Calculating integral using pdf's $\int_0^1x^r(1-x)^{n-r}dx$

Question

I am trying to find the expectation of an order statistic and have reduced the question to finding the following integral $$\int_0^1x^r(1-x)^{n-r}dx$$

I know what this integral equals but can't seem to show it. I would suspect that I have to try and express it in terms of a pdf which should cancel once I integrate?

Answer 1

That is given by the Beta function

$$\int_0^1x^r(1-x)^{n-r}dx =B(r+1, n-r+1)=\frac{\Gamma(r+1)\Gamma(n-r+1)}{\Gamma(n+2)}\\=\frac{r!(n-r)!}{(n+1)!}= \frac{1}{(n+1){n\choose r}} $$

Answer 2

Answering the question raised by the OP as a comment regarding if the limits of integration on the integral are changed from $[0,1]$ to $[a,b]$ ($b > a$) to give the new integral of $$I = \int_a^b x^r (1 - x)^{n - r} \, dx.$$

Recalling the definition for the incomplete Beta function $$\text{B}(x; a,b) = \int_0^x t^{a - 1} (1 - t)^{b - 1} \, dt,$$ then \begin{align*} I &= \int_a^b x^r (1 - x)^{n - r} \, dx\\ &= \int_0^b x^r (1 - x)^{n - r} \, dx + \int_a^0 x^r (1 - x)^{n - r} \, dx\\ &= \int_0^b x^r (1 - x)^{n - r} \, dx - \int_0^a x^r (1 - x)^{n - r} \, dx\\ &= \text{B}(b; r + 1, n - r + 1) - \text{B}(a; r + 1, n - r + 1). \end{align*}