# If a function is discontinuous, why doesn't it contradict the Fundamental Theorem of Calculus

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

• Because the ftc only claims to hold for continuous integrands – Vercassivelaunos Dec 8 '20 at 19:17
• The standard version of the FTC requires that $F$ be continuous on $[0,2]$. But you have a discontinuity at $1$. – bjorn93 Dec 8 '20 at 19:42
• 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. – Arturo Magidin Dec 8 '20 at 19:47

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.