int_{-2}^2 (4-x^2)^{frac{5}{2}} dx = (in pi) | vedclass

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(B) Since $f(x) = (4-x^2)^{5/2}$ is an even function, we have $I = 2 \int_0^2 (4-x^2)^{5/2} dx$. Let $x = 2 \sin 	heta$, then $dx = 2 \cos 	heta d

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(B) Since $f(x) = (4-x^2)^{5/2}$ is an even function, we have $I = 2 \int_0^2 (4-x^2)^{5/2} dx$. Let $x = 2 \sin 	heta$, then $dx = 2 \cos 	heta d	heta$. When $x = 0, 	heta = 0$ and when $x = 2, 	heta = \frac{\pi}{2}$. $I = 2 \int_0^{\pi/2} (4 - 4 \sin^2 	heta)^{5/2} (2 \cos 	heta) d	heta$. $I = 2 \int_0^{\pi/2} (4 \cos^2 	heta)^{5/2} (2 \cos 	heta) d	heta = 2 \int_0^{\pi/2} (32 \cos^5 	heta) (2 \cos 	heta) d	heta$. $I = 128 \int_0^{\pi/2} \cos^6 	heta d	heta$. Using Wallis' formula, $\int_0^{\pi/2} \cos^n 	heta d	heta = \frac{(n-1)(n-3)...(1)}{n(n-2)...(2)} 	imes \frac{\pi}{2}$ for even $n$. $I = 128 	imes \frac{5 	imes 3 	imes 1}{6 	imes 4 	imes 2} 	imes \frac{\pi}{2} = 128 	imes \frac{15}{48} 	imes \frac{\pi}{2} = 128 	imes \frac{5}{16} 	imes \frac{\pi}{2} = 8 	imes 5 	imes \frac{\pi}{2} = 20 \pi$.

$\int_{-2}^2 (4-x^2)^{\frac{5}{2}} dx = $ (in $\pi$)

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