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I was thinking about this question while walking home today and can't seem to prove or come up with a counterexample myself.

Let $f:[a,b]\rightarrow\mathbb{R}$ be a continuous function, $f(x),$ differentiable everywhere on $(a,b)$, except for one point $\xi$ in $(a,b)$ where $f'(\xi)$ does not exist. Since $f$ is continuous everywhere and differentiable everywhere on $(a,b)$ except at $\xi$, $f'$ is gauge integrable on $[a,b]$. Am I correct that $\int_a^bf'(x)\;\mathrm{d}x=f(b)-f(a)$ despite that nasty point $\xi$ in $(a,b)$? And could this value be interpreted as the "signed" area under the curve from high school calculus?

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  • $\begingroup$ What makes you thing $f'$ is integrable? $\endgroup$ – Kavi Rama Murthy Dec 13 '18 at 23:52
  • $\begingroup$ I thought that $f'$ is gauge integrable because $f$ is differentiable everywhere except at a countable number of points? $\endgroup$ – user626213 Dec 14 '18 at 0:42
  • $\begingroup$ I do not know what 'gauge integrable' means but I think $f'$ need not be integrable in any sense. Note that $f'$ need not be bounded. $\endgroup$ – Kavi Rama Murthy Dec 14 '18 at 5:32
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If $f(x)=x\sin (\frac 1 x)$ ($f(0)=0$) then $f$ satisfies your hypothesis but $f$ does not satisfy the Fundamental Theorem of Calculus . If it did satisfy this on each interval contained in $[-1,1]$ the it would be absolutely continuous and hence of bounded variation on $[-1,1]$ but it is not of bounded variation.

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