# Inverse of the fundamental theorem of calculus

The fundumental theorem of calculus states:
if :
1) $$f$$ was integrable over an interval like $$[a,b]$$.
2) $$f$$ was continuous at $$x=c$$, $$a.
3) $$F(x)= \int_a ^x{f(t)}$$.

then:
$$F'(c)=f(c)$$

I wonder if there is any function with following properties :

1) $$f$$ was integrable over an interval like $$[a,b]$$.
2) $$f$$ was continuous over $$[a,b]$$ but at $$x=c$$, where isn't continuous.
3) $$F(x)=\int_a^x{f(t)}$$, and $$F$$ is differentiable over $$[a,b]$$, and $$F'(x)=f(x)$$

Note:

if there was any function like $$f$$ with above properties, then, $$L_1$$ and/or $$L_2$$ wouldn't exist ((event in the cases $$L_1=\pm\infty$$ and/or $$L_2=\pm\infty$$)), where:
$$L_1=\lim_{x \rightarrow c^-} f(x) \quad L_2=\lim_{x \rightarrow c^+}f(x)$$
any response, would be appreciated.

I think the following function covers the properties you need.

$$f(x) = \begin{cases} 2x\sin(\frac{1}{x})-\cos(\frac{1}{x}), x \neq 0 \\ 0, x=0 \end{cases}$$

It is continuous in $$[a,b]$$ where $$a$$ is a negative solution of the equation $$2x\sin(\frac{1}{x})-\cos(\frac{1}{x})=0$$ and $$b$$ a positive number, except from $$x=c=0$$ (that's where it is not continuous).

It is integrable over $$[a,b]$$.

$$F(x)=\int_a^x f(t)dt =\begin{cases} x^2\sin(\frac{1}{x}), x \neq 0 \\ 0, x=0 \end{cases}$$

and $$F'(x)=f(x)$$ for every $$x \in [a,b]$$.

As you have noted, $$L_1$$ and $$L_2$$ do not exist.