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

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(B) The area $A$ is given by the integral of $|y| dx$ from $x = 0$ to $x = \frac{3\pi}{2}$.$A = \int_{0}^{\frac{3\pi}{2}} |\cos x| dx$We know that $\c

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$3$

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(B) The area $A$ is given by the integral of $|y| dx$ from $x = 0$ to $x = \frac{3\pi}{2}$.$A = \int_{0}^{\frac{3\pi}{2}} |\cos x| dx$We know that $\cos x > 0$ for $x \in [0, \frac{\pi}{2}]$ and $\cos x So,$A = \int_{0}^{\frac{\pi}{2}} \cos x dx + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} (-\cos x) dx$$A = [\sin x]_{0}^{\frac{\pi}{2}} - [\sin x]_{\frac{\pi}{2}}^{\frac{3\pi}{2}}$$A = (\sin \frac{\pi}{2} - \sin 0) - (\sin \frac{3\pi}{2} - \sin \frac{\pi}{2})$$A = (1 - 0) - (-1 - 1)$$A = 1 - (-2) = 1 + 2 = 3$ square units.Thus,the correct option is $B$.

The area of the region bounded by the curve $y = \cos x$ between $x = 0$ and $x = \frac{3\pi}{2}$ is . . . . . . square units.

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