The area bounded by the curves y = cos x and | vedclass

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(D) The area $A$ is given by the integral of the absolute difference between the two functions over the interval $[0, \frac{3\pi}{2}]$.$A = \int_{0}^{

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$4\sqrt{2} - 2$

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(D) The area $A$ is given by the integral of the absolute difference between the two functions over the interval $[0, \frac{3\pi}{2}]$.$A = \int_{0}^{\frac{3\pi}{2}} |\cos x - \sin x| dx$We find the intersection points where $\cos x = \sin x$,which occurs at $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$ within the given range.$A = \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) dx + \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (\sin x - \cos x) dx + \int_{\frac{5\pi}{4}}^{\frac{3\pi}{2}} (\cos x - \sin x) dx$Evaluating the integrals:$1$) $\int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) dx = [\sin x + \cos x]_{0}^{\frac{\pi}{4}} = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1) = \sqrt{2} - 1$$2$) $\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (\sin x - \cos x) dx = [-\cos x - \sin x]_{\frac{\pi}{4}}^{\frac{5\pi}{4}} = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) = \sqrt{2} + \sqrt{2} = 2\sqrt{2}$$3$) $\int_{\frac{5\pi}{4}}^{\frac{3\pi}{2}} (\cos x - \sin x) dx = [\sin x + \cos x]_{\frac{5\pi}{4}}^{\frac{3\pi}{2}} = (-1 + 0) - (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) = -1 + \sqrt{2}$Total Area $A = (\sqrt{2} - 1) + 2\sqrt{2} + (\sqrt{2} - 1) = 4\sqrt{2} - 2$.

The area bounded by the curves $y = \cos x$ and $y = \sin x$ between the ordinates $x = 0$ and $x = \frac{3\pi}{2}$ is:

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