Answer
$\pi^2 \leq \int_{\pi}^{2\pi} (x-2sin~x)~dx \leq \frac{5\pi^2}{3}+\sqrt{3}~\pi$
Work Step by Step
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:
$\pi(2\pi-\pi) \leq \int_{\pi}^{2\pi} (x-2sin~x)~dx \leq (\frac{5\pi}{3}+\sqrt{3})(2\pi-\pi)$
$\pi^2 \leq \int_{\pi}^{2\pi} (x-2sin~x)~dx \leq \frac{5\pi^2}{3}+\sqrt{3}~\pi$
