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Math Help - finding f for a definite integral

  1. #1
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    finding f for a definite integral

    Hi I feel very lost.

    Question is if f is a differentiable function such that f(x) is never zero and

    definite integral (I will use "S") b=x; a=0 f(t)dt = (f(x))^2 for all x, find f.

    So my first thought is to let y = f(x) so S b=x; a=0 f(t) dt = y^2

    then S b=x; a=0 2y = y^2 but I still don't have a clue what f is and if I jumped too far with the 2y

    Please help. I can't find an example anywhere in my textbook on how to find f.

    Thanks

    This calculus beginner
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  2. #2
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    \begin{gathered}<br />
  g(x) = \int_0^x {f(t)dt}  = f^2 (x) \hfill \\<br />
  g'(x) = f(x) = 2f(x)f'(x)\; \Rightarrow \;f'(x) = \frac{1}<br />
{2} \hfill \\ \end{gathered}

    Can you carry on?
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  3. #3
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    Sorry I am not completely sure.

    2f(x) f'(x) = f(x) so the only thing that would cancel f'(x) out is 1/2 so I get that but I am not sure what f is yet.

    Yep the I have not found the calculus light switch yet.

    Thanks for your help.
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  4. #4
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    (The point here is that you don't have to be at all sure what f(x) is to determine f'(x) - and if that's a point you already got then you weren't asking for further help so forgive the following attempt.)

    Anyway, just in case a picture helps...

    The fundamental theroem tells us...



    ... where the vertical line differentiates downwards (anti-differentiates up) with respect to x.

    And the problem says that \int _0^x f(t) dt = [f(x)]^2, so we're entitled to say...



    But [f(x)]^2 is a composite function, requiring the chain-rule, which we can depict so...



    ... where straight continuous lines differentiate downwards (anti-diff up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (which is the inner function of the composite and hence subject to the chain rule).

    So we get to...



    ... the bottom row to be solved for f ' (x) as you discussed.


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    Last edited by tom@ballooncalculus; September 21st 2009 at 02:17 AM. Reason: tiny typo
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