The area bounded by the curve y = x sin x and | vedclass

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(D) The required area $A$ is given by the sum of the absolute values of the integrals over the intervals where the function is positive and negative.$

Vedclass · Accepted Answer

$4\pi \, 	ext{sq. units}$

Vedclass · Answer

(D) The required area $A$ is given by the sum of the absolute values of the integrals over the intervals where the function is positive and negative.$A = \int_0^{\pi} x \sin x \, dx + \left| \int_{\pi}^{2\pi} x \sin x \, dx ight|$Using integration by parts,$\int x \sin x \, dx = -x \cos x + \sin x$.For the first part: $\int_0^{\pi} x \sin x \, dx = [-x \cos x + \sin x]_0^{\pi} = (-\pi \cos \pi + \sin \pi) - (0) = \pi$.For the second part: $\int_{\pi}^{2\pi} x \sin x \, dx = [-x \cos x + \sin x]_{\pi}^{2\pi} = (-2\pi \cos 2\pi + \sin 2\pi) - (-\pi \cos \pi + \sin \pi) = (-2\pi) - (\pi) = -3\pi$.The absolute value is $|-3\pi| = 3\pi$.Total area $A = \pi + 3\pi = 4\pi \, 	ext{sq. units}$.

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

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