The area (in sq. units) of the region bounded | vedclass

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$10$

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(D) The curves are $y=4|\cos x|$ and $y=-|\cos x|$.Since $|\cos x| \ge 0$ for all $x$,the upper curve is $y=4|\cos x|$ and the lower curve is $y=-|\cos x|$.The area $A$ is given by the integral $\int_{-\pi/2}^{\pi/2} [4|\cos x| - (-|\cos x|)] dx$.$A = \int_{-\pi/2}^{\pi/2} 5|\cos x| dx$.Since $\cos x \ge 0$ for $x \in [-\pi/2, \pi/2]$,we have $|\cos x| = \cos x$.$A = 5 \int_{-\pi/2}^{\pi/2} \cos x dx$.Using the property of even functions,$A = 5 	imes 2 \int_{0}^{\pi/2} \cos x dx$.$A = 10 [\sin x]_{0}^{\pi/2} = 10(1 - 0) = 10$ sq. units.

The area (in sq. units) of the region bounded by the curves $y=4|\cos x|$ and $y=-|\cos x|$ from $x=-\frac{\pi}{2}$ to $x=\frac{\pi}{2}$ is

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