The area bounded by the curve y = k sin x bet | vedclass

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Question

(A) The area $A$ bounded by the curve $y = f(x)$ from $x = a$ to $x = b$ is given by $A = \int_{a}^{b} |f(x)| \, dx$. Here,$f(x) = k \sin x$,$a = \pi$

Vedclass · Accepted Answer

$2k$ sq. unit

Vedclass · Answer

(A) The area $A$ bounded by the curve $y = f(x)$ from $x = a$ to $x = b$ is given by $A = \int_{a}^{b} |f(x)| \, dx$. Here,$f(x) = k \sin x$,$a = \pi$,and $b = 2\pi$. In the interval $[\pi, 2\pi]$,$\sin x$ is negative. Therefore,the area is $A = \int_{\pi}^{2\pi} |k \sin x| \, dx = \int_{\pi}^{2\pi} -k \sin x \, dx$. $A = -k [-\cos x]_{\pi}^{2\pi} = k [\cos x]_{\pi}^{2\pi}$. $A = k (\cos(2\pi) - \cos(\pi)) = k (1 - (-1)) = k(1 + 1) = 2k$. Thus,the required area is $2k$ sq. unit.

The area bounded by the curve $y = k \sin x$ between $x = \pi$ and $x = 2\pi$ is:

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