Area bounded by the curve y = sin x between x | vedclass

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(B) The curve is $y = \sin x$. The area bounded by the curve between $x = 0$ and $x = 2\pi$ is given by the integral of the absolute value of the func

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) The curve is $y = \sin x$. The area bounded by the curve between $x = 0$ and $x = 2\pi$ is given by the integral of the absolute value of the function.Required Area $= \int_0^{2\pi} |\sin x| \, dx$Since $\sin x \ge 0$ for $x \in [0, \pi]$ and $\sin x \le 0$ for $x \in [\pi, 2\pi]$,we split the integral:Required Area $= \int_0^{\pi} \sin x \, dx + \int_{\pi}^{2\pi} (-\sin x) \, dx$$= [-\cos x]_0^{\pi} + [\cos x]_{\pi}^{2\pi}$$= -(\cos \pi - \cos 0) + (\cos 2\pi - \cos \pi)$$= -(-1 - 1) + (1 - (-1))$$= -(-2) + (2) = 2 + 2 = 4 \, sq. 	ext{ unit}$.

Area bounded by the curve $y = \sin x$ between $x = 0$ and $x = 2\pi$ is ......... $sq. \text{ unit}$.

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