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Math Help - Integration reduction formulae

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    Integration reduction formulae

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

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

    and the integral becomes

    \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 u = (1 - x)^{n-1} so du = -(n - 1)(1-x)^{n-2}

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

    and the integral becomes

    \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]

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


    Also note that the "stuff" before each integral involves x(1 - x). What is going to happen when you substitute x = 0 or x = 1? It will become 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 n + 1 integrals, and each time the numerator has the decreasing power of n multiplied to it, and each time the denominator has the increasing power of m multiplied to it.

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

    \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 x = 0 and x = 1 you will end up with

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