Joan Thompson

2021-12-20

Find c such that fave $=f\left(c\right),f\left(x\right)={\left(x-3\right)}^{2},\left[2,5\right]$

alexandrebaud43

Average value of the function f(x) over the interval $\left[a,b\right]$ is given by
${f}_{avg}=\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)dx$
Average value of $f\left(x\right)={\left(x-3\right)}^{2}$ ovthe interval [2,5] is given by
${f}_{avg}=\frac{1}{5-2}{\int }_{2}^{5}{\left(x-3\right)}^{2}dx$
Substitute$x-3=u$ and $dx=du$
Limits of Integration will change from ${\int }_{-}{2}^{5}\to {\int }_{-}2-{3}^{5-3}={\int }_{-1}^{2}$
$=\frac{1}{3}{\int }_{-1}^{2}{u}^{2}du$
$=\frac{1}{3}{\left[\begin{array}{c}\frac{{u}^{3}}{3}\end{array}\right]}_{-1}^{2}$
$=\frac{1}{3}\left[\begin{array}{c}\frac{8}{3}+\frac{1}{3}\end{array}\right]=1$

Orlando Paz

Average value of the function f(x) over the interval [a,b] is given by
${f}_{avg}=\frac{1}{b-a}{\int }_{a}^{b}{f}_{\left(x\right)}dx$
Average value of $f\left(x\right)={\left(x-3\right)}^{2}$ over the interval $\left(2,5\right]$ is given by
${f}_{avg}=\frac{1}{5-2}{\int }_{2}^{5}\left(x-3{\right)}^{2}dx$
$=\frac{1}{3}{\left[\begin{array}{c}\frac{\left(x-3{\right)}^{3}}{3}\end{array}\right]}_{2}^{5}$
$=\frac{1}{3}\left[\begin{array}{c}\frac{8}{3}+\frac{1}{3}\end{array}\right]=\frac{1}{3}×\frac{9}{3}=1$

Do you have a similar question?