The area (in sq. units) of the region bounded | vedclass

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(A) The area $A$ is given by the integral of the absolute difference between the two functions from $x=0$ to $x=\frac{\pi}{2}$.$A = \int_{0}^{\frac{\p

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

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(A) The area $A$ is given by the integral of the absolute difference between the two functions from $x=0$ to $x=\frac{\pi}{2}$.$A = \int_{0}^{\frac{\pi}{2}} |\sin x - \cos x| \, dx$.The curves $\sin x$ and $\cos x$ intersect at $x=\frac{\pi}{4}$ in the interval $[0, \frac{\pi}{2}]$.For $0 \le x \le \frac{\pi}{4}$,$\cos x \ge \sin x$. For $\frac{\pi}{4} \le x \le \frac{\pi}{2}$,$\sin x \ge \cos x$.Thus,$A = \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\sin x - \cos x) \, dx$.Evaluating the first integral: $[\sin x + \cos x]_{0}^{\frac{\pi}{4}} = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1) = \sqrt{2} - 1$.Evaluating the second integral: $[-\cos x - \sin x]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = (-0 - 1) - (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) = -1 + \sqrt{2}$.Total area $A = (\sqrt{2} - 1) + (\sqrt{2} - 1) = 2(\sqrt{2} - 1)$ sq. units.

The area (in sq. units) of the region bounded by the lines $x=0, x=\frac{\pi}{2}$ and $f(x)=\sin x, g(x)=\cos x$ is

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