Lesson

 

Notes

 

Questions & Solutions

For questions 1-6, find the average function value in the given interval.

\(\textbf{1)}\) \(f(x)=5x^2+8x-10 \,\, [0,3]\)

Show Average Function Value

The average function value is \(17\)

Show Work

\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x) \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3-0}\int_{0}^{3}\left(5x^2+8x-10\right) \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3}\left(\frac{5x^3}{3}+\frac{8x^2}{2}-10x\right) \, \Big|_{0}^{3}\)
\(\,\,\,f_{avg}=\frac{1}{3}\left(\frac{5(3)^3}{3}+\frac{8(3)^2}{2}-10(3) -\left(\frac{5(0)^3}{3}+\frac{8(0)^2}{2}-10(0)\right)\right)\)
\(\,\,\,f_{avg}=\frac{1}{3}\left(\frac{135}{3}+\frac{72}{2}-30 -\left(0\right)\right)\)
\(\,\,\,f_{avg}=\frac{1}{3}\left(45+36-30\right)\)
\(\,\,\,f_{avg}=\frac{1}{3}\left(51\right)\)
\(\,\,\,\)The average function value is \(17\)

 

\(\textbf{2)}\) \(f(x)=\sin ⁡x \,\,\, [0,2π]\)

Show Average Function Value

The average function value is \(0\)

Show Work

\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x) \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2\pi-0}\int_{0}^{2\pi}\sin x \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2\pi}\left(-\cos x \, \Big|_{0}^{2\pi}\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2\pi}\left(-\cos 2\pi – \left(-\cos 0\right)\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2\pi}\left(-1 – \left(-1\right)\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{2\pi}\left(0\right)\)
\(\,\,\,\)The average function value is \(0\)

 

\(\textbf{3)}\) \(f(x)=x \,\,\, [0,4]\)

Show Average Function Value

The average function value is \(2\)

Show Work

\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x) \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4-0}\int_{0}^{4}x \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4} \cdot \frac{x^2}{2}\Big|_{0}^{4}\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4} \cdot \left[ \frac{(4)^2}{2}-\frac{(0)^2}{2} \right]\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4} \cdot \left[\frac{16}{2}-0\right]\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4} \cdot \left[8\right]\)
\(\,\,\,\)The average function value is \(2\)

 

\(\textbf{4)}\) \(f(x)=3x^2+1 \,\,\, [3,5]\)

Show Average Function Value

The average function value is \(50\)

Show Work

\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x) \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{5-3}\int_{3}^{5}3x^2+1 \, dx\)
\(\,\,\,f_{avg}= \displaystyle\frac{1}{2}\left[x^3+x \Big|_{3}^{5} \right]\)
\(\,\,\,f_{avg}= \displaystyle\frac{1}{2}\left[\left((5)^3+(5)\right)-\left((3)^3+(3)\right)\right] \)
\(\,\,\,f_{avg}= \displaystyle\frac{1}{2}\left[\left(130\right)-\left(30\right)\right] \)
\(\,\,\,f_{avg}= \displaystyle\frac{1}{2}\cdot(100) \)
\(\,\,\,\)The average function value is \(50\)

 

\(\textbf{5)}\) \(f(x)=\frac{1}{x} \,\,\, [1,4]\)

Show Average Function Value

The average function value is \(\displaystyle\frac{\ln{4}}{3} \approx 0.462098\)

Show Work

\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x) \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{4-1}\int_{1}^{4}\frac{1}{x} \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3} \cdot \left[\ln{x} \Big|_{1}^{4}\right]\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3} \cdot \left[\ln{4}-\ln{1} \right]\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{3} \cdot \left[\ln{4}-0 \right]\)
\(\,\,\,\)The average function value is \(\displaystyle\frac{\ln{4}}{3} \approx 0.462098\)

 

\(\textbf{6)}\) \(f(x)=x^5-x \,\,\, [0,1]\)

Show Average Function Value

The average function value is \(-\displaystyle \frac{1}{3}\)

Show Work

\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x) \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{1-0}\int_{0}^{1}x^5-x \, dx\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{1}\left(\frac{x^6}{6}-\frac{x^2}{2} \, \Big|_{0}^{1}\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{(1)^6}{6}-\frac{(1)^2}{2}-\left(\frac{(0)^6}{6}-\frac{(0)^2}{2}\right)\)
\(\,\,\,f_{avg}=\displaystyle\frac{1}{6}-\frac{1}{2}-\left(0\right)\)
\(\,\,\,\)The average function value is \(-\displaystyle \frac{1}{3}\)

 

Challenge Problems

\(\textbf{7)}\) Find \(c\) such that \(f(c)=f_{avg}\) of \(f(x)=x^2\) on \([0,4]\)

Show Answer

The answer is \(c= \frac{4\sqrt{3}}{3}≈ 2.309\)

Show Work

\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x) \, dx\)
\(\,\,\,f(c)=\displaystyle\frac{1}{4-0}\int_{0}^{4}x^2 \, dx\)
\(\,\,\,f(c)=\displaystyle\frac{1}{4} \left[\frac{x^{2+1}}{2+1} \Big|_0^4\right]\)
\(\,\,\,f(c)=\displaystyle\frac{1}{4} \left[\frac{x^{3}}{3} \Big|_0^4\right]\)
\(\,\,\,f(c)=\displaystyle\frac{1}{4} \left[\frac{(4)^{3}}{3}-\frac{(0)^{3}}{3}\right]\)
\(\,\,\,f(c)=\displaystyle\frac{1}{4} \left[\frac{64}{3}-0\right]\)
\(\,\,\,f(c)=\displaystyle \frac{16}{3}\)
\(\,\,\,c^2=\displaystyle \frac{16}{3}\)
\(\,\,\,c=\displaystyle \pm \sqrt{\frac{16}{3}}\)
\(\,\,\,c= \pm \frac{4\sqrt{3}}{3}≈ \pm 2.309\)
\(\,\,\,-2.309\) is not in \([0,4]\) so exclude it.
\(\,\,\,\)The answer is \(c= \frac{4\sqrt{3}}{3}≈ 2.309\)

 

\(\textbf{8)}\) The average value of \(f(x)\) over the interval \([3,9]\) is \(11\).
Find \(\displaystyle\int_{3}^{9}f(x) \, dx\)

Show Answer

The answer is \(66\)

Show Work

\(\,\,\,f_{avg}=\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x) \, dx\)
\(\,\,\,11=\displaystyle\frac{1}{9-3}\int_{3}^{9}f(x) \, dx\)
\(\,\,\,11=\displaystyle\frac{1}{6}\int_{3}^{9}f(x) \, dx\)
\(\,\,\,66=\displaystyle\int_{3}^{9}f(x) \, dx\)
\(\,\,\,\)The answer is \(66\)

 

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In Summary

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.