# Show that $\int_{0}^{1}x^{m}(1-x)^{n}dx=\int_{0}^{1}x^{n}(1-x)^{m}dx$

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.

• You can use the substitution $t = 1-x$. – Sangchul Lee Aug 8 '13 at 5:31

Hint: Make the change of variables $1-x=t$.
• I have $t = 1-x$. Then $dx = -dt$, $(1-x)^{n}=t^{n}$ and $x^{m}=(1-t)^{m}$, so I get $$-\int_{0}^{1}(1-t)^{m}t^{n}dt=\int_{0}^{1}x^{n}(1-x)^{m}dx$$but there is this minus sign, what do I do with it? – Fazzolini Aug 8 '13 at 5:40
• @Fazzolini You must also adjust the bounds of integration after the substitution; you should actually have $$- \int_1^0 ...$$ – user61527 Aug 8 '13 at 5:44
• @Mhenni: (+1) Is this integral related to Beta function $B(m,n)$? Thanks. – mrs Aug 8 '13 at 9:15