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Q.
The area bounded by the curve y = x sin x and the x-axis between x = 0 and x = 2$\pi$ is
Application of Integrals
Solution:
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 $= \int^{\pi}_{0} x \sin xdx + \int^{2\pi}_{\pi} \left(-x \sin x \right)dx$
$ = - x \cos x\bigg|^{\pi}_0 - \int^{\pi}_{0} - \cos x dx -$
$ \left[-x \cos x \bigg|^{2\pi}_{\pi} + \int^{2\pi}_{\pi} \cos x dx \right]$
$= - 2\pi \cos \pi + 2 \sin \pi + 2\pi \cos\pi - \sin 2\pi$
$= 2\pi + 2\pi(1) -0= 4\pi $