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

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(B) Given curves are $y_1 = \sin x + \cos x$ and $y_2 = |\cos x - \sin x|$ for $x \in [0, \pi/2]$.We know that $\cos x - \sin x \ge 0$ for $x \in [0,

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

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(B) Given curves are $y_1 = \sin x + \cos x$ and $y_2 = |\cos x - \sin x|$ for $x \in [0, \pi/2]$.We know that $\cos x - \sin x \ge 0$ for $x \in [0, \pi/4]$ and $\cos x - \sin x Thus,$y_2 = \cos x - \sin x$ for $x \in [0, \pi/4]$ and $y_2 = \sin x - \cos x$ for $x \in [\pi/4, \pi/2]$.The required area $A$ is given by $\int_0^{\pi/2} |y_1 - y_2| dx$.For $x \in [0, \pi/4]$:$y_1 - y_2 = (\sin x + \cos x) - (\cos x - \sin x) = 2 \sin x$.For $x \in [\pi/4, \pi/2]$:$y_1 - y_2 = (\sin x + \cos x) - (\sin x - \cos x) = 2 \cos x$.Therefore,the area $A = \int_0^{\pi/4} 2 \sin x \, dx + \int_{\pi/4}^{\pi/2} 2 \cos x \, dx$.$A = 2[-\cos x]_0^{\pi/4} + 2[\sin x]_{\pi/4}^{\pi/2}$.$A = 2\left( -\frac{1}{\sqrt{2}} - (-1) \right) + 2\left( 1 - \frac{1}{\sqrt{2}} \right)$.$A = 2\left( 1 - \frac{1}{\sqrt{2}} \right) + 2\left( 1 - \frac{1}{\sqrt{2}} \right) = 4\left( 1 - \frac{1}{\sqrt{2}} \right) = 4\left( \frac{\sqrt{2}-1}{\sqrt{2}} \right) = 2\sqrt{2}(\sqrt{2}-1)$.

The area enclosed by the curves $y=\sin x+\cos x$ and $y=|\cos x-\sin x|$ over the interval $\left[0, \frac{\pi}{2}\right]$ is

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