If int_{0}^{2}(sqrt{2x}-sqrt{2x-x^{2}}) dx = | vedclass

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(C) Let $LHS = \int_{0}^{2}(\sqrt{2x}-\sqrt{2x-x^{2}}) dx$. Evaluating the first part: $\int_{0}^{2}\sqrt{2x} dx = \sqrt{2} [\frac{x^{3/2}}{3/2}]_{0}^

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$\int_{0}^{1}(1-\sqrt{1-y^{2}}) dy$

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(C) Let $LHS = \int_{0}^{2}(\sqrt{2x}-\sqrt{2x-x^{2}}) dx$. Evaluating the first part: $\int_{0}^{2}\sqrt{2x} dx = \sqrt{2} [\frac{x^{3/2}}{3/2}]_{0}^{2} = \sqrt{2} \cdot \frac{2}{3} \cdot 2\sqrt{2} = \frac{8}{3}$. Evaluating the second part: $\int_{0}^{2}\sqrt{1-(x-1)^{2}} dx$. Let $x-1 = \sin 	heta$,then $dx = \cos 	heta d	heta$. The integral becomes $\int_{-\pi/2}^{\pi/2} \cos^{2} 	heta d	heta = \frac{\pi}{2}$. So,$LHS = \frac{8}{3} - \frac{\pi}{2}$. Now,evaluating the $RHS$ terms: $\int_{0}^{1}(1-\sqrt{1-y^{2}}-\frac{y^{2}}{2}) dy = [y - \frac{1}{2}(y\sqrt{1-y^{2}} + \sin^{-1} y) - \frac{y^{3}}{6}]_{0}^{1} = 1 - \frac{\pi}{4} - \frac{1}{6} = \frac{5}{6} - \frac{\pi}{4}$. $\int_{1}^{2}(2-\frac{y^{2}}{2}) dy = [2y - \frac{y^{3}}{6}]_{1}^{2} = (4 - \frac{8}{6}) - (2 - \frac{1}{6}) = 2 - \frac{7}{6} = \frac{5}{6}$. Thus,$LHS = \frac{5}{6} - \frac{\pi}{4} + \frac{5}{6} + I = \frac{5}{3} - \frac{\pi}{4} + I$. Equating $LHS$ and $RHS$: $\frac{8}{3} - \frac{\pi}{2} = \frac{5}{3} - \frac{\pi}{4} + I$. $I = \frac{3}{3} - \frac{\pi}{4} = 1 - \frac{\pi}{4} = \int_{0}^{1}(1-\sqrt{1-y^{2}}) dy$.

If $\int_{0}^{2}(\sqrt{2x}-\sqrt{2x-x^{2}}) dx = \int_{0}^{1}(1-\sqrt{1-y^{2}}-\frac{y^{2}}{2}) dy + \int_{1}^{2}(2-\frac{y^{2}}{2}) dy + I$,then $I = \dots$

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