Answer
$\dfrac{2}{\pi} $
Work Step by Step
We need to use the fundamental Theorem of calculus. $\overline{f}=\dfrac{1}{b-a}\int_a^b f(x) dx $
Now, $\overline{f}=\dfrac{1}{\pi-0}\int_0^{\pi} \sin x dx\\=-\dfrac{1}{\pi}[\cos x ]_0^{\pi} \\=-\dfrac{1}{\pi}[\cos \pi-\cos (0) ]\\=-\dfrac{1}{\pi}[-1-1 ]\\=\dfrac{2}{\pi} $
