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So the problem is if: $\int_0^x f(t)\, dt = f(x) $ then f(x) is identically zero.

So far I've tried an approach with the mean value theorem and I end up with the equation: $f(x) = xf(a)$ for some $a$ in $[0, x]$ for all $x$.

And that's as far as I got. I think the mean value theorem is the right approach to this, but I don't know what to do much after that. I also think an approach would be Riemann sums, but I didn't get too far with that either. So any ideas?

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    As suggested above, try the FTC, and notice that the only functions that are derivatives of themselves are $ce^x$, with $c$ a constant (try proving this).2012-04-19

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Roughly.. $\int_0^x f(t) dt = f(x)$ implies that $f'(x) = f(x)$ Hence, $ f(x) = c e^x \tag{1}$ But $\int_0^x f(t) dt = f(x) - f(0).$ So, $f(0) = 0.$ Substitute in $(1)$.. what is $c$?

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    Is it okay to differentiate $f$ to obtain $f'(x)=f(x)$ just because $f$ is continuous? Isn't Differentiability $\implies $ Continuity and not the other way round?2016-06-04