x的n次方乘(1-x)的n次方在0到1区间积分怎么计算？

Question

如题所述

Answer 1

I(n,n)
=∫(0->1) x^n .(1-x)^n dx
=[1/(n+1) ]∫(0->1) (1-x)^n d x^(n+1)
=[1/(n+1) ] [ x^(n+1). (1-x)^n ]|(0->1) +[n/(n+1) ]∫(0->1) x^(n+1) .(1-x)^(n-1) dx
=0 +[n/(n+1) ]∫(0->1) x^(n+1) .(1-x)^(n-1) dx
=[n/(n+1) ] I (n+1, n-1)
=[n/(n+1) ] [ (n-1)/(n+2)] I(n+2, n-2)
=[n/(n+1) ] [ (n-1)/(n+2)] ....[ 1/(2n) ]I(2n,0)
=[(n!)^2/(2n)!] I(2n,0)
=[(n!)^2/(2n)!] ∫(0->1) x^(2n) dx
=(n!)^2/(2n+1)!

Answer 2

x的n次方×(1-x)的m次方在【0,1】的定积分,令t=1-x,则x=1-t,dx=-dt 原式变为-t的m次方×(1-t)的n次方在[1,0]的定积分,即为t的m次方×(1-t)的n次方在[0,1]的定积分,将t换为x原式即得证.