# Prove two integrals are the same using U substitution

1. Nov 14, 2008

### v0id19

1. The problem statement, all variables and given/known data
If a and b are positive numbers, show that $$\int_0^1 x^a*(1-x)^b\,dx = \int_0^1 x^b*(1-x)^a\,dx$$ using only U substitution.

2. Relevant equations
Just U substitution and the given equation--I can't use multiplication rules or anything like that; otherwise it would be easy.

3. The attempt at a solution
I tried to set U=(1-x) and I end up with $$\int_0^1 (1-U)^a*(U)^B\,dx$$ for the right side, but that doesn't seem to get me anywhere. I know I somehow need to switch the places of the x and (1-x) but I can't seem to get around going in a circle and ending up with what I started with.

2. Nov 14, 2008

### Pere Callahan

Multiplication of real numbers is commutative, so

$$(1-U)^a*U^b=U^b*(1-U)^a$$

Or does * denote something different from multiplication of real numbers?

If you let U = 1-x you also have dx=-dU. Morover your integration bounds change. You need to be more careful in applying the u substitution rule.

3. Nov 14, 2008

### v0id19

yeah I just realized that, thanks
$$\int_1^0 (1-U)^a*(U)^b\,(-du) = \int_0^1 (1-U)^a*(U)^b\,(du)$$ but I still get stuck there. Why would switching the order of (U-1)a and Ub help?

4. Nov 14, 2008

### Pere Callahan

Haven't you just shown

$$\int_0^1{x^a(1-x)^bdx}=\int_0^1{u^b(1-u)^adu}$$
?

How does this relate to the original question you are asked to prove? Remeber, that it doesn't make a difference how you call the integration variable. x, u, whatever, it is only a label.

5. Nov 14, 2008

### v0id19

Oh. Wow. That makes sense!

Thanks a bunch!!