## Calculus: Early Transcendentals 8th Edition

$\pi^2 \leq \int_{\pi}^{2\pi} (x-2sin~x)~dx \leq \frac{5\pi^2}{3}+\sqrt{3}~\pi$
On the interval $\pi \leq x \leq 2\pi$: $\pi \leq (x-2sin~x) \leq \frac{5\pi}{3}+\sqrt{3}$ Therefore, by Property 8: $\pi(2\pi-\pi) \leq \int_{\pi}^{2\pi} (x-2sin~x)~dx \leq (\frac{5\pi}{3}+\sqrt{3})(2\pi-\pi)$ $\pi^2 \leq \int_{\pi}^{2\pi} (x-2sin~x)~dx \leq \frac{5\pi^2}{3}+\sqrt{3}~\pi$