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Math Help - How to show that it's equal (Intergral)

  1. #1
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    Question How 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
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  2. #2
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    Quote Originally Posted by noppawit View Post
    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
    There are numerous approaches to show that each integral is equal to \frac{n! m!}{(n + m + 1)!}. Here is one approach:

    Using integration by parts:


    \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:


    \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:

    \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 \int_{0}^{1} x^n (1 - x)^m \, dx has the same value.
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  3. #3
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    If you use the substitution:

    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.
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