Answer
\[x=\frac{a}{\sqrt{3}}\]
Work Step by Step
\[\begin{align}
& f\left( x \right)=1-\frac{{{x}^{2}}}{{{a}^{2}}}\text{ on }\left[ 0,a \right],\text{ }a\text{ is a positive real number} \\
& \text{The average is given by} \\
& {{f}_{avg}}=\frac{1}{b-a}\int_{a}^{b}{f\left( x \right)dx} \\
& \text{Therefore,} \\
& {{f}_{avg}}=\frac{1}{a-0}\int_{0}^{a}{\left( 1-\frac{{{x}^{2}}}{{{a}^{2}}} \right)dx} \\
& {{f}_{avg}}=\frac{1}{a}\int_{0}^{a}{\left( 1-\frac{{{x}^{2}}}{{{a}^{2}}} \right)dx} \\
& \text{Integrating} \\
& {{f}_{avg}}=\frac{1}{a}\left[ x-\frac{{{x}^{3}}}{3{{a}^{2}}} \right]_{0}^{a} \\
& {{f}_{avg}}=\frac{1}{a}\left[ a-\frac{{{a}^{3}}}{3{{a}^{2}}} \right]-\frac{1}{a}\left[ 0-\frac{{{0}^{3}}}{3{{a}^{2}}} \right] \\
& {{f}_{avg}}=\frac{1}{a}\left[ a-\frac{a}{3} \right] \\
& {{f}_{avg}}=\frac{2}{3} \\
& \text{The point at which the given function equals its average value is} \\
& 1-\frac{{{x}^{2}}}{{{a}^{2}}}=\frac{2}{3} \\
& \text{Solve for }x \\
& \frac{{{x}^{2}}}{{{a}^{2}}}=1-\frac{2}{3} \\
& \frac{{{x}^{2}}}{{{a}^{2}}}=\frac{1}{3} \\
& {{x}^{2}}=\frac{{{a}^{2}}}{3} \\
& x=\pm \frac{a}{\sqrt{3}} \\
& a\text{ is a positive real number},\text{ then} \\
& x=\frac{a}{\sqrt{3}} \\
\end{align}\]
