Let I = int_{pi / 4}^{pi / 3} frac{sin x}{x} | vedclass

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(A) Consider the function $f(x) = \frac{\sin x}{x}$.For $x \in [\frac{\pi}{4}, \frac{\pi}{3}]$,the derivative $f'(x) = \frac{x \cos x - \sin x}{x^2} =

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$\frac{\sqrt{3}}{8} \leq I \leq \frac{\sqrt{2}}{6}$

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(A) Consider the function $f(x) = \frac{\sin x}{x}$.For $x \in [\frac{\pi}{4}, \frac{\pi}{3}]$,the derivative $f'(x) = \frac{x \cos x - \sin x}{x^2} = \frac{\cos x (x - 	an x)}{x^2}$.Since $	an x > x$ for $x \in (0, \frac{\pi}{2})$,$f'(x) Therefore,$f(\frac{\pi}{3}) \leq f(x) \leq f(\frac{\pi}{4})$ for all $x \in [\frac{\pi}{4}, \frac{\pi}{3}]$.Calculating the values: $f(\frac{\pi}{3}) = \frac{\sin(\pi/3)}{\pi/3} = \frac{\sqrt{3}/2}{\pi/3} = \frac{3\sqrt{3}}{2\pi}$ and $f(\frac{\pi}{4}) = \frac{\sin(\pi/4)}{\pi/4} = \frac{1/\sqrt{2}}{\pi/4} = \frac{4}{\pi\sqrt{2}} = \frac{2\sqrt{2}}{\pi}$.Integrating the inequality $\int_{\pi/4}^{\pi/3} f(\frac{\pi}{3}) dx \leq \int_{\pi/4}^{\pi/3} f(x) dx \leq \int_{\pi/4}^{\pi/3} f(\frac{\pi}{4}) dx$:$\frac{3\sqrt{3}}{2\pi} (\frac{\pi}{3} - \frac{\pi}{4}) \leq I \leq \frac{2\sqrt{2}}{\pi} (\frac{\pi}{3} - \frac{\pi}{4})$.$\frac{3\sqrt{3}}{2\pi} (\frac{\pi}{12}) \leq I \leq \frac{2\sqrt{2}}{\pi} (\frac{\pi}{12})$.$\frac{\sqrt{3}}{8} \leq I \leq \frac{\sqrt{2}}{6}$.

Let $I = \int_{\pi / 4}^{\pi / 3} \frac{\sin x}{x} dx$. Then

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