The area of the region bounded by the curve y | vedclass

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(C) The area $A$ is given by the integral of the absolute value of the function: $A = \int_{0}^{3\pi/2} |\cos x| \, dx$. Since $\cos x$ is positive in

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$3$

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(C) The area $A$ is given by the integral of the absolute value of the function: $A = \int_{0}^{3\pi/2} |\cos x| \, dx$. Since $\cos x$ is positive in $[0, \pi/2]$ and negative in $[\pi/2, 3\pi/2]$,we split the integral: $A = \int_{0}^{\pi/2} \cos x \, dx + \int_{\pi/2}^{3\pi/2} -\cos x \, dx$. Evaluating the first part: $[\sin x]_{0}^{\pi/2} = \sin(\pi/2) - \sin(0) = 1 - 0 = 1$. Evaluating the second part: $-[\sin x]_{\pi/2}^{3\pi/2} = -(\sin(3\pi/2) - \sin(\pi/2)) = -(-1 - 1) = -(-2) = 2$. Total area $A = 1 + 2 = 3$ square units.

The area of the region bounded by the curve $y = \cos x$,$x = 0$,and $x = \frac{3\pi}{2}$ is:

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