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

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(A) The area $A$ is given by the integral of the absolute value of the function over the interval $[0, \pi]$.$A = \int_{0}^{\pi} |\cos x| \, dx$Since

Vedclass · Accepted Answer

$2$

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(A) The area $A$ is given by the integral of the absolute value of the function over the interval $[0, \pi]$.$A = \int_{0}^{\pi} |\cos x| \, dx$Since $\cos x \ge 0$ for $x \in [0, \pi/2]$ and $\cos x \le 0$ for $x \in [\pi/2, \pi]$,we split the integral:$A = \int_{0}^{\pi/2} \cos x \, dx + \int_{\pi/2}^{\pi} (-\cos x) \, dx$$A = [\sin x]_{0}^{\pi/2} - [\sin x]_{\pi/2}^{\pi}$$A = (\sin(\pi/2) - \sin(0)) - (\sin(\pi) - \sin(\pi/2))$$A = (1 - 0) - (0 - 1)$$A = 1 + 1 = 2$ sq. units.Therefore,the correct option is $A$.

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

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