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I am trying to show that: $$\int_{0}^{1}x^{m}(1-x)^{n}dx=\int_{0}^{1}x^{n}(1-x)^{m}dx$$ For any positive integers $n$ and $m$. Which is true if I try to evaluate it numerically.

I tried to use the binomial theorem, but then I end up with: $$ \int_{0}^{1} \sum_{i=0}^{n} \begin{bmatrix} n \\ i \end{bmatrix}x^{m+n-i} dx = \int_{0}^{1} \sum_{i=0}^{m} \begin{bmatrix} m \\ i \end{bmatrix}x^{m+n-i} dx $$

There is a term $x^{m+n-i}$ on both sides which looks like there should be something I can do with it to simplify but then I am stuck.

This is an exercise following Apostol's Calculus I Chapter 5 where integration by substitution is discussed, but it is not clear how to use integration by substitution here.

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    $\begingroup$ You can use the substitution $t = 1-x$. $\endgroup$ – Sangchul Lee Aug 8 '13 at 5:31
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Hint: Make the change of variables $1-x=t$.

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  • $\begingroup$ I have $t = 1-x$. Then $dx = -dt$, $(1-x)^{n}=t^{n}$ and $x^{m}=(1-t)^{m}$, so I get $$-\int_{0}^{1}(1-t)^{m}t^{n}dt=\int_{0}^{1}x^{n}(1-x)^{m}dx$$but there is this minus sign, what do I do with it? $\endgroup$ – Fazzolini Aug 8 '13 at 5:40
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    $\begingroup$ @Fazzolini You must also adjust the bounds of integration after the substitution; you should actually have $$- \int_1^0 ...$$ $\endgroup$ – user61527 Aug 8 '13 at 5:44
  • $\begingroup$ I forgot to change the bounds, thank you! $\endgroup$ – Fazzolini Aug 8 '13 at 5:46
  • $\begingroup$ @Mhenni: (+1) Is this integral related to Beta function $B(m,n)$? Thanks. $\endgroup$ – mrs Aug 8 '13 at 9:15
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    $\begingroup$ @DonAntonio: I see. Thanks for clarification. $\endgroup$ – mrs Aug 8 '13 at 10:51

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