Originally Posted by

**SlipEternal** Ok, I get it now. The dt was throwing me. With dx, it actually seems fairly trivial, so I may be missing something.

Base case: n = 0

$\displaystyle \int_0^1(1-x)^y~dx = \left. -\dfrac{(1-x)^{y+1}}{y+1} \right]_0^1 = \dfrac{1}{y+1} = \dfrac{0!}{y+0+1}$

So, it satisfies the base case. Suppose it is true for all nonnegative integers less than $\displaystyle n>0$. I want to show it is true for $\displaystyle n$.

The integration by parts is very straightforward: $\displaystyle u = x^n, dv = (1-x)^y~dx \Rightarrow du = nx^{n-1}~dx, v = -\dfrac{(1-x)^{y+1}}{y+1}$.

Now, we have:

$\displaystyle \int_0^1 x^n(1-x)^y~dx = \left[ -\dfrac{x^n(1-x)^{y+1}}{y+1} \right]_0^1 + \dfrac{n}{y+1}\int_0^1 x^{n-1}(1-x)^{y+1}~dx$

The first term is zero at both 1 and 0. For the second term, since $\displaystyle y+1 \in \mathbb{R}$ and $\displaystyle n-1$ is a nonnegative integer less than $\displaystyle n>0$, so by the induction assumption, we can apply the hypothesis.