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Let $\mathrm{I}=\int_{\pi / 4}^{\pi / 3} \frac{\sin \mathrm{x}}{\mathrm{x}} \mathrm{dx}$. Then
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Verified Answer
The correct answer is:
$\frac{\sqrt{3}}{8} \leq \mathrm{I} \leq \frac{\sqrt{2}}{6}$
$f(x)=\frac{\sin x}{x}$ (decreasing function)
$\Rightarrow f\left(\frac{\pi}{3}\right) < f(x) < f\left(\frac{\pi}{4}\right)$
$\Rightarrow \frac{\sqrt{3}}{2} \times \frac{3}{\pi} \int_{\pi / 4}^{\pi / 3} d x < I < \frac{1}{\sqrt{2}} \times \frac{4}{\pi} \times \int_{\pi / 4}^{\pi / 3} d x$
$\Rightarrow \frac{\sqrt{3}}{8} < I < \frac{\sqrt{2}}{6}$
$\Rightarrow f\left(\frac{\pi}{3}\right) < f(x) < f\left(\frac{\pi}{4}\right)$
$\Rightarrow \frac{\sqrt{3}}{2} \times \frac{3}{\pi} \int_{\pi / 4}^{\pi / 3} d x < I < \frac{1}{\sqrt{2}} \times \frac{4}{\pi} \times \int_{\pi / 4}^{\pi / 3} d x$
$\Rightarrow \frac{\sqrt{3}}{8} < I < \frac{\sqrt{2}}{6}$
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