Question

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

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

Answer 1

Answer: (C)

Given: curve y = x sinx
Limit at x-axis between x = 0 and x = 2$\pi$
y = x sinx is + ve when 0 < x < $\pi$
and y = x sinx is - ve when $\pi$ < x < 2$\pi$
$\therefore$ Area $={\int }_0^{\pi }x\sin xdx+{\int }_{\pi }^{2\pi }\left(-x\sin x\right)dx$
$=-x\cos x{|}_0^{\pi }-{\int }_0^{\pi }-\cos xdx-$
$\left[-x\cos x{|}_{\pi }^{2\pi }+{\int }_{\pi }^{2\pi }\cos xdx\right]$
$=-2\pi \cos \pi +2\sin \pi +2\pi \cos \pi -\sin 2\pi$
$=2\pi +2\pi (1)-0=4\pi$