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Prove two integrals are the same using U substitution

  1. Nov 14, 2008 #1
    1. The problem statement, all variables and given/known data
    If a and b are positive numbers, show that [tex]\int_0^1 x^a*(1-x)^b\,dx = \int_0^1 x^b*(1-x)^a\,dx[/tex] 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 [tex]\int_0^1 (1-U)^a*(U)^B\,dx[/tex] 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. jcsd
  3. Nov 14, 2008 #2
    Multiplication of real numbers is commutative, so

    [tex]
    (1-U)^a*U^b=U^b*(1-U)^a
    [/tex]

    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.
     
  4. Nov 14, 2008 #3
    yeah I just realized that, thanks
    [tex]\int_1^0 (1-U)^a*(U)^b\,(-du) = \int_0^1 (1-U)^a*(U)^b\,(du)[/tex] but I still get stuck there. Why would switching the order of (U-1)a and Ub help?
     
  5. Nov 14, 2008 #4
    Haven't you just shown

    [tex]
    \int_0^1{x^a(1-x)^bdx}=\int_0^1{u^b(1-u)^adu}
    [/tex]
    ?

    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.
     
  6. Nov 14, 2008 #5
    Oh. Wow. That makes sense!

    Thanks a bunch!! :biggrin:
     
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