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My book states, that if we apply the "mean value theorem in integral form" on $f'$, then follows: $\int_a^b{f'} = (b-a)f'(c)$. If we compare this with the "canonical mean value theorem": $f(b) - f(a) = (b-a)f'(c)$, then this suggests that $\int_a^b{f'} = f(b) - f(a)$. It also states that although the former suggests the latter, it doesn't prove the latter.

My question is why isn't this a proof?

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I think it's a circular argument, since this theorem depends FFTC – dwarandae Jan 16 '13 at 23:20
@dwarandae what is fftc? – xcrypt Jan 16 '13 at 23:23
The $c$s are only the same if you know the latter equality holds. If you just know both LHS are equal to the same RHS at some $c$, those could be different $c$s. – gnometorule Jan 16 '13 at 23:23
@xcrypt First Fundamentel Theorem of Calculus, sorry. – dwarandae Jan 16 '13 at 23:24
You can establish the mean value theorem using Rolle's and Heine-Borel theorems. After that, you can prove the Fundamental Theorem of Calculus using the mean value theorem and Darboux sums. This is what Rudin's Principles does. – Ayman Hourieh Jan 16 '13 at 23:37

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