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Suppose that $f$ and $g$ are differentiable functions satisfying $$\int_{0}^{f(x)} fg = g(f(x)).$$ Prove that $g(0) = 0$.

We apply the fundamental theorem of calculus to get $f(f(x))g(f(x)) = g'(f(x))f'(x)$. I am not sure what to do from here, but I think finding out what value of $x$ gives $f(x)= 0$ will help.

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  • $\begingroup$ Do not dislike a question, if someone does not understand the answer. $\endgroup$
    – Arbuja
    Commented May 13, 2016 at 16:55

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This isn't true. A simple counterexample is when $f$ and $g$ are both identically equal to one.

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  • $\begingroup$ @user19405892 Work it out! If $f$ and $g$ are both identically equal to $1$ then $\int_0^{f(x)}fg\,dt=\int_0^1 dt=1=g(f(x))$. But $g(0)\ne0$. $\endgroup$ Commented May 13, 2016 at 16:52
  • $\begingroup$ $\int_0^{f(x)}fg=\int_0^11=1=g(f(x))$, but clearly $g(0)\neq0$. $\endgroup$
    – Jason
    Commented May 13, 2016 at 16:52
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    $\begingroup$ Nice counter punch. $\endgroup$
    – mvw
    Commented May 13, 2016 at 16:52
  • $\begingroup$ This was a question in my textbook, so I didn't think it was wrong at first. $\endgroup$ Commented May 13, 2016 at 16:52

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