Question

Let $I=\underset{\pi /4}{\overset{\pi /3}{\int }}\frac{\sin x}{x}dx$. Then

• A. $\frac{1}{2}\le I\le 1$
• B. $4\le I\le 2\sqrt{30}$
• C. $\frac{\sqrt{3}}{8}\le I\le \frac{\sqrt{2}}{6}$
• D. $1\le I\le \frac{2\sqrt{3}}{\sqrt{2}}$

Answer 1

The correct option is C $\frac{\sqrt{3}}{8}\le I\le \frac{\sqrt{2}}{6}$

$I=\underset{\pi /4}{\overset{\pi /3}{\int }}\frac{\sin x}{x}dx$
We know that,
$\frac{\sin x}{x}$ is a decreasing function in the interval $[\frac{\pi }{4},\frac{\pi }{3}]$

$\therefore (\frac{\pi }{3}-\frac{\pi }{4})\frac{\sin x}{x}{|}_{\frac{\pi }{3}}\le I\le (\frac{\pi }{3}-\frac{\pi }{4})\frac{\sin x}{x}{|}_{\frac{\pi }{4}}\Rightarrow \frac{\pi }{12}\times \frac{\sin \frac{\pi }{3}}{\frac{\pi }{3}}\le I\le \frac{\pi }{12}\times \frac{\sin \frac{\pi }{4}}{\frac{\pi }{4}}$
$\Rightarrow \frac{\sqrt{3}}{8}\le I\le \frac{\sqrt{2}}{6}$