The area bounded by the curve y=cos x,the lin | vedclass

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(A) Given,$y=\cos x$.The equation of the line joining $(-\pi/4, 1/\sqrt{2})$ and $(0,2)$ is:$y - 2 = \frac{2 - 1/\sqrt{2}}{0 - (-\pi/4)} (x - 0)$$y -

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$\left(\frac{4+\sqrt{2}}{8}\right) \pi-\sqrt{2}$

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(A) Given,$y=\cos x$.The equation of the line joining $(-\pi/4, 1/\sqrt{2})$ and $(0,2)$ is:$y - 2 = \frac{2 - 1/\sqrt{2}}{0 - (-\pi/4)} (x - 0)$$y - 2 = \frac{(2\sqrt{2}-1)/\sqrt{2}}{\pi/4} x$$y = \frac{4(2\sqrt{2}-1)}{\pi\sqrt{2}} x + 2 = \frac{4(2-\sqrt{2}/2)}{\pi} x + 2 = \frac{8-2\sqrt{2}}{\pi} x + 2$.Due to symmetry,the area of the shaded region is twice the area between the line $y = \frac{8-2\sqrt{2}}{\pi} x + 2$ and the curve $y = \cos x$ from $x=0$ to $x=\pi/4$.Area $= 2 \int_{0}^{\pi/4} \left( \left( \frac{8-2\sqrt{2}}{\pi} x + 2 \right) - \cos x \right) dx$$= 2 \left[ \frac{8-2\sqrt{2}}{\pi} \cdot \frac{x^2}{2} + 2x - \sin x \right]_{0}^{\pi/4}$$= 2 \left[ \frac{4-\sqrt{2}}{\pi} \cdot \frac{\pi^2}{16} + 2(\pi/4) - \sin(\pi/4) \right]$$= 2 \left[ \frac{(4-\sqrt{2})\pi}{16} + \frac{\pi}{2} - \frac{1}{\sqrt{2}} \right]$$= \frac{(4-\sqrt{2})\pi}{8} + \pi - \frac{2}{\sqrt{2}}$$= \left( \frac{4-\sqrt{2}+8}{8} \right) \pi - \sqrt{2} = \left( \frac{12-\sqrt{2}}{8} \right) \pi - \sqrt{2}$.Re-evaluating the slope calculation: The line joining $(0,2)$ and $(\pi/4, 1/\sqrt{2})$ has slope $m = \frac{1/\sqrt{2} - 2}{\pi/4 - 0} = \frac{1-2\sqrt{2}}{\sqrt{2}} \cdot \frac{4}{\pi} = \frac{4-8\sqrt{2}}{\pi\sqrt{2}} = \frac{2\sqrt{2}-8}{\pi}$.Thus $y = \frac{2\sqrt{2}-8}{\pi} x + 2$.Area $= 2 \int_{0}^{\pi/4} (2 + \frac{2\sqrt{2}-8}{\pi} x - \cos x) dx = 2 [2x + \frac{2\sqrt{2}-8}{\pi} \frac{x^2}{2} - \sin x]_0^{\pi/4}$$= 2 [2(\pi/4) + \frac{\sqrt{2}-4}{\pi} \frac{\pi^2}{16} - 1/\sqrt{2}] = 2 [\pi/2 + \frac{(\sqrt{2}-4)\pi}{16} - 1/\sqrt{2}] = \pi + \frac{(\sqrt{2}-4)\pi}{8} - \sqrt{2} = \frac{8\pi + \sqrt{2}\pi - 4\pi}{8} - \sqrt{2} = \frac{4+\sqrt{2}}{8}\pi - \sqrt{2}$.

The area bounded by the curve $y=\cos x$,the line joining $(-\pi / 4, \cos (-\pi / 4))$ and $(0,2)$ and the line joining $(\pi / 4, \cos (\pi / 4))$ and $(0,2)$ is

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