Average function value tells us the mean height of a function over a given interval. Instead of averaging a list of numbers, calculus uses a definite integral to average all the function values continuously from \(x=a\) to \(x=b\). The formula is \(f_{avg}=\displaystyle\frac{1}{b-a}\int_a^b f(x)\,dx\). This page also includes related variations where you use the average value to find an integral or find a value \(c\) where \(f(c)=f_{avg}\).
Lesson
Notes
Practice Problems
For questions 1-20, find the average function value in the given interval.
\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x) \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2\pi-0}\int_{0}^{2\pi}\sin x \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2\pi}\left(-\cos x \, \Big|_{0}^{2\pi}\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2\pi}\left(-\cos 2\pi – \left(-\cos 0\right)\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2\pi}\left(-1 – \left(-1\right)\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2\pi}\left(0\right)\)
\(\,\,\,\)The average function value is \(0\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x) \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{1-0}\int_{0}^{1}x^5-x \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{1}\left(\frac{x^6}{6}-\frac{x^2}{2} \, \Big|_{0}^{1}\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{(1)^6}{6}-\frac{(1)^2}{2}-\left(\frac{(0)^6}{6}-\frac{(0)^2}{2}\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{6}-\frac{1}{2}-\left(0\right)\)
\(\,\,\,\)The average function value is \(-\displaystyle \frac{1}{3}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_a^b f(x)\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{5-1}\int_1^5(2x+6)\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4}\left[x^2+6x\right]_1^5\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4}\left((25+30)-(1+6)\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4}(48)\)
\(\,\,\,\)The average function value is \(12\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3-0}\int_0^3(x^2+2x)\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3}\left[\frac{x^3}{3}+x^2\right]_0^3\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3}\left(\frac{27}{3}+9\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3}(18)\)
\(\,\,\,\)The average function value is \(6\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2-0}\int_0^2(4-x^2)\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2}\left[4x-\frac{x^3}{3}\right]_0^2\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2}\left(8-\frac{8}{3}\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2}\left(\frac{16}{3}\right)\)
\(\,\,\,\)The average function value is \(\displaystyle\frac{8}{3}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\ln4-0}\int_0^{\ln4}e^x\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\ln4}\left[e^x\right]_0^{\ln4}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\ln4}\left(e^{\ln4}-e^0\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\ln4}(4-1)\)
\(\,\,\,\)The average function value is \(\displaystyle\frac{3}{\ln 4}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\pi-0}\int_0^\pi\cos x\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\pi}\left[\sin x\right]_0^\pi\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\pi}\left(\sin\pi-\sin0\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\pi}(0-0)\)
\(\,\,\,\)The average function value is \(0\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{9-0}\int_0^9\sqrt{x}\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{9}\int_0^9x^{1/2}\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{9}\left[\frac{2}{3}x^{3/2}\right]_0^9\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{9}\left(\frac{2}{3}\cdot 27\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{18}{9}\)
\(\,\,\,\)The average function value is \(2\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2-1}\int_1^2\frac{1}{x^2}\,dx\)
\(\,\,\,f_{avg}=\displaystyle\int_1^2x^{-2}\,dx\)
\(\,\,\,f_{avg}=\displaystyle\left[-x^{-1}\right]_1^2\)
\(\,\,\,f_{avg}=\displaystyle\left(-\frac{1}{2}\right)-(-1)\)
\(\,\,\,\)The average function value is \(\displaystyle\frac{1}{2}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2-0}\int_0^2(6x^2-4x+1)\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2}\left[2x^3-2x^2+x\right]_0^2\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2}\left(16-8+2\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2}(10)\)
\(\,\,\,\)The average function value is \(5\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{e-1}\int_1^e\ln x\,dx\)
\(\,\,\,\int \ln x\,dx=x\ln x-x\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{e-1}\left[x\ln x-x\right]_1^e\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{e-1}\left((e\cdot1-e)-(1\cdot0-1)\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{e-1}(0-(-1))\)
\(\,\,\,\)The average function value is \(\displaystyle\frac{1}{e-1}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3-0}\int_0^3\frac{2}{x+1}\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3}\left[2\ln|x+1|\right]_0^3\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3}\left(2\ln4-2\ln1\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{2\ln4}{3}\)
\(\,\,\,\)The average function value is \(\displaystyle\frac{2\ln 4}{3}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2-(-2)}\int_{-2}^{2}|x|\,dx\)
\(\,\,\,\int_{-2}^{2}|x|\,dx=2\int_0^2x\,dx\)
\(\,\,\,\int_{-2}^{2}|x|\,dx=2\left[\frac{x^2}{2}\right]_0^2\)
\(\,\,\,\int_{-2}^{2}|x|\,dx=4\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4}(4)\)
\(\,\,\,\)The average function value is \(1\)
\(\textbf{18)}\) \(f(x)=\sec^2 x \,\,\, \left[0,\frac{\pi}{4}\right]\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\frac{\pi}{4}-0}\int_0^{\pi/4}\sec^2 x\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{4}{\pi}\left[\tan x\right]_0^{\pi/4}\)
\(\,\,\,f_{avg}=\displaystyle\frac{4}{\pi}\left(\tan\frac{\pi}{4}-\tan0\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{4}{\pi}(1-0)\)
\(\,\,\,\)The average function value is \(\displaystyle\frac{4}{\pi}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\ln2-0}\int_0^{\ln2}3e^{2x}\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\ln2}\left[\frac{3}{2}e^{2x}\right]_0^{\ln2}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\ln2}\left(\frac{3}{2}e^{2\ln2}-\frac{3}{2}e^0\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\ln2}\left(\frac{3}{2}\cdot4-\frac{3}{2}\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\ln2}\left(\frac{9}{2}\right)\)
\(\,\,\,\)The average function value is \(\displaystyle\frac{9}{2\ln 2}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{1-0}\int_0^1\frac{x}{x^2+1}\,dx\)
\(\,\,\,u=x^2+1\)
\(\,\,\,du=2x\,dx\)
\(\,\,\,\frac{1}{2}du=x\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2}\int_1^2\frac{1}{u}\,du\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2}\left[\ln|u|\right]_1^2\)
\(\,\,\,f_{avg}=\displaystyle\frac{\ln2}{2}\)
\(\,\,\,\)The average function value is \(\displaystyle\frac{\ln2}{2}\)
Challenge Problems
\(\textbf{21)}\) Find \(c\) such that \(f(c)=f_{avg}\) of \(f(x)=x^2\) on \([0,4]\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3-0}\int_0^3(x^2+1)\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3}\left[\frac{x^3}{3}+x\right]_0^3\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3}\left(9+3\right)\)
\(\,\,\,f_{avg}=4\)
\(\,\,\,f(c)=c^2+1\)
\(\,\,\,c^2+1=4\)
\(\,\,\,c^2=3\)
\(\,\,\,c=\pm\sqrt{3}\)
\(\,\,\,-\sqrt{3}\) is not in \([0,3]\) so exclude it.
\(\,\,\,\)The answer is \(c=\sqrt{3}\)
In calculus, the average function value is a concept that is used to determine the average or mean value of a function over a given interval. It is an important concept in mathematics and has numerous applications in various fields, including physics, engineering, and economics.
