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Are there any cases which the First Fundamental Theorem of Calculus would fail?

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General idea is good, but too many words, hiding essential idea. Let $f(x)$ be as in your definition. Then $F(x)=\int_0^x f(t)\,dt$ exists and is $0$ for all $x$. Certainly $F(x)$ is differentiable for all $x$, but $F'(1)=0$ while $f(1)=1$. (You were asked to prove or disprove $A$ and $B$. You have disproved it if you show $B$ fails.) –  André Nicolas Apr 5 '12 at 23:11
    
Your function is certainly Riemann integrable (only countably many discontinuities). The problem with $F^\prime(1/2)$ is immediate. If $f(x)$ were in addition continuous though, the statement would hold. But isn't $f(1/2) = 1/2$? –  Patrick Apr 5 '12 at 23:20
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A true theorem never fails. –  Christian Blatter Apr 6 '12 at 17:41

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More simply take $f(x)=\cases{-1,& $0\le x<1/2$\cr 1,& $1/2\le x\le 1$}$. Then $F'(1/2)$ does not exist (write down the definition of the derivative as a limit of a difference quotient to see this).

For your example, I don't quite follow your argument, but it seems you've mentioned the necessary ingredients: $F$ is identically $0$ (this essentially follows from the proof that your $f$ is integrable), so $F'$ is identically $0$. But $F'(x)\ne f(x)$ for any rational number $x$.

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