TITLE: Show that $\int_{0}^{1}x^{m}(1-x)^{n}dx=\int_{0}^{1}x^{n}(1-x)^{m}dx$
QUESTION [5 upvotes]: I am trying to show that:
$$\int_{0}^{1}x^{m}(1-x)^{n}dx=\int_{0}^{1}x^{n}(1-x)^{m}dx$$
For any positive integers $n$ and $m$. Which is true if I try to evaluate it numerically.
I tried to use the binomial theorem, but then I end up with:
$$
\int_{0}^{1}
\sum_{i=0}^{n}
\begin{bmatrix}
n \\ 
i
\end{bmatrix}x^{m+n-i}
dx
=
\int_{0}^{1}
\sum_{i=0}^{m}
\begin{bmatrix}
m \\ 
i
\end{bmatrix}x^{m+n-i}
dx
$$
There is a term $x^{m+n-i}$ on both sides which looks like there should be something I can do with it to simplify but then I am stuck.
This is an exercise following Apostol's Calculus I Chapter 5 where integration by substitution is discussed, but it is not clear how to use integration by substitution here.

REPLY [7 votes]: Hint: Make the change of variables $1-x=t$.