1. ## Integration reduction formulae

Show that $\displaystyle \displaystyle\int_0^1{x^m(1-x)^n \ dx=\frac{m!n!}{(m+n+1)!}$, for all integers $\displaystyle m,n\geq 0$

2. Let $\displaystyle u = (1 - x)^n$ so $\displaystyle du = -n(1 - x)^{n - 1}$

and $\displaystyle dv = x^m$ so $\displaystyle v = \frac{x^{m + 1}}{m + 1}$

and the integral becomes

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

Now let $\displaystyle u = (1 - x)^{n-1}$ so $\displaystyle du = -(n - 1)(1-x)^{n-2}$

and $\displaystyle dv = x^{m + 1}$ so $\displaystyle v = \frac{x^{m+2}}{m+2}$

and the integral becomes

$\displaystyle \frac{x^{m + 1}(1 - x)^n}{m + 1} +\frac{n}{m+1} \int{x^{m+1}(1-x)^{n-1}\,dx} = \frac{x^{m + 1}(1 - x)^n}{m + 1} + \frac{n}{m + 1}\left[\frac{x^{m + 2}(1 - x)^n}{m + 2} + \frac{n-1}{m + 2}\int{x^{m + 2}(1 - x)^{n-2}\,dx}\right]$

$\displaystyle = \frac{x^{m + 1}(1 - x)^n}{m + 1} + \frac{nx^{m + 2}(1 - x)^n}{(m + 2)(m + 1)} + \frac{n(n - 1)}{(m + 2)(m + 1)}\int{x^{m + 2}(1 - x)^{n - 2}\,dx}$

Notice a couple of things. Each successive integral has the powers of $\displaystyle m$ increasing by $\displaystyle 1$, while the powers of $\displaystyle n$ are decreasing by $\displaystyle 1$. So that means eventually the powers of $\displaystyle n$ will reduce to $\displaystyle 0$, meaning there will be a "final" integral. This means you will have had to take $\displaystyle n + 1$ integrals. Therefore the power of $\displaystyle m$ will become $\displaystyle m + n + 1$.

Also note that the "stuff" before each integral involves $\displaystyle x(1 - x)$. What is going to happen when you substitute $\displaystyle x = 0$ or $\displaystyle x = 1$? It will become $\displaystyle 0$. So that means the ONLY stuff worth worrying about is the stuff in the final integral.

Now let's look at the coefficient of each integral. We will be taking $\displaystyle n + 1$ integrals, and each time the numerator has the decreasing power of $\displaystyle n$ multiplied to it, and each time the denominator has the increasing power of $\displaystyle m$ multiplied to it.

So after the $\displaystyle n + 1$ integrals you will get as your integral's coefficient...

$\displaystyle \frac{n(n-1)(n-2)\dots 3\cdot 2\cdot 1}{(m + n + 1)(m + n)(m + n - 1)\dots (m + 2)(m + 1)} = \frac{n!}{\frac{(m + n + 1)!}{m!}} = \frac{m!n!}{(m + n + 1)!}$.

So after having taken all the integrals and substituted in $\displaystyle x = 0$ and $\displaystyle x = 1$ you will end up with

$\displaystyle \int_0^1{x^m(1 - x)^n\,dx}=\frac{m!n!}{(m + n + 1)!}$.