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(B) The area $A$ is given by the integral $\int_{0}^{2\pi} |\cos x| \, dx$.We divide the interval $[0, 2\pi]$ based on the sign of $\cos x$:$1$. For $

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$4$

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(B) The area $A$ is given by the integral $\int_{0}^{2\pi} |\cos x| \, dx$.We divide the interval $[0, 2\pi]$ based on the sign of $\cos x$:$1$. For $x \in [0, \pi/2]$,$\cos x \ge 0$.$2$. For $x \in [\pi/2, 3\pi/2]$,$\cos x \le 0$.$3$. For $x \in [3\pi/2, 2\pi]$,$\cos x \ge 0$.Thus,the area is:$A = \int_{0}^{\pi/2} \cos x \, dx + \int_{\pi/2}^{3\pi/2} (-\cos x) \, dx + \int_{3\pi/2}^{2\pi} \cos x \, dx$Calculating each part:$\int_{0}^{\pi/2} \cos x \, dx = [\sin x]_{0}^{\pi/2} = 1 - 0 = 1$$\int_{\pi/2}^{3\pi/2} -\cos x \, dx = -[\sin x]_{\pi/2}^{3\pi/2} = -(-1 - 1) = 2$$\int_{3\pi/2}^{2\pi} \cos x \, dx = [\sin x]_{3\pi/2}^{2\pi} = 0 - (-1) = 1$Total area $A = 1 + 2 + 1 = 4 \, 	ext{sq. units}$.

The area between the curve $y = \cos x$ and the $x$-axis for $0 \le x \le 2\pi$ is:

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