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.
\(\textbf{1)}\) \(f(x)=5x^2+8x-10 \,\, [0,3]\)The average function value is \(17\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x) \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3-0}\int_{0}^{3}\left(5x^2+8x-10\right) \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3}\left(\frac{5x^3}{3}+\frac{8x^2}{2}-10x\right) \, \Big|_{0}^{3}\)
\(\,\,\,f_{avg}=\frac{1}{3}\left(\frac{5(3)^3}{3}+\frac{8(3)^2}{2}-10(3) -\left(\frac{5(0)^3}{3}+\frac{8(0)^2}{2}-10(0)\right)\right)\)
\(\,\,\,f_{avg}=\frac{1}{3}\left(\frac{135}{3}+\frac{72}{2}-30 -\left(0\right)\right)\)
\(\,\,\,f_{avg}=\frac{1}{3}\left(45+36-30\right)\)
\(\,\,\,f_{avg}=\frac{1}{3}\left(51\right)\)
\(\,\,\,\)The average function value is \(17\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x) \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3-0}\int_{0}^{3}\left(5x^2+8x-10\right) \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3}\left(\frac{5x^3}{3}+\frac{8x^2}{2}-10x\right) \, \Big|_{0}^{3}\)
\(\,\,\,f_{avg}=\frac{1}{3}\left(\frac{5(3)^3}{3}+\frac{8(3)^2}{2}-10(3) -\left(\frac{5(0)^3}{3}+\frac{8(0)^2}{2}-10(0)\right)\right)\)
\(\,\,\,f_{avg}=\frac{1}{3}\left(\frac{135}{3}+\frac{72}{2}-30 -\left(0\right)\right)\)
\(\,\,\,f_{avg}=\frac{1}{3}\left(45+36-30\right)\)
\(\,\,\,f_{avg}=\frac{1}{3}\left(51\right)\)
\(\,\,\,\)The average function value is \(17\)
\(\textbf{2)}\) \(f(x)=\sin x \,\,\, [0,2\pi]\) The average function value is \(0\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x) \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2\pi-0}\int_{0}^{2\pi}\sin x \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2\pi}\left(-\cos x \, \Big|_{0}^{2\pi}\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2\pi}\left(-\cos 2\pi – \left(-\cos 0\right)\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2\pi}\left(-1 – \left(-1\right)\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2\pi}\left(0\right)\)
\(\,\,\,\)The average function value is \(0\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x) \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2\pi-0}\int_{0}^{2\pi}\sin x \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2\pi}\left(-\cos x \, \Big|_{0}^{2\pi}\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2\pi}\left(-\cos 2\pi – \left(-\cos 0\right)\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2\pi}\left(-1 – \left(-1\right)\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2\pi}\left(0\right)\)
\(\,\,\,\)The average function value is \(0\)
\(\textbf{3)}\) \(f(x)=x \,\,\, [0,4]\) The average function value is \(2\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x) \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4-0}\int_{0}^{4}x \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4} \cdot \frac{x^2}{2}\Big|_{0}^{4}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4} \cdot \left[ \frac{(4)^2}{2}-\frac{(0)^2}{2} \right]\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4} \cdot \left[\frac{16}{2}-0\right]\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4} \cdot \left[8\right]\)
\(\,\,\,\)The average function value is \(2\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x) \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4-0}\int_{0}^{4}x \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4} \cdot \frac{x^2}{2}\Big|_{0}^{4}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4} \cdot \left[ \frac{(4)^2}{2}-\frac{(0)^2}{2} \right]\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4} \cdot \left[\frac{16}{2}-0\right]\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4} \cdot \left[8\right]\)
\(\,\,\,\)The average function value is \(2\)
\(\textbf{4)}\) \(f(x)=3x^2+1 \,\,\, [3,5]\) The average function value is \(50\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x) \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{5-3}\int_{3}^{5}3x^2+1 \, dx\)
\(\,\,\,f_{avg}= \displaystyle\frac{1}{2}\left[x^3+x \Big|_{3}^{5} \right]\)
\(\,\,\,f_{avg}= \displaystyle\frac{1}{2}\left[\left((5)^3+(5)\right)-\left((3)^3+(3)\right)\right] \)
\(\,\,\,f_{avg}= \displaystyle\frac{1}{2}\left[\left(130\right)-\left(30\right)\right] \)
\(\,\,\,f_{avg}= \displaystyle\frac{1}{2}\cdot(100) \)
\(\,\,\,\)The average function value is \(50\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x) \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{5-3}\int_{3}^{5}3x^2+1 \, dx\)
\(\,\,\,f_{avg}= \displaystyle\frac{1}{2}\left[x^3+x \Big|_{3}^{5} \right]\)
\(\,\,\,f_{avg}= \displaystyle\frac{1}{2}\left[\left((5)^3+(5)\right)-\left((3)^3+(3)\right)\right] \)
\(\,\,\,f_{avg}= \displaystyle\frac{1}{2}\left[\left(130\right)-\left(30\right)\right] \)
\(\,\,\,f_{avg}= \displaystyle\frac{1}{2}\cdot(100) \)
\(\,\,\,\)The average function value is \(50\)
\(\textbf{5)}\) \(f(x)=\frac{1}{x} \,\,\, [1,4]\) The average function value is \(\displaystyle\frac{\ln{4}}{3} \approx 0.462098\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x) \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4-1}\int_{1}^{4}\frac{1}{x} \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3} \cdot \left[\ln{x} \Big|_{1}^{4}\right]\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3} \cdot \left[\ln{4}-\ln{1} \right]\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3} \cdot \left[\ln{4}-0 \right]\)
\(\,\,\,\)The average function value is \(\displaystyle\frac{\ln{4}}{3} \approx 0.462098\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x) \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4-1}\int_{1}^{4}\frac{1}{x} \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3} \cdot \left[\ln{x} \Big|_{1}^{4}\right]\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3} \cdot \left[\ln{4}-\ln{1} \right]\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3} \cdot \left[\ln{4}-0 \right]\)
\(\,\,\,\)The average function value is \(\displaystyle\frac{\ln{4}}{3} \approx 0.462098\)
\(\textbf{6)}\) \(f(x)=x^5-x \,\,\, [0,1]\) The average function value is \(-\displaystyle \frac{1}{3}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x) \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{1-0}\int_{0}^{1}x^5-x \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{1}\left(\frac{x^6}{6}-\frac{x^2}{2} \, \Big|_{0}^{1}\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{(1)^6}{6}-\frac{(1)^2}{2}-\left(\frac{(0)^6}{6}-\frac{(0)^2}{2}\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{6}-\frac{1}{2}-\left(0\right)\)
\(\,\,\,\)The average function value is \(-\displaystyle \frac{1}{3}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x) \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{1-0}\int_{0}^{1}x^5-x \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{1}\left(\frac{x^6}{6}-\frac{x^2}{2} \, \Big|_{0}^{1}\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{(1)^6}{6}-\frac{(1)^2}{2}-\left(\frac{(0)^6}{6}-\frac{(0)^2}{2}\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{6}-\frac{1}{2}-\left(0\right)\)
\(\,\,\,\)The average function value is \(-\displaystyle \frac{1}{3}\)
\(\textbf{7)}\) \(f(x)=2x+6 \,\,\, [1,5]\) The average function value is \(12\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_a^b f(x)\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{5-1}\int_1^5(2x+6)\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4}\left[x^2+6x\right]_1^5\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4}\left((25+30)-(1+6)\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4}(48)\)
\(\,\,\,\)The average function value is \(12\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_a^b f(x)\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{5-1}\int_1^5(2x+6)\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4}\left[x^2+6x\right]_1^5\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4}\left((25+30)-(1+6)\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4}(48)\)
\(\,\,\,\)The average function value is \(12\)
\(\textbf{8)}\) \(f(x)=x^2+2x \,\,\, [0,3]\) The average function value is \(6\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3-0}\int_0^3(x^2+2x)\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3}\left[\frac{x^3}{3}+x^2\right]_0^3\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3}\left(\frac{27}{3}+9\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3}(18)\)
\(\,\,\,\)The average function value is \(6\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3-0}\int_0^3(x^2+2x)\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3}\left[\frac{x^3}{3}+x^2\right]_0^3\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3}\left(\frac{27}{3}+9\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3}(18)\)
\(\,\,\,\)The average function value is \(6\)
\(\textbf{9)}\) \(f(x)=4-x^2 \,\,\, [0,2]\) The average function value is \(\displaystyle\frac{8}{3}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2-0}\int_0^2(4-x^2)\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2}\left[4x-\frac{x^3}{3}\right]_0^2\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2}\left(8-\frac{8}{3}\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2}\left(\frac{16}{3}\right)\)
\(\,\,\,\)The average function value is \(\displaystyle\frac{8}{3}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2-0}\int_0^2(4-x^2)\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2}\left[4x-\frac{x^3}{3}\right]_0^2\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2}\left(8-\frac{8}{3}\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2}\left(\frac{16}{3}\right)\)
\(\,\,\,\)The average function value is \(\displaystyle\frac{8}{3}\)
\(\textbf{10)}\) \(f(x)=e^x \,\,\, [0,\ln 4]\) The average function value is \(\displaystyle\frac{3}{\ln 4}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\ln4-0}\int_0^{\ln4}e^x\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\ln4}\left[e^x\right]_0^{\ln4}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\ln4}\left(e^{\ln4}-e^0\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\ln4}(4-1)\)
\(\,\,\,\)The average function value is \(\displaystyle\frac{3}{\ln 4}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\ln4-0}\int_0^{\ln4}e^x\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\ln4}\left[e^x\right]_0^{\ln4}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\ln4}\left(e^{\ln4}-e^0\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\ln4}(4-1)\)
\(\,\,\,\)The average function value is \(\displaystyle\frac{3}{\ln 4}\)
\(\textbf{11)}\) \(f(x)=\cos x \,\,\, [0,\pi]\) The average function value is \(0\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\pi-0}\int_0^\pi\cos x\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\pi}\left[\sin x\right]_0^\pi\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\pi}\left(\sin\pi-\sin0\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\pi}(0-0)\)
\(\,\,\,\)The average function value is \(0\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\pi-0}\int_0^\pi\cos x\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\pi}\left[\sin x\right]_0^\pi\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\pi}\left(\sin\pi-\sin0\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\pi}(0-0)\)
\(\,\,\,\)The average function value is \(0\)
\(\textbf{12)}\) \(f(x)=\sqrt{x} \,\,\, [0,9]\) The average function value is \(2\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{9-0}\int_0^9\sqrt{x}\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{9}\int_0^9x^{1/2}\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{9}\left[\frac{2}{3}x^{3/2}\right]_0^9\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{9}\left(\frac{2}{3}\cdot 27\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{18}{9}\)
\(\,\,\,\)The average function value is \(2\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{9-0}\int_0^9\sqrt{x}\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{9}\int_0^9x^{1/2}\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{9}\left[\frac{2}{3}x^{3/2}\right]_0^9\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{9}\left(\frac{2}{3}\cdot 27\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{18}{9}\)
\(\,\,\,\)The average function value is \(2\)
\(\textbf{13)}\) \(f(x)=\frac{1}{x^2} \,\,\, [1,2]\) The average function value is \(\displaystyle\frac{1}{2}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2-1}\int_1^2\frac{1}{x^2}\,dx\)
\(\,\,\,f_{avg}=\displaystyle\int_1^2x^{-2}\,dx\)
\(\,\,\,f_{avg}=\displaystyle\left[-x^{-1}\right]_1^2\)
\(\,\,\,f_{avg}=\displaystyle\left(-\frac{1}{2}\right)-(-1)\)
\(\,\,\,\)The average function value is \(\displaystyle\frac{1}{2}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2-1}\int_1^2\frac{1}{x^2}\,dx\)
\(\,\,\,f_{avg}=\displaystyle\int_1^2x^{-2}\,dx\)
\(\,\,\,f_{avg}=\displaystyle\left[-x^{-1}\right]_1^2\)
\(\,\,\,f_{avg}=\displaystyle\left(-\frac{1}{2}\right)-(-1)\)
\(\,\,\,\)The average function value is \(\displaystyle\frac{1}{2}\)
\(\textbf{14)}\) \(f(x)=6x^2-4x+1 \,\,\, [0,2]\) The average function value is \(5\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2-0}\int_0^2(6x^2-4x+1)\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2}\left[2x^3-2x^2+x\right]_0^2\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2}\left(16-8+2\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2}(10)\)
\(\,\,\,\)The average function value is \(5\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2-0}\int_0^2(6x^2-4x+1)\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2}\left[2x^3-2x^2+x\right]_0^2\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2}\left(16-8+2\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2}(10)\)
\(\,\,\,\)The average function value is \(5\)
\(\textbf{15)}\) \(f(x)=\ln x \,\,\, [1,e]\) The average function value is \(\displaystyle\frac{1}{e-1}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{e-1}\int_1^e\ln x\,dx\)
\(\,\,\,\int \ln x\,dx=x\ln x-x\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{e-1}\left[x\ln x-x\right]_1^e\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{e-1}\left((e\cdot1-e)-(1\cdot0-1)\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{e-1}(0-(-1))\)
\(\,\,\,\)The average function value is \(\displaystyle\frac{1}{e-1}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{e-1}\int_1^e\ln x\,dx\)
\(\,\,\,\int \ln x\,dx=x\ln x-x\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{e-1}\left[x\ln x-x\right]_1^e\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{e-1}\left((e\cdot1-e)-(1\cdot0-1)\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{e-1}(0-(-1))\)
\(\,\,\,\)The average function value is \(\displaystyle\frac{1}{e-1}\)
\(\textbf{16)}\) \(f(x)=\frac{2}{x+1} \,\,\, [0,3]\) The average function value is \(\displaystyle\frac{2\ln 4}{3}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3-0}\int_0^3\frac{2}{x+1}\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3}\left[2\ln|x+1|\right]_0^3\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3}\left(2\ln4-2\ln1\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{2\ln4}{3}\)
\(\,\,\,\)The average function value is \(\displaystyle\frac{2\ln 4}{3}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3-0}\int_0^3\frac{2}{x+1}\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3}\left[2\ln|x+1|\right]_0^3\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3}\left(2\ln4-2\ln1\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{2\ln4}{3}\)
\(\,\,\,\)The average function value is \(\displaystyle\frac{2\ln 4}{3}\)
\(\textbf{17)}\) \(f(x)=|x| \,\,\, [-2,2]\) The average function value is \(1\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2-(-2)}\int_{-2}^{2}|x|\,dx\)
\(\,\,\,\int_{-2}^{2}|x|\,dx=2\int_0^2x\,dx\)
\(\,\,\,\int_{-2}^{2}|x|\,dx=2\left[\frac{x^2}{2}\right]_0^2\)
\(\,\,\,\int_{-2}^{2}|x|\,dx=4\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4}(4)\)
\(\,\,\,\)The average function value is \(1\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2-(-2)}\int_{-2}^{2}|x|\,dx\)
\(\,\,\,\int_{-2}^{2}|x|\,dx=2\int_0^2x\,dx\)
\(\,\,\,\int_{-2}^{2}|x|\,dx=2\left[\frac{x^2}{2}\right]_0^2\)
\(\,\,\,\int_{-2}^{2}|x|\,dx=4\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4}(4)\)
\(\,\,\,\)The average function value is \(1\)
\(\textbf{18)}\) \(f(x)=\sec^2 x \,\,\, \left[0,\frac{\pi}{4}\right]\) The average function value is \(\displaystyle\frac{4}{\pi}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\frac{\pi}{4}-0}\int_0^{\pi/4}\sec^2 x\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{4}{\pi}\left[\tan x\right]_0^{\pi/4}\)
\(\,\,\,f_{avg}=\displaystyle\frac{4}{\pi}\left(\tan\frac{\pi}{4}-\tan0\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{4}{\pi}(1-0)\)
\(\,\,\,\)The average function value is \(\displaystyle\frac{4}{\pi}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\frac{\pi}{4}-0}\int_0^{\pi/4}\sec^2 x\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{4}{\pi}\left[\tan x\right]_0^{\pi/4}\)
\(\,\,\,f_{avg}=\displaystyle\frac{4}{\pi}\left(\tan\frac{\pi}{4}-\tan0\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{4}{\pi}(1-0)\)
\(\,\,\,\)The average function value is \(\displaystyle\frac{4}{\pi}\)
\(\textbf{19)}\) \(f(x)=3e^{2x} \,\,\, [0,\ln 2]\) The average function value is \(\displaystyle\frac{9}{2\ln 2}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\ln2-0}\int_0^{\ln2}3e^{2x}\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\ln2}\left[\frac{3}{2}e^{2x}\right]_0^{\ln2}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\ln2}\left(\frac{3}{2}e^{2\ln2}-\frac{3}{2}e^0\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\ln2}\left(\frac{3}{2}\cdot4-\frac{3}{2}\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\ln2}\left(\frac{9}{2}\right)\)
\(\,\,\,\)The average function value is \(\displaystyle\frac{9}{2\ln 2}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\ln2-0}\int_0^{\ln2}3e^{2x}\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\ln2}\left[\frac{3}{2}e^{2x}\right]_0^{\ln2}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\ln2}\left(\frac{3}{2}e^{2\ln2}-\frac{3}{2}e^0\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\ln2}\left(\frac{3}{2}\cdot4-\frac{3}{2}\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{\ln2}\left(\frac{9}{2}\right)\)
\(\,\,\,\)The average function value is \(\displaystyle\frac{9}{2\ln 2}\)
\(\textbf{20)}\) \(f(x)=\frac{x}{x^2+1} \,\,\, [0,1]\) The average function value is \(\displaystyle\frac{\ln2}{2}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{1-0}\int_0^1\frac{x}{x^2+1}\,dx\)
\(\,\,\,u=x^2+1\)
\(\,\,\,du=2x\,dx\)
\(\,\,\,\frac{1}{2}du=x\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2}\int_1^2\frac{1}{u}\,du\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2}\left[\ln|u|\right]_1^2\)
\(\,\,\,f_{avg}=\displaystyle\frac{\ln2}{2}\)
\(\,\,\,\)The average function value is \(\displaystyle\frac{\ln2}{2}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{1-0}\int_0^1\frac{x}{x^2+1}\,dx\)
\(\,\,\,u=x^2+1\)
\(\,\,\,du=2x\,dx\)
\(\,\,\,\frac{1}{2}du=x\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2}\int_1^2\frac{1}{u}\,du\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2}\left[\ln|u|\right]_1^2\)
\(\,\,\,f_{avg}=\displaystyle\frac{\ln2}{2}\)
\(\,\,\,\)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 \([0,4]\) The answer is \(c= \frac{4\sqrt{3}}{3}≈ 2.309\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x) \, dx\)
\(\,\,\,f(c)=\displaystyle\frac{1}{4-0}\int_{0}^{4}x^2 \, dx\)
\(\,\,\,f(c)=\displaystyle\frac{1}{4} \left[\frac{x^{3}}{3} \Big|_0^4\right]\)
\(\,\,\,f(c)=\displaystyle\frac{1}{4} \left[\frac{64}{3}-0\right]\)
\(\,\,\,f(c)=\displaystyle \frac{16}{3}\)
\(\,\,\,c^2=\displaystyle \frac{16}{3}\)
\(\,\,\,c=\displaystyle \pm \sqrt{\frac{16}{3}}\)
\(\,\,\,c= \pm \frac{4\sqrt{3}}{3}≈ \pm 2.309\)
\(\,\,\,-2.309\) is not in \([0,4]\) so exclude it.
\(\,\,\,\)The answer is \(c= \frac{4\sqrt{3}}{3}≈ 2.309\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x) \, dx\)
\(\,\,\,f(c)=\displaystyle\frac{1}{4-0}\int_{0}^{4}x^2 \, dx\)
\(\,\,\,f(c)=\displaystyle\frac{1}{4} \left[\frac{x^{3}}{3} \Big|_0^4\right]\)
\(\,\,\,f(c)=\displaystyle\frac{1}{4} \left[\frac{64}{3}-0\right]\)
\(\,\,\,f(c)=\displaystyle \frac{16}{3}\)
\(\,\,\,c^2=\displaystyle \frac{16}{3}\)
\(\,\,\,c=\displaystyle \pm \sqrt{\frac{16}{3}}\)
\(\,\,\,c= \pm \frac{4\sqrt{3}}{3}≈ \pm 2.309\)
\(\,\,\,-2.309\) is not in \([0,4]\) so exclude it.
\(\,\,\,\)The answer is \(c= \frac{4\sqrt{3}}{3}≈ 2.309\)
\(\textbf{22)}\) The average value of \(f(x)\) over the interval \([3,9]\) is \(11\). Find \(\displaystyle\int_{3}^{9}f(x) \, dx\) The answer is \(66\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x) \, dx\)
\(\,\,\,11=\displaystyle\frac{1}{9-3}\int_{3}^{9}f(x) \, dx\)
\(\,\,\,11=\displaystyle\frac{1}{6}\int_{3}^{9}f(x) \, dx\)
\(\,\,\,66=\displaystyle\int_{3}^{9}f(x) \, dx\)
\(\,\,\,\)The answer is \(66\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x) \, dx\)
\(\,\,\,11=\displaystyle\frac{1}{9-3}\int_{3}^{9}f(x) \, dx\)
\(\,\,\,11=\displaystyle\frac{1}{6}\int_{3}^{9}f(x) \, dx\)
\(\,\,\,66=\displaystyle\int_{3}^{9}f(x) \, dx\)
\(\,\,\,\)The answer is \(66\)
\(\textbf{23)}\) Find \(c\) such that \(f(c)=f_{avg}\) of \(f(x)=x\) on \([2,8]\) The answer is \(c=5\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{8-2}\int_2^8x\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{6}\left[\frac{x^2}{2}\right]_2^8\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{6}\left(\frac{64}{2}-\frac{4}{2}\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{6}(30)\)
\(\,\,\,f_{avg}=5\)
\(\,\,\,f(c)=c\)
\(\,\,\,c=5\)
\(\,\,\,\)The answer is \(c=5\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{8-2}\int_2^8x\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{6}\left[\frac{x^2}{2}\right]_2^8\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{6}\left(\frac{64}{2}-\frac{4}{2}\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{6}(30)\)
\(\,\,\,f_{avg}=5\)
\(\,\,\,f(c)=c\)
\(\,\,\,c=5\)
\(\,\,\,\)The answer is \(c=5\)
\(\textbf{24)}\) The average value of \(f(x)\) over the interval \([2,10]\) is \(7\). Find \(\displaystyle\int_2^{10}f(x)\,dx\) The answer is \(56\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_a^b f(x)\,dx\)
\(\,\,\,7=\displaystyle\frac{1}{10-2}\int_2^{10}f(x)\,dx\)
\(\,\,\,7=\displaystyle\frac{1}{8}\int_2^{10}f(x)\,dx\)
\(\,\,\,56=\displaystyle\int_2^{10}f(x)\,dx\)
\(\,\,\,\)The answer is \(56\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_a^b f(x)\,dx\)
\(\,\,\,7=\displaystyle\frac{1}{10-2}\int_2^{10}f(x)\,dx\)
\(\,\,\,7=\displaystyle\frac{1}{8}\int_2^{10}f(x)\,dx\)
\(\,\,\,56=\displaystyle\int_2^{10}f(x)\,dx\)
\(\,\,\,\)The answer is \(56\)
\(\textbf{25)}\) Find \(c\) such that \(f(c)=f_{avg}\) of \(f(x)=x^2+1\) on \([0,3]\) The answer is \(c=\sqrt{3}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3-0}\int_0^3(x^2+1)\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3}\left[\frac{x^3}{3}+x\right]_0^3\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3}\left(9+3\right)\)
\(\,\,\,f_{avg}=4\)
\(\,\,\,f(c)=c^2+1\)
\(\,\,\,c^2+1=4\)
\(\,\,\,c^2=3\)
\(\,\,\,c=\pm\sqrt{3}\)
\(\,\,\,-\sqrt{3}\) is not in \([0,3]\) so exclude it.
\(\,\,\,\)The answer is \(c=\sqrt{3}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3-0}\int_0^3(x^2+1)\,dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3}\left[\frac{x^3}{3}+x\right]_0^3\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3}\left(9+3\right)\)
\(\,\,\,f_{avg}=4\)
\(\,\,\,f(c)=c^2+1\)
\(\,\,\,c^2+1=4\)
\(\,\,\,c^2=3\)
\(\,\,\,c=\pm\sqrt{3}\)
\(\,\,\,-\sqrt{3}\) is not in \([0,3]\) so exclude it.
\(\,\,\,\)The answer is \(c=\sqrt{3}\)
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