Faulty application of the Fundamental Theorem of Calculus to $f(x) = 0$ for $x\ne 0$, $f(0)=1$ I think I have given a fallacious proof but I can't seem to find what is wrong with it. Suppose $f : \mathbb{R} \rightarrow \mathbb{R}$ has the property that $\forall a,b \in \mathbb{R}. \int_a^b f(t)dt = b - a$. Consider the following "proof" that f(x) = 1. Define $g(x) = \int_0^x f(t)dt$. Then from above we know g(x) = x. Further by the fundamental theorem of calculus $g'(x) = f(x) = 1$. But then consider $$f(x) = \begin{cases} 0 & \text{if $x=0$} \\\ 1 & \text{if $ x \not = 0$}\end{cases}$$ Doesn't this function have the above property?

The fundamental theorem of calculus states that:

> If $f:[a,b]\to\mathbb R$ (say) is a **continuous** function, and we define:
>
> $$ F(x)=\int_a^x f(t)dt $$
>
> then $F'(x)=f(x)$.

I have put one of the words in the statement in **bold** , which is the clue to seeing why the FTC does not apply to your function. Indeed, the example you provide show why the hypothesis in bold is necessary to derive the FTC.