Question

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

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

Answer 1

The correct option is D $4\pi$

$A={\int }_0^{\pi }xsinxdx+{\int }_0^{2\pi }x(-sinx)dxA={[-xcosx+sinx]}_0^{\pi }-{[-xcosx+sinx]}_{\pi }^{2\pi }A=(-\pi (-1\left)\right)-\left[\right(-2\pi +0)-(-\pi (-1)+0\left)\right]A=\pi -[-2\pi -\pi ]A=4\pi$