Mathematics - Area Under Curves Question with Solution | TestHub

On $[0,2\pi ],y={\sin }^{-1}|\sin x|$ is the triangle wave $x,\pi -x,x-\pi ,2\pi -x$ on successive quarter pieces, so by symmetry the area is $4{\int }_0^{\pi /2}\left|x-x^2\right|dx=4{\int }_0^1\left(x-x^2\right)dx+4{\int }_1^{\pi /2}\left(x^2-x\right)dx=4\left(\frac{1}{6}\right)+4\left(\frac{{\pi }^3}{24}-\frac{{\pi }^2}{8}+\frac{1}{6}\right)=$ $\frac{4}{3}+\frac{{\pi }^2(\pi -3)}{6}$.