Answer
Average value $=\displaystyle \frac{2}{\pi}$
x$\approx 0.881$
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 = $\displaystyle \frac{1}{(\pi/2)-0}\int_{0}^{\pi/2}\cos xdx$
$=[\displaystyle \frac{2}{\pi}\sin x]_{0}^{\pi/2}$
$=\displaystyle \frac{2}{\pi}(\sin\frac{\pi}{2}-\sin 0)$
$=\displaystyle \frac{2}{\pi}(1-0)$
$=\displaystyle \frac{2}{\pi}$
To find x where f(x)=average value, solve (on [0,$\pi$] )
$\displaystyle \cos x=\frac{2}{\pi}$
using a graphing calculator, (see below)
x$\approx 0.881$
