Originally Posted by
harish21 Since
$\displaystyle \displaystyle \int_0^1 x^{\alpha - 1} \;(1-x)^{\beta -1} dx$ is a Beta function with $\displaystyle B(\alpha,\beta) = \dfrac{\Gamma(\alpha) \; \Gamma(\beta)}{\Gamma(\alpha + \beta)}$
Also, $\displaystyle \Gamma(\alpha) = (\alpha-1)!$
therefore,
you can write:
$\displaystyle \displaystyle \int_0^1 x^m \; (1-x)^n \;dx = \int_0^1 x^{(m+1)-1} \; (1-x)^{(n+1)-1} \;dx$,
which is a beta function $\displaystyle B(m+1, n+1)$
$\displaystyle B(m+1, n+1) = \dfrac{\Gamma(m+1) \; \Gamma(n+1)}{\Gamma(m+n+2)}$
Now finish it off!