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

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Question

(C) The area $A$ is given by the definite integral of the function $y = \sin \left(\frac{x}{3}\right)$ from $x = 0$ to $x = 3\pi$.$A = \int_0^{3 \pi}

Vedclass · Accepted Answer

$6 	ext{ sq. units}$

Vedclass · Answer

(C) The area $A$ is given by the definite integral of the function $y = \sin \left(\frac{x}{3}ight)$ from $x = 0$ to $x = 3\pi$.$A = \int_0^{3 \pi} \sin \left(\frac{x}{3}ight) dx$We know that the integral of $\sin(kx)$ is $-\frac{1}{k} \cos(kx)$. Here $k = \frac{1}{3}$,so the integral is $-3 \cos \left(\frac{x}{3}ight)$.$A = \left[ -3 \cos \left(\frac{x}{3}ight) ight]_0^{3 \pi}$$A = -3 \left[ \cos \left(\frac{3 \pi}{3}ight) - \cos \left(\frac{0}{3}ight) ight]$$A = -3 [ \cos(\pi) - \cos(0) ]$Since $\cos(\pi) = -1$ and $\cos(0) = 1$:$A = -3 [ -1 - 1 ] = -3 [ -2 ] = 6 	ext{ sq. units}$.

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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