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I've found this question on a paper and I have some difficulty understanding what exactly I'm asked to do.

For each of the following functions $f$, if $F(x)=\int_{0}^{x} {f}$, at which points $x$ does $F'(x)=f(x)$ ?

  • $ f(x) = \begin{cases} 0, & x \le 1 \\ 1, & x > 1 \end{cases}$

  • $f(x)=...$

  • $f(x)=...$

But how can this not hold true for all integrable functions $f$, since by differentiating we get:

$F'(x)=(x)' \times f(x)-(0)' \times f(0)=f(x)$

Zero Pancakes
  • 1,409
  • Try integrating the first one (the one with the step discontinuity for $f$). Can you draw the picture of $F(x)$? Then see what happens when you try to differentiate at certain points. – Michael Jul 06 '17 at 14:06
  • Clearly $ F'(x) =f(x) $ for all $x\neq 1$. And $F'(1)$ does not exist. All this is a direct consequence of Fundamental Theorem of Calculus. – Paramanand Singh Jul 06 '17 at 14:53
  • I have discussed the Fundamental Theorem of Calculus in this answer https://math.stackexchange.com/a/1900844/72031 and you may have a look at it. – Paramanand Singh Jul 06 '17 at 14:56
  • @ParamanandSingh I don't see how the FFTC clarifies if $F$ is differentiable at the points of discontinuity of $f$. How can we get information about this from the FFTC? – Zero Pancakes Jul 06 '17 at 15:03
  • FTC does not say anything about points where $f$ is discontinuous. But the typical proof of FTC helps to know what happens when $f$ has a simple discontinuity (that is left and right limits of $f$ exist). Thus we have the result that if left hand limit $\lim_{x\to c^{-}} f(x) $ exists then left hand derivative of $F$ at $c$ is equal to this limit and similarly for right hand limits and right hand derivative. – Paramanand Singh Jul 06 '17 at 15:07

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