Chapter 6.3, Problem 11E

To determine

To find: The value of $x$ .

Answer to Problem 11E

The value of $x$ is $1$ .

Explanation of Solution

Given information: $y=x^2-1,\left[0,\sqrt{3}\right]$

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

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

Since $f\left(x\right)=x^2-1,a=0\&b=\sqrt{3}$ ,

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

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

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

Now, solving for $x$ ,

  $\begin{array}{l}\therefore x^2-1=0\\ x^2=1\\ x=\pm 1\end{array}$

Therefore, $x=1$ because $x\in \left[0,\sqrt{3}\right]$