Question

Find the area bounded by $y=x|\sin x|$ and $x-$axis between $x=0,x=2\pi$.

Answer 1

We have,

$y=x|\sin x|$

we know that

$0<x<2\pi ,\sin x>0$

$\pi <x<2\pi ,\sin x<0$

Therefore,

Area$={\int }_0^{\pi }\underset{I}{x}\underset{II}{\sin }xdx-{\int }_{\pi }^{2\pi }\underset{I}{x}\underset{II}{\sin }xdx$

$={\left[x\right(-\cos x\left)\right]}_0^{\pi }-{\int }_0^{\pi }1(-\cos x)dx-\left[\right[x(-\cos x){]}_{\pi }^{2\pi }-{\int }_{\pi }^{2\pi }1(-\cos 3x)dx]$

$=-[\pi \cos \pi -0]+\left(\sin x{)}_0^{\pi }\right[-[2\pi \cos 2\pi -\pi \cos \pi ]+\left(\sin x{)}_{\pi }^{2\pi }\right]$

$=-[\pi x-1]+\sin \pi -\sin 0-\left[\right[-2\pi \times 1+\pi x-1]+\sin 2\pi -\sin \pi ]$

$=-\pi +0-0+2\pi +\pi +0-0$

$=2x$.