Chapter 6.3, Problem 12E

To determine

To find: The value of $x$ .

Answer to Problem 12E

The values of $x$ are $-\frac{3}{2}\&\sqrt{3}$ .

Explanation of Solution

Given information: $y=-\frac{x^2}{2},\left[0,3\right]$

Formula used: $f_{avg}=\frac{1}{b-a}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$

Calculation: The given interval is $\left[0,3\right]$ .

Since $f\left(x\right)=-\frac{x^2}{2},a=0\&b=3$ ,

Now, find the $f_{avg}$ by applying the formula $f_{avg}=\frac{1}{b-a}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$ ,

  $\begin{array}{l}{f}_{avg}=\frac{1}{3-0}{\displaystyle \underset{0}{\overset{3}{\int }}f\left(-\frac{x^2}{2}\right)dx}\\ =\frac{1}{3}{\left[-\frac{x^3}{6}\right]}_0^3\\ =\frac{1}{3}{\left[-\frac{x^3}{6}\right]}_0^3\\ =\frac{1}{3}\left[-\frac{3^3}{6}+\frac{0^3}{6}\right]\\ =\frac{1}{3}\left[-\frac{9}{2}\right]\\ =-\frac{3}{2}\end{array}$

Since, $f\left(x\right)=f_{avg}$

Now, solving for $x$ ,

  $\begin{array}{l}\therefore -\frac{x}{2}=-\frac{3}{2}\\ x^2=3\\ x=\pm \sqrt{3}\end{array}$

Therefore, $x=\sqrt{3}$ because $x\in \left[0,3\right]$

Hence, the solutions are $-\frac{3}{2}\&\sqrt{3}$ .