The area of the region bounded by y=cos x,x=0 | vedclass

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(C) The area $A$ is given by the integral of $|y|$ from $x=0$ to $x=\pi$. Since $y = \cos x$ is positive in $[0, \pi/2]$ and negative in $[\pi/2, \pi]

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) The area $A$ is given by the integral of $|y|$ from $x=0$ to $x=\pi$. Since $y = \cos x$ is positive in $[0, \pi/2]$ and negative in $[\pi/2, \pi]$,the area is: $A = \int_0^{\pi/2} \cos x \, dx + \left| \int_{\pi/2}^{\pi} \cos x \, dx ight|$ $A = [\sin x]_0^{\pi/2} + |[\sin x]_{\pi/2}^{\pi}|$ $A = (\sin(\pi/2) - \sin(0)) + |\sin(\pi) - \sin(\pi/2)|$ $A = (1 - 0) + |0 - 1|$ $A = 1 + |-1| = 1 + 1 = 2 	ext{ sq. units.}$

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

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