Find the area bounded by the curve y=sin x be | vedclass

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(B) The graph of $y=\sin x$ is shown in the figure. The area bounded by the curve between $x=0$ and $x=2 \pi$ is given by the sum of the areas of the

Vedclass · Accepted Answer

$4 	ext{ sq. units}$

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(B) The graph of $y=\sin x$ is shown in the figure. The area bounded by the curve between $x=0$ and $x=2 \pi$ is given by the sum of the areas of the two loops.$	ext{Required Area} = \int_{0}^{\pi} \sin x \, dx + \left| \int_{\pi}^{2 \pi} \sin x \, dx ight|$$= [-\cos x]_{0}^{\pi} + \left| [-\cos x]_{\pi}^{2 \pi} ight|$$= (-\cos \pi - (-\cos 0)) + |(-\cos 2 \pi - (-\cos \pi))|$$= (-(-1) + 1) + |(-1 - (1))|$$= (1 + 1) + |-2|$$= 2 + 2 = 4 	ext{ sq. units}$.

Find the area bounded by the curve $y=\sin x$ between $x=0$ and $x=2 \pi$.

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