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Homework Help: Derivative/Integral proof

  1. Jan 28, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that if f(x) is continuous and [tex]f(x) = \int_0^x f(x) dx[/tex], then f(x) = 0.

    2. Relevant equations



    3. The attempt at a solution

    If [tex]f(x) = \int_0^x f(x) dx[/tex], then by integrating by the FTC we have f'(x) = f(x). Thus the only solution to this equation will have the form [tex]f(x) = ce^x[/tex] for some constant c. Now, [tex]f(x) = \int_0^x f(x) dx = f(x) - f(0) [/tex], implying that f(0 = 0. So since we know the solution to the equation will be [tex] f(x) = ce^x[/tex] then we have [tex] 0 = f(0) = ce^0 = c[/tex], implying that c = 0. Thus f(x) = 0. QED

    Is this correct?
     
  2. jcsd
  3. Jan 29, 2009 #2

    Mark44

    Staff: Mentor

    I'm confused by two things:
    1. Your use of x in one of the limits of integration and as the dummy variable in the integral. It would be better to use different variables.
    2. Your use of ' (as in dx'). Is this supposed to mean the derivative with respect to x of the definite integral?
    Based on these points, I believe you are saying that
    [tex]f(x) = \frac{d}{dx}\int_0^x f(t)dt[/tex]

    Now maybe I've missed something in how I've interpreted your problem, but the equation just above is true for every function f that is continuous on [0, b], and where 0 <= x <=b, per the FTC, so it does not follow that f(x) is identically 0.
     
  4. Jan 29, 2009 #3
    Oops, let me retype the question:

    Prove if f(x) is continuous and [tex]f(x) = \int_0^x f(x) dx[/tex] then f(x) = 0.

    That's the question, and the mark beside the dx was only a comma, it wasn't meant to denote the derivative of the integral. So now that I fixed the question, isn't f'(x) = f(x)? And so f(x) = ce^x, but 0 f(0) = ce^0 = c and so c = 0 and f(x) = 0.
     
  5. Jan 29, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You still may be causing some confusion by having the dummy variable of integration be the same as the limit, but I get what you are saying. I think that proof is ok. As f is the integral of a continuous function it's differentiable.
     
  6. Jan 29, 2009 #5

    Mark44

    Staff: Mentor

    Same here.
     
  7. Jan 29, 2009 #6
    Ugh, I meant to change the variable of integration to t but I forgot! Sorry, and thanks for checking my work.
     
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