The result is that
$$\int_{0}^1 x^m (1-x)^n dx = \frac{m!n!}{(m+n+1)!}.$$
I think we could just do this by repeated integration by parts... but I could have also done this:
$$\int_{0}^1 x^m(1-x)^n dx = \int_{0}^1 x^m \left[\sum_{k=0}^{n}\begin{pmatrix}n \\k \end{pmatrix}(-1)^kx^k\right]dx\\ = \int_{0}^1 \sum_{k=0}^n \begin{pmatrix}n\\k\end{pmatrix}(-1)^k x^{k+m} dx\\ = \sum_{k=0}^n \int_0^1 \begin{pmatrix}n\\k\end{pmatrix}(-1)^kx^{k+m}dx \qquad \text{assuming I can interchange the integral and sum... why??} \\ = \sum_{k=0}^n \begin{pmatrix}n\\k \end{pmatrix} (-1)^k \frac{1}{k+m+1}.$$
Putting it into a calculator, it gives the original result. Is there some interpretation for this, and how would I have shown the result? It looks like some sort of taylor series...
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3Since there are only finite number of terms involved, interchange at will! Also, see: https://en.wikipedia.org/wiki/Beta_function – Aryabhata Aug 01 '17 at 06:51
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I dont know why you want combinatorial approach as the integral is a beta function – Archis Welankar Aug 01 '17 at 06:52
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1btw, for a proof, see: https://math.stackexchange.com/questions/86542/prove-binomnk-1-n1-int-01xk1-xn-kdx-for-0-leq-k-le. It has probabilistic interpretations... – Aryabhata Aug 01 '17 at 07:02
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@ArchisWelankar I didn't know of the function before posting this. – similarityinvariance1 Aug 01 '17 at 22:49
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@Aryabhata Thank you, I will check those links out. – similarityinvariance1 Aug 01 '17 at 22:50