<s>[INST] Why does $\int_0^1\binom{2n}{n}x^n(1-x)^ndF(x)=\frac{1}{2n+1}$? Why does $$\int_0^1\binom{2n}{n}x^n(1-x)^ndF(x)=\frac{1}{2n+1},$$
Where $F(x)$ is uniform over the interval $[0,1]$?

It's not clear from simple factoring:
$$\binom{2n}{n}\int_0^1x^n(1-x)^ndF(x)$$
$$=\frac{(2n)!}{n!^2}\int_0^1x^n(1-x)^ndF(x).$$
 [/INST] By the binomial theorem,
\begin{align*}
\sum_{k=0}^{n} \left( \int_{0}^{1} \binom{n}{k} x^k(1-x)^{n-k} \, dx \right) z^k
&= \int_{0}^{1} (1 + (z-1)x)^n \, dx \\
&= \frac{1}{n+1} \cdot \frac{z^{n+1} - 1}{z-1} \\
&= \sum_{k=0}^{n} \frac{z^k}{n+1}.
\end{align*}
So we have
$$ \int_{0}^{1} \binom{n}{k} x^k(1-x)^{n-k} \, dx = \frac{1}{n+1} $$
for all $0 \leq k \leq n$.
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