Answer
$\frac{\pi}{12} \leq \int_{\pi/4}^{\pi/3} tan~x~dx \leq \frac{\sqrt{3}~\pi}{12}$
Work Step by Step
On the interval $\frac{\pi}{4} \leq x \leq \frac{\pi}{3}$:
$1 \leq tan~x \leq \sqrt{3}$
Therefore, by the Comparison Property 8:
$1(\frac{\pi}{3}-\frac{\pi}{4}) \leq \int_{\pi/4}^{\pi/3} tan~x~dx \leq \sqrt{3}(\frac{\pi}{3}-\frac{\pi}{4})$
$1(\frac{\pi}{12}) \leq \int_{\pi/4}^{\pi/3} tan~x~dx \leq \sqrt{3}(\frac{\pi}{12})$
$\frac{\pi}{12} \leq \int_{\pi/4}^{\pi/3} tan~x~dx \leq \frac{\sqrt{3}~\pi}{12}$
