The area bounded by the curve y = sin left(fr | vedclass

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(C) The required area is given by the integral $\int_{0}^{3\pi} y \, dx = \int_{0}^{3\pi} \sin \left(\frac{x}{3}\right) \, dx$. Let $t = \frac{x}{3}$,

Vedclass · Accepted Answer

$6$

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(C) The required area is given by the integral $\int_{0}^{3\pi} y \, dx = \int_{0}^{3\pi} \sin \left(\frac{x}{3}\right) \, dx$. Let $t = \frac{x}{3}$,then $dx = 3 \, dt$. When $x = 0$,$t = 0$. When $x = 3\pi$,$t = \pi$. Substituting these into the integral,we get: Area $= \int_{0}^{\pi} \sin(t) \cdot 3 \, dt = 3 \int_{0}^{\pi} \sin(t) \, dt$. $= 3 [-\cos(t)]_{0}^{\pi}$. $= -3 [\cos(\pi) - \cos(0)]$. $= -3 [-1 - 1] = -3(-2) = 6$. Thus,the area is $6$ square units.

The area bounded by the curve $y = \sin \left(\frac{x}{3}\right)$,the $x$-axis,and the lines $x = 0$ and $x = 3\pi$ is

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