int_0^2 x^{frac{5}{2}} sqrt{2-x} , dx = | vedclass

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(C) Let $I = \int_0^2 x^{5/2} \sqrt{2-x} \, dx$. Substitute $x = 2 \sin^2 	heta$, then $dx = 4 \sin 	heta \cos 	heta \, d	heta$. When $x=0, 	heta

Vedclass · Accepted Answer

$\frac{5 \pi}{8}$

Vedclass · Answer

(C) Let $I = \int_0^2 x^{5/2} \sqrt{2-x} \, dx$. Substitute $x = 2 \sin^2 	heta$, then $dx = 4 \sin 	heta \cos 	heta \, d	heta$. When $x=0, 	heta=0$ and when $x=2, 	heta=\pi/2$. $I = \int_0^{\pi/2} (2 \sin^2 	heta)^{5/2} \sqrt{2 - 2 \sin^2 	heta} \cdot (4 \sin 	heta \cos 	heta) \, d	heta$ $I = \int_0^{\pi/2} (2^{5/2} \sin^5 	heta) (\sqrt{2} \cos 	heta) (4 \sin 	heta \cos 	heta) \, d	heta$ $I = \int_0^{\pi/2} (2^{5/2} \cdot 2^{1/2} \cdot 4) \sin^6 	heta \cos^2 	heta \, d	heta$ $I = 32 \int_0^{\pi/2} \sin^6 	heta \cos^2 	heta \, d	heta$ Using Wallis's formula $\int_0^{\pi/2} \sin^m 	heta \cos^n 	heta \, d	heta = \frac{[(m-1)(m-3)...][(n-1)(n-3)...]}{(m+n)(m+n-2)...} \cdot \frac{\pi}{2}$ (if $m, n$ are even): $I = 32 \cdot \frac{(5 \cdot 3 \cdot 1) \cdot (1)}{(8 \cdot 6 \cdot 4 \cdot 2)} \cdot \frac{\pi}{2}$ $I = 32 \cdot \frac{15}{384} \cdot \frac{\pi}{2} = 32 \cdot \frac{5}{128} \cdot \frac{\pi}{2} = \frac{5 \pi}{8}$.

$\int_0^2 x^{\frac{5}{2}} \sqrt{2-x} \, dx =$

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