TITLE: Integral with binomial coefficient
QUESTION [5 upvotes]: Is it possible to evaluate this integral without using the gamma function
$$ \int_0^1 {a \choose b}x^b(1-x)^{a-b} dx$$
It looks a little like part of binomial theorem, but I don't have an idea how to manage with that. The solution should be $\frac{b}{a+1}$.

REPLY [3 votes]: Consider the generating function $F(y) = \sum_{b=0}^a  y^b \int_0^1 \binom{a}{b} x^b (1-x)^{a-b} \mathrm{d}x$. By the binomial theorem we have  
$$F(y) = \int_0^1 (1-(1-y)x)^a \mathrm{d} x = \frac{1}{a+1} \frac{1-y^{a+1}}{1-y} = \frac{1}{a+1}(1+y+\cdots+y^a) $$
Now take the coefficient of $y^b$ to get $\int_0^1 \binom{a}{b} x^b (1-x)^{a-b} \mathrm{d}x = 1/(a+1)$.