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

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(A) The area bounded by the curve $y = \cos x$ from $x = 0$ to $x = \pi$ is given by the integral of the absolute value of the function: $Area = \int_

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$2$ sq. units

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(A) The area bounded by the curve $y = \cos x$ from $x = 0$ to $x = \pi$ is given by the integral of the absolute value of the function: $Area = \int_{0}^{\pi} |\cos x| dx$ Since $\cos x \geq 0$ for $x \in [0, \pi/2]$ and $\cos x \leq 0$ for $x \in [\pi/2, \pi]$,we split the integral: $Area = \int_{0}^{\pi/2} \cos x dx + \int_{\pi/2}^{\pi} (-\cos x) dx$ $Area = [\sin x]_{0}^{\pi/2} - [\sin x]_{\pi/2}^{\pi}$ $Area = (\sin(\pi/2) - \sin 0) - (\sin \pi - \sin(\pi/2))$ $Area = (1 - 0) - (0 - 1) = 1 + 1 = 2$ sq. units.

The area of the region bounded by the curve $y = \cos x$,$x = 0$,and $x = \pi$ is

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