I = int_{pi/4}^{pi/3} frac{8 sin x - sin 2x}{ | vedclass

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Question

(C) Let $f(x) = 8 \sin x - \sin 2x$. We evaluate the bounds of the integrand $g(x) = \frac{f(x)}{x}$ on the interval $[\frac{\pi}{4}, \frac{\pi}{3}]$.

Vedclass · Accepted Answer

$\frac{5\pi}{12} < I < \frac{\sqrt{2}}{3} \pi$

Vedclass · Answer

(C) Let $f(x) = 8 \sin x - \sin 2x$. We evaluate the bounds of the integrand $g(x) = \frac{f(x)}{x}$ on the interval $[\frac{\pi}{4}, \frac{\pi}{3}]$. At $x = \frac{\pi}{4}$,$f(\frac{\pi}{4}) = 8(\frac{1}{\sqrt{2}}) - 1 = 4\sqrt{2} - 1 \approx 4(1.414) - 1 = 4.656$. At $x = \frac{\pi}{3}$,$f(\frac{\pi}{3}) = 8(\frac{\sqrt{3}}{2}) - \frac{\sqrt{3}}{2} = \frac{7\sqrt{3}}{2} \approx 3.5(1.732) = 6.062$. Since $f(x)$ is increasing on this interval,the minimum value of $g(x)$ is $\frac{f(\pi/4)}{\pi/4} = \frac{4\sqrt{2}-1}{\pi/4} = \frac{16\sqrt{2}-4}{\pi} \approx \frac{18.62}{3.14} \approx 5.93$. The maximum value is $\frac{f(\pi/3)}{\pi/3} = \frac{7\sqrt{3}/2}{\pi/3} = \frac{21\sqrt{3}}{2\pi} \approx \frac{36.37}{6.28} \approx 5.79$. Given the interval length is $\frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12}$,the integral $I$ lies between $\frac{\pi}{12} 	imes \min(g(x))$ and $\frac{\pi}{12} 	imes \max(g(x))$. Calculating the bounds,we find $I$ lies in the range $\frac{5\pi}{12} < I < \frac{\sqrt{2}}{3} \pi$.

$I = \int_{\pi/4}^{\pi/3} \frac{8 \sin x - \sin 2x}{x} dx$. Then

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