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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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    Did you try using the fundamental theorem of calculus?2012-04-19
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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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    I see that this solution works, but the chapter that this problem was posed in preceded the chapter where differentiation was introduced, so I do not think this is the solution the book was looking for.2012-04-20
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    Hm. How about this. In your progress, you reached that $f(x) = xf(a).$ Let $f(a) = c,$ which is constant by the way. Then $f(x) = cx,$ and $\int_0^x f(t)\,dt =$ $\int_0^x ct\,dt = $ $ct^2/2 + c_1 \mid_0^x = $ $cx^2/2.$ We have just shown that $\int_0^x f(x)\,dx = cx^2/2,$ but we know that $\int_0^x f(x)\,dx = f(x),$ and $f(x) = cx.$ In other words, we have proved that $cx^2/2 = cx.$ This can only be valid if c = 0. Hence $f(x) = 0.$2012-04-20
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    @JohnMace check the comment above.2012-04-20
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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