int_3^5(x-3)^3(5-x)^5 d x= | vedclass

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(A) Let $I = \int_3^5 (x-3)^3 (5-x)^5 dx$.Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$, we have $I = \int_3^5 (5-x)^3 (x-3)^5 dx$.Alte

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$\frac{64}{63}$

Vedclass · Answer

(A) Let $I = \int_3^5 (x-3)^3 (5-x)^5 dx$.Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$, we have $I = \int_3^5 (5-x)^3 (x-3)^5 dx$.Alternatively, use the substitution $x-3 = 2t$, so $x=3+2t$ and $dx=2dt$.When $x=3, t=0$ and when $x=5, t=1$.$I = \int_0^1 (2t)^3 (5-(3+2t))^5 (2dt) = \int_0^1 8t^3 (2-2t)^5 (2dt) = 16 	imes 2^5 \int_0^1 t^3 (1-t)^5 dt$.$I = 512 \int_0^1 t^3 (1-t)^5 dt$.Using the Beta function $\int_0^1 x^{m-1} (1-x)^{n-1} dx = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)} = \frac{(m-1)!(n-1)!}{(m+n-1)!}$.Here $m-1=3 \implies m=4$ and $n-1=5 \implies n=6$.$I = 512 	imes \frac{3! 	imes 5!}{9!} = 512 	imes \frac{6 	imes 120}{362880} = 512 	imes \frac{720}{362880} = 512 	imes \frac{1}{504} = \frac{512}{504} = \frac{64}{63}$.

$\int_3^5(x-3)^3(5-x)^5 d x=$

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