Area of the region bounded by y = cos x,x = - | vedclass

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(D) The area $A$ is given by the integral of the absolute value of the function over the given interval: $A = \int_{-\pi/2}^{\pi/2} |\cos x| \, dx$. S

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$2$

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(D) The area $A$ is given by the integral of the absolute value of the function over the given interval: $A = \int_{-\pi/2}^{\pi/2} |\cos x| \, dx$. Since $\cos x \geq 0$ for $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,we have: $A = \int_{-\pi/2}^{\pi/2} \cos x \, dx$. Evaluating the integral: $A = [\sin x]_{-\pi/2}^{\pi/2} = \sin(\frac{\pi}{2}) - \sin(-\frac{\pi}{2})$. $A = 1 - (-1) = 1 + 1 = 2$. Thus,the area is $2$ square units.

Area of the region bounded by $y = \cos x$,$x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$ is . . . . . . .

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