The value of int_0^pi left| sin x - frac{2x}{ | vedclass

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(B) Let $I = \int_0^\pi \left| \sin x - \frac{2x}{\pi} \right| dx$. Consider the function $f(x) = \sin x - \frac{2x}{\pi}$. At $x = \frac{\pi}{2}$,$f(

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$\frac{\pi}{2}$

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(B) Let $I = \int_0^\pi \left| \sin x - \frac{2x}{\pi} \right| dx$. Consider the function $f(x) = \sin x - \frac{2x}{\pi}$. At $x = \frac{\pi}{2}$,$f(\frac{\pi}{2}) = \sin(\frac{\pi}{2}) - \frac{2(\pi/2)}{\pi} = 1 - 1 = 0$. For $0 \le x \le \frac{\pi}{2}$,$\sin x \ge \frac{2x}{\pi}$,so $|\sin x - \frac{2x}{\pi}| = \sin x - \frac{2x}{\pi}$. For $\frac{\pi}{2} \le x \le \pi$,$\sin x \le \frac{2x}{\pi}$,so $|\sin x - \frac{2x}{\pi}| = \frac{2x}{\pi} - \sin x$. Thus,$I = \int_0^{\pi/2} (\sin x - \frac{2x}{\pi}) dx + \int_{\pi/2}^{\pi} (\frac{2x}{\pi} - \sin x) dx$. Evaluating the integrals: $I = [-\cos x - \frac{x^2}{\pi}]_0^{\pi/2} + [\frac{x^2}{\pi} + \cos x]_{\pi/2}^{\pi}$. $I = [(-\cos(\frac{\pi}{2}) - \frac{(\pi/2)^2}{\pi}) - (-\cos(0) - 0)] + [(\frac{\pi^2}{\pi} + \cos(\pi)) - (\frac{(\pi/2)^2}{\pi} + \cos(\frac{\pi}{2}))]$. $I = [(0 - \frac{\pi}{4}) - (-1)] + [(\pi - 1) - (\frac{\pi}{4} + 0)]$. $I = 1 - \frac{\pi}{4} + \pi - 1 - \frac{\pi}{4} = \pi - \frac{\pi}{2} = \frac{\pi}{2}$.

The value of $\int_0^\pi \left| \sin x - \frac{2x}{\pi} \right| dx$ is

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