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Let $I=\int_{\pi / 4}^{\pi / 3} \frac{\sin x}{x} d x$. Then
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Verified Answer
The correct answer is:
$\frac{\sqrt{3}}{8} \leq 1 \leq \frac{\sqrt{2}}{6}$
We have,
$$
I=\int_{\pi / 4}^{\pi / 3} \frac{\sin x}{x} d x
$$
since, $\frac{\sin x}{x}$ is a decreasing function.
$$
\begin{array}{c}
\frac{\pi}{12} \times \frac{\sin \pi / 3}{\pi / 3} \leq 1 \leq \frac{\pi}{12} \times \frac{\sin \pi / 4}{\pi / 4} \\
\frac{\sqrt{3}}{8} \leq I \leq \frac{\sqrt{2}}{6}
\end{array}
$$
$$
I=\int_{\pi / 4}^{\pi / 3} \frac{\sin x}{x} d x
$$
since, $\frac{\sin x}{x}$ is a decreasing function.
$$
\begin{array}{c}
\frac{\pi}{12} \times \frac{\sin \pi / 3}{\pi / 3} \leq 1 \leq \frac{\pi}{12} \times \frac{\sin \pi / 4}{\pi / 4} \\
\frac{\sqrt{3}}{8} \leq I \leq \frac{\sqrt{2}}{6}
\end{array}
$$
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