Answer
\[x={{\sin }^{-1}}\left( \frac{2}{\pi } \right)\]
Work Step by Step
\[\begin{align}
& f\left( x \right)=\frac{\pi }{4}\sin x\text{ on }\left[ 0,\pi \right] \\
& \text{The average is given by} \\
& {{f}_{avg}}=\frac{1}{b-a}\int_{a}^{b}{f\left( x \right)dx} \\
& \text{Therefore,} \\
& {{f}_{avg}}=\frac{1}{\pi -0}\int_{0}^{\pi }{\left( \frac{\pi }{4}\sin x \right)dx} \\
& {{f}_{avg}}=\frac{1}{4}\int_{0}^{\pi }{\sin xdx} \\
& \text{Integrating} \\
& {{f}_{avg}}=\frac{1}{4}\left[ -\cos x \right]_{0}^{\pi } \\
& {{f}_{avg}}=\frac{1}{4}\left[ -\cos \pi +\cos 0 \right] \\
& {{f}_{avg}}=\frac{1}{4}\left[ 2 \right] \\
& {{f}_{avg}}=\frac{1}{2} \\
& \text{The point at which the given function equals its average value is} \\
& \frac{\pi }{4}\sin x=\frac{1}{2} \\
& \text{Solve for }x \\
& \sin x=\frac{2}{\pi } \\
& \text{For the interval }\left[ 0,\pi \right] \\
& \sin x=\frac{2}{\pi }\Rightarrow x={{\sin }^{-1}}\left( \frac{2}{\pi } \right) \\
\end{align}\]
