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Does the second fundamental theorem of calculus (SFTC) really require that

$$ F'(x)=f(x) \text{ for all x in } [a, b]? $$

As an example, the sign function has a jump discontinuity and therefore thus not have an antiderivative. However, disregarding the point where the discontinuity lies, its antiderivative is the absolute value function.

It seems to work to calculate the area between the curve of the sign function and the x-axis using the SFTC with the absolute value function as the antiderivative, considering any interval. Is this just a coincidence and hence a case where the SFTC cannot be applied, or should the aforementioned requirement rather be something like

$$F'(x)=f(x) \text{ for all x in } [a, b] \text{ except for a finite set of points}.$$

I've only worked with Riemann integrals.

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  • $\begingroup$ Remember that the area below the $x$-axis is counted negatively. $|x|$ is not an antiderivative for the sign function. $\endgroup$ – Dzoooks Jun 8 '18 at 12:25
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    $\begingroup$ <pedantic mode>If "$F'(x) = f(x)$ for all $x \in [a,b]$" is part of the statement of the theorem, then the theorem requires that. </pedantic mode> Either the SFTC or a generalisation of it also holds in more general situations. Then the statement of the conditions becomes more involved. But if $F$ is continuous, differentiable at all but finitely many points and $F'(x) = f(x)$ holds wherever $F$ is differentiable, and $f$ is Riemann integrable over $[a,b]$, then $F(y) = F(a) + \int_a^y f(x)\,dx$ holds for all $y \in [a,b]$. And one can generalise further. $\endgroup$ – Daniel Fischer Jun 8 '18 at 12:31
  • $\begingroup$ @Dzoooks I meant that $|x|$ is the antiderivative for the sign function except for at one point ($x=0$). $\endgroup$ – gblomqvist Jun 8 '18 at 13:31
  • $\begingroup$ @DanielFischer Oh, I see. Thanks! $\endgroup$ – gblomqvist Jun 8 '18 at 13:31
  • $\begingroup$ Well the statement as well as the proof of SFTC requires the condition $F'=f$ on whole of $[a, b] $ and further that $f$ is Riemann integrable on $[a, b] $ (surprisingly every derivative is not Riemann integrable and hence this extra hypothesis is needed). $\endgroup$ – Paramanand Singh Jun 9 '18 at 2:56
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In your example of the sign function: Here is an antiderivative (except at one point):

$$F(x) = \begin{cases} 17 - x,& x<0\\ 4+x,& x>0 \end{cases}$$

We can see that $F'(x) = f(x)$ except at $x=0$. But $F(1)-F(-1) \ne \int_{-1}^1 f(x)\;dx$.

As Daniel wrote, if we require that $F$ is continuous and that $f$ is Riemann integrable, then your conclusion is OK.

But note: there are examples where $F'(x) = f(x)$ everywhere but $f$ is not Riemann integrable; so in those cases we cannot write $F(b)-F(a) = \int_a^b f(x)\;dx$

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    $\begingroup$ You will learn more on this when you do Lebesgue integration some day. $\endgroup$ – GEdgar Jun 8 '18 at 13:11
  • $\begingroup$ Great counterexample. Thank you! $\endgroup$ – gblomqvist Jun 8 '18 at 13:27

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