Question

Find the value of:

${\int }_0^{2\pi }\left|\sin x\right|dx$

Answer 1

$\begin{array}{c}\left|\sin x\right|=\left\{\begin{array}{c}\sin x0\le x\le \pi \\ -\sin x\pi \le x\le 2\pi \end{array}\right\}\\ \therefore {\int }_0^{2\pi }\left|\sin x\right|dx={\int }_0^{\pi }\left|\sin x\right|dx+{\int }_{\pi }^{2\pi }\left|\sin x\right|dx\\ ={\int }_0^{\pi }\sin xdx-{\int }_0^{2\pi }\sin xdx\\ ={\left[-\cos x\right]}_0^{\pi }+{\left[\cos x\right]}_0^{2\pi }\\ =2+2\\ =4\end{array}$