Answer
$\frac{\pi^2}{2} +2$
Work Step by Step
Set up the integration to find the area under the shaded region on the interval $[0,\pi]$
$\int_0^{\pi} (x+sinx)dx$, Integrate the integrand
$[\frac{1}{2}x^2 -cosx]_0^{\pi} $, Evaluate the definite integral
$(\frac{\pi^2}{2} +1) - (0-1)$, Simplify
$\frac{\pi^2}{2} +2$
