int_0^a x(2ax - x^2)^{3/2} dx = | vedclass

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(C) Let $x = a(1 - \cos 	heta)$. Then $dx = a \sin 	heta d	heta$. When $x=0, 	heta=0$ and when $x=a, 	heta=\pi/2$. The expression $2ax - x^2 = 2a

Vedclass · Accepted Answer

$a^5 [\frac{3\pi}{16} - \frac{1}{5}]$

Vedclass · Answer

(C) Let $x = a(1 - \cos 	heta)$. Then $dx = a \sin 	heta d	heta$. When $x=0, 	heta=0$ and when $x=a, 	heta=\pi/2$. The expression $2ax - x^2 = 2a^2(1 - \cos 	heta) - a^2(1 - \cos 	heta)^2 = a^2(1 - \cos^2 	heta) = a^2 \sin^2 	heta$. Substituting these into the integral: $\int_0^a x(2ax - x^2)^{3/2} dx = \int_0^{\pi/2} a(1 - \cos 	heta) (a^2 \sin^2 	heta)^{3/2} a \sin 	heta d	heta$ $= a^5 \int_0^{\pi/2} (1 - \cos 	heta) \sin^4 	heta d	heta$ $= a^5 [\int_0^{\pi/2} \sin^4 	heta d	heta - \int_0^{\pi/2} \sin^4 	heta \cos 	heta d	heta]$ Using Wallis' formula for the first part: $\int_0^{\pi/2} \sin^4 	heta d	heta = \frac{3}{4} 	imes \frac{1}{2} 	imes \frac{\pi}{2} = \frac{3\pi}{16}$. For the second part,let $u = \sin 	heta$,then $du = \cos 	heta d	heta$: $\int_0^1 u^4 du = [\frac{u^5}{5}]_0^1 = \frac{1}{5}$. Thus,the result is $a^5 [\frac{3\pi}{16} - \frac{1}{5}]$.

$\int_0^a x(2ax - x^2)^{3/2} dx = $

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