[ {a,b} \right].\)

    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 \(f\left( x \right)\) on \(\left[ {a,b} \right]\) 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

    Click a problem to see the solution.

    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 \(\int\limits_1^5 {f\left( x \right)dx}?\)

    Example 2

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

    Example 3

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

    Example 4

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

    Example 5

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

    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.

    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 \(\left[ {0,3} \right].\)

    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 \(\left[ {0,9} \right].\)

    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.

    Example 10

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

    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 \(\left[ {0, \large{\frac{\pi }{2}}\normalsize} \right].\)

    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

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

    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

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

    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:

    \[{{{\bar T}_{\left[ {6,12} \right]}} = \frac{1}{{12 – 6}}\int\limits_6^{12} {T\left( t \right)dt} }={ \frac{1}{6}\int\limits_6^{12} {\left[ {20 – 5\cos \left( {\frac{{\pi t}}{{12}}} \right)} \right]dt} }={ \frac{1}{6}\left. {\left[ {20t – 5 \cdot \frac{{12}}{\pi }\sin \left( {\frac{{\pi t}}{{12}}} \right)} \right]} \right|_6^{12} }={ \frac{1}{6}\left[ {\left( {240 – \frac{{60}}{\pi }\sin \pi } \right) }\right.}-{\left.{ \left( {120 – \frac{{60}}{\pi }\sin \frac{\pi }{2}} \right)} \right] }={ \frac{1}{6}\left( {120 – \frac{{60}}{\pi }} \right) }={ 20 – \frac{{10}}{\pi } \approx 16.8^\text{ o}C}\]

    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,

    \[{\frac{2}{{{{\left( {c + 1} \right)}^2}}} = \frac{1}{2},}\;\; \Rightarrow {{\left( {c + 1} \right)^2} = 4,}\;\; \Rightarrow {c + 1 = \pm 2,}\;\; \Rightarrow {{c_1} = 1,\;}\kern0pt{{c_2} = – 1.}\]

    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

    \[{{\left( {c + 2} \right)^2} = 49,}\;\; \Rightarrow {c + 2 = \pm 7,}\;\; \Rightarrow {{c_1} = – 9,\;}\kern0pt{{c_2} = 5.}\]

    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

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