Q. Area bounded by the curve $y = x\sin x$ and $x$-axis between $x = 0$ and $x = 2\pi$ is
A) $2\pi$ sq unit
B) $3\pi$ sq unit
C) $4\pi$ sq unit
D) $5\pi$ sq unit
Explanation
(c)
Required area $= \int_{0}^{\pi}{x\sin{x\ dx + \left| \int_{\pi}^{2\pi}{x\sin{x\ dx}} \right|}}$
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$= \{ - x{\cos{x + \sin{x\}}}}_{0}^{\pi} + |\{ - x{\cos{x + \sin{x\}}}}_{\pi}^{2\pi}|$
$= (\pi + 0) - (0 + 0) + |( - 2\pi + 0) - (\pi + 0)$
$= \pi + 3\pi$
$= 4\pi$ sq unit
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.