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 the absolute value of the function over the interval $[\frac{\pi}{2}, \frac{3\pi}{2}]$.$A = \int_{\frac{\

Vedclass · Accepted Answer

$2$

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(B) The area $A$ is given by the integral of the absolute value of the function over the interval $[\frac{\pi}{2}, \frac{3\pi}{2}]$.$A = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} |\cos x| \, dx$In the interval $[\frac{\pi}{2}, \frac{3\pi}{2}]$,$\cos x$ is negative in the interval $(\frac{\pi}{2}, \frac{3\pi}{2}]$.Thus,$|\cos x| = -\cos x$.$A = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} -\cos x \, dx$$A = -[\sin x]_{\frac{\pi}{2}}^{\frac{3\pi}{2}}$$A = -[\sin(\frac{3\pi}{2}) - \sin(\frac{\pi}{2})]$$A = -[-1 - 1]$$A = -[-2] = 2$ sq. units.Therefore,the correct option is $B$.

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

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