If int_0^a sqrt{frac{a-x}{x}} dx = frac{k}{2} | vedclass

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(A) Let $I = \int_0^a \sqrt{\frac{a-x}{x}} dx$. Substitute $x = a \sin^2 	heta$,then $dx = 2a \sin 	heta \cos 	heta d	heta$. When $x = 0$,$	heta

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$\pi a$

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(A) Let $I = \int_0^a \sqrt{\frac{a-x}{x}} dx$. Substitute $x = a \sin^2 	heta$,then $dx = 2a \sin 	heta \cos 	heta d	heta$. When $x = 0$,$	heta = 0$. When $x = a$,$	heta = \frac{\pi}{2}$. $I = \int_0^{\frac{\pi}{2}} \sqrt{\frac{a - a \sin^2 	heta}{a \sin^2 	heta}} (2a \sin 	heta \cos 	heta) d	heta$ $I = \int_0^{\frac{\pi}{2}} \frac{\cos 	heta}{\sin 	heta} (2a \sin 	heta \cos 	heta) d	heta$ $I = 2a \int_0^{\frac{\pi}{2}} \cos^2 	heta d	heta$ Using the identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$,we get: $I = 2a \int_0^{\frac{\pi}{2}} \frac{1 + \cos 2	heta}{2} d	heta = a \int_0^{\frac{\pi}{2}} (1 + \cos 2	heta) d	heta$ $I = a \left[ 	heta + \frac{\sin 2	heta}{2} ight]_0^{\frac{\pi}{2}} = a \left( \frac{\pi}{2} + 0 - 0 - 0 ight) = \frac{a\pi}{2}$. Given $\int_0^a \sqrt{\frac{a-x}{x}} dx = \frac{k}{2}$,we have $\frac{a\pi}{2} = \frac{k}{2}$. Therefore,$k = \pi a$.

If $\int_0^a \sqrt{\frac{a-x}{x}} dx = \frac{k}{2}$,then $k = $

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