Question

The area bounded by $y=x\left|\sin x\right|$ and $X$-axis between $x=0$ and $x=2\pi$ is

• A. $2\pi$ sq. units
• B. $3\pi$ sq. units
• C. $\pi$ sq. units
• D. $4\pi$ sq. units

Answer 1

The correct option is D $4\pi$ sq. units

$y=\{\begin{array}{c}x\sin x,\text{if}x\sin x\ge 0\text{i.e.}0\le x\le x\\ -x\sin x>0,\text{if}x\sin x<0\text{i.e.}\pi <x\le 2\pi \end{array}$
Therefore, required area $={\int }_0^{\pi }x\sin xdx+{\int }_0^{2\pi }-x\sin xdx$
$={\left[x(-\cos x)\right]}_0^{\pi }-{\int }_0^{\pi }(-\cos x)dx-{\left[x(-\cos x)\right]}_0^{2\pi }+{\int }_0^{\pi }(-\cos x)dx=4\pi$ sq. units