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The second part of the fundamental theorem of calculus is stated in wipedia (http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Second_part) as:

"Let $f$ and $F$ be real-valued functions defined on a closed interval $[a, b]$ such that the derivative of $F$ is $f$. That is, $f$ and $F$ are functions such that for all $x$ in $[a, b]$, $F'(x) = f(x).$ If f is Riemann integrable on $[a, b]$ then $\int_a^b f(x)\,dx = F(b) - F(a).$ The Second part is somewhat stronger than the Corollary because it does not assume that $f$ is continuous."

Can anyone give an example where the second part can be applied, but not the first? ($f$ not continuous). My intuition says that when $F$ is differentiable everywhere on $[a, b]$, its derivative is continuous so such an example cannot exist.

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  • $\begingroup$ as you said, $f$ should not be continuous. So a good example is the Heaviside function. And as interval you can choose e.g $[-1,1]$. However related to your question are so called absolutely continuous functions. But maybe this is too early for you $\endgroup$ – Quickbeam2k1 Aug 25 '13 at 20:34
  • $\begingroup$ @Quickbeam2k1 Then $F$ is not differentiable everywhere. A derivative satisfies the intermediate value theorem. $\endgroup$ – Daniel Fischer Aug 25 '13 at 20:38
  • $\begingroup$ @Quickbeam2k1 - But if we integrate the heaviside function, the left and right derivatives at 0 do not equal and therefore $F$ is not differentiable. Or am I missing something in the definitions? $\endgroup$ – Leo Aug 25 '13 at 20:39
  • $\begingroup$ ? You can apply the second part. The primitive is the "positive part" function. Maybe I misundertsood your question? $\endgroup$ – Quickbeam2k1 Aug 25 '13 at 20:40
  • $\begingroup$ Sorry i thought you ment something different with the second part. my bad $\endgroup$ – Quickbeam2k1 Aug 25 '13 at 20:42
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No, for instance $$\begin{cases} x^2\sin\frac{1}{x}, & x \ne 0, \\0, & x=0 \end{cases}$$ is everywhere differentiable. Its derivative is discontinuous, but Riemann integrable on every finite interval.

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  • $\begingroup$ Isn't the derivative "equal" infinity at some points? which means this function is not differentiable at these points. $\endgroup$ – Leo Aug 25 '13 at 20:57
  • $\begingroup$ No, its derivative is $$\begin{cases} 2x \sin \frac{1}{x} - \cos \frac{1}{x}, & x \ne 0, \\ 0, & x = 0. \end{cases}$$ $\endgroup$ – njguliyev Aug 25 '13 at 21:00

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