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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.