The area of the region A = {(x, y) : |cos x - | vedclass

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Question

(D) The given region is $|\cos x - \sin x| \leq y \leq \sin x$ for $0 \leq x \leq \frac{\pi}{2}$.First,find the intersection point of $\cos x - \sin x

Vedclass · Accepted Answer

$\sqrt{5} - 2\sqrt{2} + 1$

Vedclass · Answer

(D) The given region is $|\cos x - \sin x| \leq y \leq \sin x$ for $0 \leq x \leq \frac{\pi}{2}$.First,find the intersection point of $\cos x - \sin x = \sin x$:$\Rightarrow 	an x = \frac{1}{2}$.Let $\psi = 	an^{-1}(\frac{1}{2})$. Then $	an \psi = \frac{1}{2}$,$\sin \psi = \frac{1}{\sqrt{5}}$,and $\cos \psi = \frac{2}{\sqrt{5}}$.The area is given by $\int_{\psi}^{\pi/2} (\sin x - |\cos x - \sin x|) dx$.We split the integral at $x = \frac{\pi}{4}$ where $\cos x = \sin x$:$Area = \int_{\psi}^{\pi/4} (\sin x - (\cos x - \sin x)) dx + \int_{\pi/4}^{\pi/2} (\sin x - (\sin x - \cos x)) dx$$= \int_{\psi}^{\pi/4} (2\sin x - \cos x) dx + \int_{\pi/4}^{\pi/2} \cos x dx$$= [-2\cos x - \sin x]_{\psi}^{\pi/4} + [\sin x]_{\pi/4}^{\pi/2}$$= (-2\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) - (-2\cos \psi - \sin \psi) + (1 - \frac{1}{\sqrt{2}})$$= -\frac{3}{\sqrt{2}} + 2(\frac{2}{\sqrt{5}}) + \frac{1}{\sqrt{5}} + 1 - \frac{1}{\sqrt{2}}$$= \frac{5}{\sqrt{5}} - \frac{4}{\sqrt{2}} + 1 = \sqrt{5} - 2\sqrt{2} + 1$.

The area of the region $A = \{(x, y) : |\cos x - \sin x| \leq y \leq \sin x, 0 \leq x \leq \frac{\pi}{2}\}$ is:

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