int_{-1}^4 sqrt{frac{4-x}{x+1}} , dx = | vedclass

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(D) Let $I = \int_{-1}^4 \sqrt{\frac{4-x}{x+1}} \, dx$. Substitute $x = 4 \cos^2 	heta - 1 \sin^2 	heta$ is not ideal. Let $x+1 = 5 \cos^2 	heta$,t

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$\frac{5 \pi}{2}$

Vedclass · Answer

(D) Let $I = \int_{-1}^4 \sqrt{\frac{4-x}{x+1}} \, dx$. Substitute $x = 4 \cos^2 	heta - 1 \sin^2 	heta$ is not ideal. Let $x+1 = 5 \cos^2 	heta$,then $dx = -10 \cos 	heta \sin 	heta \, d	heta$. When $x = -1$,$5 \cos^2 	heta = 0 \implies 	heta = \frac{\pi}{2}$. When $x = 4$,$5 \cos^2 	heta = 5 \implies 	heta = 0$. Then $4-x = 4 - (5 \cos^2 	heta - 1) = 5 - 5 \cos^2 	heta = 5 \sin^2 	heta$. $I = \int_{\pi/2}^0 \sqrt{\frac{5 \sin^2 	heta}{5 \cos^2 	heta}} (-10 \cos 	heta \sin 	heta) \, d	heta = \int_0^{\pi/2} 	an 	heta (10 \cos 	heta \sin 	heta) \, d	heta$. $I = 10 \int_0^{\pi/2} \sin^2 	heta \, d	heta = 10 \int_0^{\pi/2} \frac{1 - \cos 2	heta}{2} \, d	heta$. $I = 5 [	heta - \frac{\sin 2	heta}{2}]_0^{\pi/2} = 5 [\frac{\pi}{2} - 0] = \frac{5\pi}{2}$.

$\int_{-1}^4 \sqrt{\frac{4-x}{x+1}} \, dx =$

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