## University Calculus: Early Transcendentals (3rd Edition)

$\pi$
We need to find the area of the rectangle with limits $0$ to $\pi$. $A=\int_0^{\pi} (1+\cos x ) dx$ This implies that $[x+\sin x]_0^{\pi}=(\pi+\sin \pi)-(0+\sin 0)$ or, $(\pi+\sin \pi)-(0+\sin 0)=\pi$ Thus, the area of the shaded region is: $2 \pi-\pi =\pi$