int_{-5}^5 x^4left(25-x^2right)^{5 / 2} d x= | vedclass

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(C) Let $I = \int_{-5}^5 x^4(25-x^2)^{5/2} dx$. Since the integrand $f(x) = x^4(25-x^2)^{5/2}$ is an even function, we have $I = 2 \int_0^5 x^4(25-x^2

Vedclass · Accepted Answer

$\frac{3\left(5^{10}\right)}{256} \pi$

Vedclass · Answer

(C) Let $I = \int_{-5}^5 x^4(25-x^2)^{5/2} dx$. Since the integrand $f(x) = x^4(25-x^2)^{5/2}$ is an even function, we have $I = 2 \int_0^5 x^4(25-x^2)^{5/2} dx$.Substitute $x = 5 \sin 	heta$, then $dx = 5 \cos 	heta d	heta$. When $x=0, 	heta=0$ and when $x=5, 	heta=\pi/2$.$I = 2 \int_0^{\pi/2} (5 \sin 	heta)^4 (25 - 25 \sin^2 	heta)^{5/2} (5 \cos 	heta) d	heta$$I = 2 \int_0^{\pi/2} 5^4 \sin^4 	heta (25 \cos^2 	heta)^{5/2} (5 \cos 	heta) d	heta$$I = 2 \int_0^{\pi/2} 5^4 \sin^4 	heta (5^5 \cos^5 	heta) (5 \cos 	heta) d	heta = 2 	imes 5^{10} \int_0^{\pi/2} \sin^4 	heta \cos^6 	heta d	heta$Using the Wallis formula $\int_0^{\pi/2} \sin^m 	heta \cos^n 	heta d	heta = \frac{(m-1)!!(n-1)!!}{(m+n)!!} 	imes \frac{\pi}{2}$ (if both $m, n$ are even):$I = 2 	imes 5^{10} 	imes \frac{(3 	imes 1) 	imes (5 	imes 3 	imes 1)}{(10 	imes 8 	imes 6 	imes 4 	imes 2)} 	imes \frac{\pi}{2}$$I = 2 	imes 5^{10} 	imes \frac{3 	imes 15}{3840} 	imes \frac{\pi}{2} = 5^{10} 	imes \frac{45}{3840} \pi = 5^{10} 	imes \frac{3}{256} \pi = \frac{3(5^{10}) \pi}{256}$.

$\int_{-5}^5 x^4\left(25-x^2\right)^{5 / 2} d x=$

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