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

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(B) The area $A$ is given by the integral of the absolute value of the function over the given interval: $A = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} |\

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The area $A$ is given by the integral of the absolute value of the function over the given interval: $A = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} |\cos x| \, dx$ In the interval $(\frac{\pi}{2}, \frac{3\pi}{2})$,$\cos x$ is negative in the interval $(\frac{\pi}{2}, \frac{3\pi}{2})$ except at the boundaries. Specifically,$\cos x \ge 0$ for $x \in [\frac{\pi}{2}, \frac{\pi}{2}]$ (not applicable) and $\cos x \le 0$ for $x \in [\frac{\pi}{2}, \frac{3\pi}{2}]$. Thus,$|\cos x| = -\cos x$ for $x \in [\frac{\pi}{2}, \frac{3\pi}{2}]$. $A = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} -\cos x \, dx$ $A = -[\sin x]_{\frac{\pi}{2}}^{\frac{3\pi}{2}}$ $A = -(\sin(\frac{3\pi}{2}) - \sin(\frac{\pi}{2}))$ $A = -(-1 - 1) = -(-2) = 2$ Therefore,the area is $2$ sq. units.

Area of the region bounded by the curve $y=\cos x$,$x=\frac{\pi}{2}$ and $x=\frac{3 \pi}{2}$ is . . . . . . sq. units.

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