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

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(D) The area $A$ is given by the integral of the absolute value of the function: $A = \int_{0}^{3\pi} |\cos x| \, dx$ Since the function $\cos x$ chan

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) The area $A$ is given by the integral of the absolute value of the function: $A = \int_{0}^{3\pi} |\cos x| \, dx$ Since the function $\cos x$ changes sign at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$ and $x = \frac{5\pi}{2}$,we split the integral: $A = \int_{0}^{\pi/2} \cos x \, dx - \int_{\pi/2}^{3\pi/2} \cos x \, dx + \int_{3\pi/2}^{5\pi/2} \cos x \, dx - \int_{5\pi/2}^{3\pi} \cos x \, dx$ Evaluating each part: $|\sin x|_{0}^{\pi/2} = |1 - 0| = 1$ $|\sin x|_{\pi/2}^{3\pi/2} = |-1 - 1| = |-2| = 2$ $|\sin x|_{3\pi/2}^{5\pi/2} = |1 - (-1)| = |2| = 2$ $|\sin x|_{5\pi/2}^{3\pi} = |0 - 1| = |-1| = 1$ Summing these values: $A = 1 + 2 + 2 + 1 = 6$ sq. units. Thus,the correct option is $D$.

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

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