Answer
$\frac{\pi}{12} \leq \int_{\pi/6}^{\pi/3} sin~x~dx \leq \frac{\sqrt{3}~\pi}{12}$
Work Step by Step
On the interval $\frac{\pi}{6} \leq x \leq \frac{\pi}{3}$:
$\frac{1}{2} \leq sin~x \leq \frac{\sqrt{3}}{2}$
Therefore, by the Comparison Property 8:
$\frac{1}{2}(\frac{\pi}{3}-\frac{\pi}{6}) \leq \int_{\pi/6}^{\pi/3} sin~x~dx \leq \frac{\sqrt{3}}{2}(\frac{\pi}{3}-\frac{\pi}{6})$
$\frac{1}{2}(\frac{\pi}{6}) \leq \int_{\pi/6}^{\pi/3} sin~x~dx \leq \frac{\sqrt{3}}{2}(\frac{\pi}{6})$
$\frac{\pi}{12} \leq \int_{\pi/6}^{\pi/3} sin~x~dx \leq \frac{\sqrt{3}~\pi}{12}$
