Question

The area bounded by curve $y=x\sin x$ and $x-$axis, between $x=0$ and $x=2\pi ,$ is equal to

• A. $2\pi$ sq. units
• B. $3\pi$ sq. units
• C. $4\pi$ sq. units
• D. $6\pi$ sq. units

Answer 1

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

The given curve is $y=x\sin x.$
Clearly, $y\ge 0$ when $0\le x\le \pi$ and $y\le 0$ when $\pi \le x\le 2\pi$

$\therefore$ Required Area $=\underset{0}{\overset{\pi }{\int }}ydx+\left|\underset{\pi }{\overset{2\pi }{\int }}ydx\right|$
$=\underset{0}{\overset{\pi }{\int }}\left(x\sin x\right)dx+\left|\underset{\pi }{\overset{2\pi }{\int }}x\sin xdx\right|$
$={[-x\cos x+\sin x]}_0^{\pi }+\left|{[-x\cos x+\sin x]}_{\pi }^{2\pi }\right|$
$=\pi +\left|\right(-2\pi )-(\pi \left)\right|$
$=\pi +|-3\pi |=\pi +3\pi =4\pi$ sq. units