The area of the region under the curve y=|sin | vedclass

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(C) The area is given by the integral $A = \int_0^{\frac{\pi}{2}} |\sin x - \cos x| \, dx$. Since $\cos x \geq \sin x$ for $0 \leq x \leq \frac{\pi}{4

Vedclass · Accepted Answer

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

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(C) The area is given by the integral $A = \int_0^{\frac{\pi}{2}} |\sin x - \cos x| \, dx$. Since $\cos x \geq \sin x$ for $0 \leq x \leq \frac{\pi}{4}$ and $\sin x \geq \cos x$ for $\frac{\pi}{4} \leq x \leq \frac{\pi}{2}$,we split the integral: $A = \int_0^{\frac{\pi}{4}} (\cos x - \sin x) \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\sin x - \cos x) \, dx$. Evaluating the first part: $[\sin x + \cos x]_0^{\frac{\pi}{4}} = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1) = \sqrt{2} - 1$. Evaluating the second part: $[-\cos x - \sin x]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = (-0 - 1) - (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) = -1 + \sqrt{2} = \sqrt{2} - 1$. Adding both parts: $A = (\sqrt{2} - 1) + (\sqrt{2} - 1) = 2(\sqrt{2} - 1)$ square units.

The area of the region under the curve $y=|\sin x-\cos x|$,$0 \leq x \leq \frac{\pi}{2}$ and above the $x$-axis,is (in square units)

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