Answer
\[{{f}_{avg}}=\frac{20}{\pi }\]
Work Step by Step
\[\begin{align}
& f\left( x \right)=10\sin x,\text{ on the interval }0\le x\le \pi \\
& \text{The average value is given by} \\
& {{f}_{avg}}=\frac{1}{b-a}\int_{a}^{b}{f\left( x \right)dx} \\
& \text{Therefore,} \\
& {{f}_{avg}}=\frac{1}{\pi -0}\int_{0}^{\pi }{10\sin xdx} \\
& {{f}_{avg}}=\frac{10}{\pi }\int_{0}^{\pi }{\sin xdx} \\
& \text{Integrating} \\
& {{f}_{avg}}=\frac{10}{\pi }\left[ -\cos x \right]_{0}^{\pi } \\
& {{f}_{avg}}=-\frac{10}{\pi }\left[ \cos x \right]_{0}^{\pi } \\
& {{f}_{avg}}=-\frac{10}{\pi }\left( \cos \pi -\cos 0 \right) \\
& {{f}_{avg}}=-\frac{10}{\pi }\left( -2 \right) \\
& {{f}_{avg}}=\frac{20}{\pi } \\
\end{align}\]
