Answer
See proof
Work Step by Step
For $\frac{\pi}{6} \leq x \leq \frac{\pi}{3}$, the sine function is bounded between:
$$\frac{1}{2}\leq \sin(x) \leq \frac{\sqrt 3 }{2}$$
so using the property:
$m(b-a)\leq \int_a^b f(x)dx\leq M(b-a)$
if $m\leq f(x)\leq M$ for $x\in [a,b]$
we have:
$\frac{1}{2}\left(\frac{\pi}{3}-\frac{\pi}{6}\right)\leq \displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\sin(x) dx \leq \frac{\sqrt 3 }{2}\left(\frac{\pi}{3}-\frac{\pi}{6}\right)$
$$\frac{\pi}{12}\leq \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\sin(x) dx\leq \frac{\sqrt 3 \pi }{12}$$
