# How to integrate $\int_0^1x^a(1-x)^bdx$

I have a question about an equation I am trying to integrate, the integral is: $$\int_0^1 x^a (1 - x)^b ~dx,$$ where $a, b > 0$.

Any assistance with this problem would be appreciated.

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Do you know the binomial theorem? –  Kevin Carlson Aug 29 '12 at 2:32
Or do you know the beta function? You need to provide us some info if you want to get useful help. –  GEdgar Aug 29 '12 at 2:40

Google "Beta function". What you have is $$B(b+1,a+1)=B(a+1,b+1)=\int_0^1x^a(1-x)^bdx=\frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+2)}$$For example, if $\,a\,,\,b\in\Bbb N\,$ , then $$B(a+1,b+1)=\frac{a!\,b!}{(a+b+1)!}$$
The integral at hand is known as Euler's integral of the first kind. It's value, as function of $a$ and $b$ is know as beta function: $$B(a+1,b+1) = \int_0^1 x^a (1-x)^b \mathrm{d} x$$ Integrating by parts, one can derive recurrence equations: $$(a+1) B(a+1, b+1) = b B(a+2,b)$$ Change of variables $x \mapsto 1-x$ gives $B(a,b) = B(b,a)$. Recurrence equations can be solved to give expression of the beta function in terms of ratio of Euler's $\Gamma$-functions: $$B(a,b) = \frac{\Gamma(a) \Gamma(b)}{\Gamma(a+b)}$$