Given that m and n are positive integers, show that

int [from 0 to 1] (x^m)(1-x)^n dx = int [from 0 to 1] (x^n)(1-x)^m dx

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- Aug 16th 2008, 05:59 PMnoppawitHow to show that it's equal (Intergral)
Given that m and n are positive integers, show that

int [from 0 to 1] (x^m)(1-x)^n dx = int [from 0 to 1] (x^n)(1-x)^m dx - Aug 16th 2008, 07:54 PMmr fantastic
There are numerous approaches to show that each integral is equal to $\displaystyle \frac{n! m!}{(n + m + 1)!}$. Here is one approach:

Using integration by parts:

$\displaystyle \beta (m, n) = \int_{0}^{1} x^m (1 - x)^n \, dx = \frac{n}{m+1} \int_{0}^{1} x^{m+1} (1 - x)^{n-1} \, dx$.

In fact, using integration by parts t times you get:

$\displaystyle \beta (m, n) = \frac{n (n-1) (n-2) \, .... \, (n - [t-1])}{(m+1) (m+2) (m+3) \, .... \, (m+t)} \int_{0}^{1} x^{m+t} (1 - x)^{n-t} \, dx$.

Let t = n:

$\displaystyle \beta (m, n) = \frac{n (n-1) (n-2) \, .... \, (1)}{(m+1) (m+2) (m+3) \, .... \, (m+t)} \int_{0}^{1} x^{m+n} \, dx$.

The final integration is simple.

From the symmetry of the answer, it's clear that $\displaystyle \int_{0}^{1} x^n (1 - x)^m \, dx$ has the same value. - Aug 17th 2008, 12:27 AMCoomast
If you use the substitution:

$\displaystyle 1-x=t$

on the left hand side, you will almost immediately find the right hand side of the equality. This might be a bit easier than mr fantastic's (perfectly valid) solution.