Answer
Average value $=\displaystyle \frac{2}{\pi}$
x$\approx$0,69,
x$\approx$2.451
Work Step by Step
If $f$ is integrable on the closed interval $[a, b]$, then the average value of $f$ on the interval is
$\displaystyle \frac{1}{b-a}\int_{a}^{b}f(x)dx$.
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Average value = $\displaystyle \frac{1}{\pi-0}\int_{0}^{\pi}\sin xdx$
$=[-\displaystyle \frac{1}{\pi}\cos x]_{0}^{\pi}$
$=-\displaystyle \frac{1}{\pi}(\cos\pi-\cos 0)$
$=-\displaystyle \frac{1}{\pi}(-1-1)$
$=\displaystyle \frac{2}{\pi}$
To find x where f(x)=average value, solve (on [0,$\pi$] )
$\displaystyle \sin x=\frac{2}{\pi}$
using a graphing calculator, (see below)
x$\approx$0,69,
x$\approx$2.451
