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Real Analysis: Integration

  1. Oct 29, 2008 #1
    1. The problem statement, all variables and given/known data
    Let f be of class C1 on [a, b], with f(a) = f(b) = 0. Show that [tex]\int_a^b xf(x)f'(x)dx[/tex] = [tex]-1/2 \int_a^b [f(x)]^2 dx[/tex].


    2. Relevant equations
    If F is an antiderivative of f, then [tex]\int_a^b f(t)dt = F(b) - F(a)[/tex]

    3. The attempt at a solution
    I'm just really not sure how to begin this one. I know that because f is of class C1 that f' is continuous. Maybe change of variables?
     
  2. jcsd
  3. Oct 29, 2008 #2
    Use integration by parts!
     
  4. Oct 29, 2008 #3

    Hurkyl

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    Less dogmatically... one of the major differences between the two things you're trying to show equal is that one has a derivative in it, and the other doesn't. And what methods do you know that can increase/decrease how derivated a part of your integrand is?
     
  5. Oct 29, 2008 #4
    [tex]\int_a^b xf(x)f'(x)dx[/tex]

    [tex] u=x, du=dx, v=\int f(x)f'(x)dx[/tex]

    [tex] v=\int f(x)f'(x)dx, t=f(x), dt=f'(x)dx=>\int tdt=\frac{1}{2}t^2=\frac{1}{2}[f(x)]^2[/tex]

    [tex]\frac{1}{2}x[f(x)]|_a^b-\frac{1}{2}\int_a^b [f(x)]^2dx=-\frac{1}{2}\int_a^b [f(x)]^2dx[/tex]

    Because:[tex]\frac{1}{2}x[f(x)]|_a^b=0[/tex]
     
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