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So for example, suppose $g(x) = \int_o^x F(t) dt$ where $$F(t) = \begin{cases} t & 0 \leq t \leq 1 \\ t - 1 & 1 < t \leq 2 \end{cases} $$

The function $g$ is not differentiable at 1. I am curious to why it doesn't contradict the Fundamental Theorem of Calculus. Some thoughts and assistance

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  • $\begingroup$ Because the ftc only claims to hold for continuous integrands $\endgroup$ – Vercassivelaunos Dec 8 '20 at 19:17
  • $\begingroup$ The standard version of the FTC requires that $F$ be continuous on $[0,2]$. But you have a discontinuity at $1$. $\endgroup$ – bjorn93 Dec 8 '20 at 19:42
  • $\begingroup$ Why does a duck not contradict the statement that "dogs cannot fly"? Because ducks are not dogs. Likewise, the FTC part 2 is a statement about continuous functions; if your function is not continuous, the FTC does not even come to the party. $\endgroup$ – Arturo Magidin Dec 8 '20 at 19:47
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Because the Fundamental Theorem of Calculus states that if $f$ is continuous, then$$x\mapsto\int_a^xf(t)\,\mathrm dt$$is differentiable. And your function $F$ is not continuous.

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