Area of the region bounded by the curve y = c | vedclass

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(A) The area $A$ is given by the integral of the absolute value of the function $y = \cos x$ from $x = -\frac{\pi}{2}$ to $x = \pi$. $A = \int_{-\frac

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$3$

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(A) The area $A$ is given by the integral of the absolute value of the function $y = \cos x$ from $x = -\frac{\pi}{2}$ to $x = \pi$. $A = \int_{-\frac{\pi}{2}}^{\pi} |\cos x| \, dx$. Since $\cos x \ge 0$ for $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ and $\cos x \le 0$ for $x \in [\frac{\pi}{2}, \pi]$,we split the integral: $A = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x \, dx + \int_{\frac{\pi}{2}}^{\pi} (-\cos x) \, dx$. Evaluating the first part: $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x \, dx = [\sin x]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \sin(\frac{\pi}{2}) - \sin(-\frac{\pi}{2}) = 1 - (-1) = 2$. Evaluating the second part: $\int_{\frac{\pi}{2}}^{\pi} -\cos x \, dx = [-\sin x]_{\frac{\pi}{2}}^{\pi} = -(\sin \pi - \sin \frac{\pi}{2}) = -(0 - 1) = 1$. Total area $A = 2 + 1 = 3$ sq. units.

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

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