The area enclosed by the curves y=sin x+cos x | vedclass

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(A) The area $A$ is given by the integral $A = \int_{0}^{\pi/2} |(\sin x + \cos x) - |\cos x - \sin x|| dx$. Since $|\cos x - \sin x| = \cos x - \sin

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$2 \sqrt{2}(\sqrt{2}-1)$

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(A) The area $A$ is given by the integral $A = \int_{0}^{\pi/2} |(\sin x + \cos x) - |\cos x - \sin x|| dx$. Since $|\cos x - \sin x| = \cos x - \sin x$ for $0 \le x \le \pi/4$ and $\sin x - \cos x$ for $\pi/4 \le x \le \pi/2$,we split the integral: $A = \int_{0}^{\pi/4} ((\sin x + \cos x) - (\cos x - \sin x)) dx + \int_{\pi/4}^{\pi/2} ((\sin x + \cos x) - (\sin x - \cos x)) dx$. Simplifying the integrands: $A = \int_{0}^{\pi/4} 2 \sin x dx + \int_{\pi/4}^{\pi/2} 2 \cos x dx$. Evaluating the integrals: $A = 2[-\cos x]_{0}^{\pi/4} + 2[\sin x]_{\pi/4}^{\pi/2}$. $A = 2(1 - \frac{1}{\sqrt{2}}) + 2(1 - \frac{1}{\sqrt{2}})$. $A = 4(1 - \frac{1}{\sqrt{2}}) = 4 - 2\sqrt{2} = 2\sqrt{2}(\sqrt{2} - 1)$.

The area enclosed by the curves $y=\sin x+\cos x$ and $y=|\cos x-\sin x|$ and the lines $x=0, x=\frac{\pi}{2}$ is:

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