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

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(C) We have $I = \int_{\pi / 4}^{\pi / 3} \frac{\sin x}{x} dx$. Since $f(x) = \frac{\sin x}{x}$ is a decreasing function on the interval $[\frac{\pi}{

Vedclass · Accepted Answer

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

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(C) We have $I = \int_{\pi / 4}^{\pi / 3} \frac{\sin x}{x} dx$. Since $f(x) = \frac{\sin x}{x}$ is a decreasing function on the interval $[\frac{\pi}{4}, \frac{\pi}{3}]$,the minimum value occurs at $x = \frac{\pi}{3}$ and the maximum value occurs at $x = \frac{\pi}{4}$. The length of the interval is $\Delta x = \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12}$. Using the property of definite integrals for a monotonic function,we have: $(	ext{length of interval}) 	imes f(	ext{upper limit}) \leq I \leq (	ext{length of interval}) 	imes f(	ext{lower limit})$. Substituting the values: $\frac{\pi}{12} 	imes \frac{\sin(\pi/3)}{\pi/3} \leq I \leq \frac{\pi}{12} 	imes \frac{\sin(\pi/4)}{\pi/4}$. $\frac{\pi}{12} 	imes \frac{\sqrt{3}/2}{\pi/3} \leq I \leq \frac{\pi}{12} 	imes \frac{1/\sqrt{2}}{\pi/4}$. $\frac{\pi}{12} 	imes \frac{3\sqrt{3}}{2\pi} \leq I \leq \frac{\pi}{12} 	imes \frac{4}{\pi\sqrt{2}}$. $\frac{3\sqrt{3}}{24} \leq I \leq \frac{4}{12\sqrt{2}}$. $\frac{\sqrt{3}}{8} \leq I \leq \frac{1}{3\sqrt{2}} = \frac{\sqrt{2}}{6}$.

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

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