Q. The area bounded by the curve, $y=x \cdot \sin x$ the $x$-axis and the lines $x=0$ and $x=2 \pi$ is

KCETKCET 2022 Report Error

$2 \pi$

$6 \pi$

$4 \pi$

$\pi$

Solution:

If $0< x< \pi, y=x \cdot \sin x$ is positive and if $\pi< x< 2 \pi$,
$y=x \cdot \sin x$ is negative. $\therefore$ the area of the region
$=\int\limits_{0}^{\pi} x \cdot \sin x d x+\int\limits_{\pi}^{2 \pi}-x \cdot \sin x d x$
$=[-x \cdot \cos x+\sin x]_{0}^{\pi}-[-x \cdot \cos x+\sin x]_{\pi}^{2 \pi} $
$=-\pi \cdot \cos \pi-(-2 \pi \cos 2 \pi+\pi \cos \pi)=4 \pi$

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