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

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Aicharya · Accepted Answer

$\begin{aligned} & \int_0^\pi\left|\sin x-\frac{2 x}{\pi}ight| \mathrm{d} x \ & =\int_0^{\frac{\pi}{2}}\left(\sin x-\frac{2 x}{\pi}ight) \mathrm{d} x+\int_{\frac{\pi}{2}}^\pi\left(\frac{2 x}{\pi}-\sin xight) \mathrm{d} x \ & =[-\cos x]_0^{\frac{\pi}{2}}-\left[\frac{x^2}{\pi}ight]_0^{\frac{\pi}{2}}+\left[\frac{x^2}{\pi}ight]_{\frac{\pi}{2}}^\pi+[\cos x]_{\frac{\pi}{2}}^\pi \ & =(0+1)-\left(\frac{\pi}{4}-0ight)+\left(\pi-\frac{\pi}{4}ight)+(-1-0) \ & =\frac{\pi}{2}\end{aligned}$

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

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