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Prove that: $$\forall n,m \in \Bbb N:\int_0^1 x^m(1-x)^n \,dx=\int_0^1(1-x)^mx^n \,dx$$

I really have no idea.

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closed as off-topic by Jack D'Aurizio, Alec Teal, jameselmore, Micah, yoknapatawpha Oct 23 '15 at 17:52

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    $\begingroup$ Slow day on the hot questions..... $\endgroup$ – Alec Teal Oct 23 '15 at 13:55
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Setting $t=1-x\iff x=1-t$, we have $\mathrm d\mkern1mu x=-\mathrm d\mkern1mu t$, hence the change of variable formula yields: $$\int_0^1 x^m(1-x)^n \,\mathrm d\mkern1mu x=\int_1^0(1-t)^m t^n (-\mathrm d\mkern1mu t)=\int_0^1t^n(1-t)^m \,\mathrm d\mkern1mu t=\int_0^1 x^n(1-x)^m \,\mathrm d\mkern1mu x. $$

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  • $\begingroup$ I don't see your last step. Why $\int_0^1 t^n(1-t)^m dt=\int_0^1 x^n(1-x)^m dx$ ? $\endgroup$ – Kelan Oct 23 '15 at 11:56
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    $\begingroup$ Because $t$ and $x$ are dummy variables. It is the same with \sum_{i=1}^5 (2i-1)=\sum_{k=1}^5 (2k-1): they're both equal to 25. In other words: theyre just place-holders. $\endgroup$ – Bernard Oct 23 '15 at 12:07
  • $\begingroup$ Is "dummy variable" the official name of it? $\endgroup$ – Ooker Oct 23 '15 at 15:16
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    $\begingroup$ One also says bound variable. $\endgroup$ – Bernard Oct 23 '15 at 15:20
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Hint: with $u = 1-x{{{{{}}}}}$.

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  • $\begingroup$ Can you tell me more? I tried to use $u=1-x,$ but I was failed. $\endgroup$ – Kelan Oct 23 '15 at 10:29

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