I realized that that FTC1 say nothing about function with domain $\mathbb R$. Then, how can I know that every continuous function has primitive function?
I made a scenario. please take a look.
let $f$ be the continuous function defined on $\mathbb R$ with no primitive function. That is, no function $F$ satisfies $F'=f$.
However, there is primitive function $F_n$ defined on $[-n,n]$ (by FTC1).
Define the set $A_n = \{(x,F_n(x)) : x\in [-n,n]\}$ then one can prove that $A_n \subset A_m$ if and only if $n<m$.
define $A=\bigcup_{n=1}^{\infty} A_n$. Finally, let $F$ be the function such that "$F(x)=y$ if and only if $(x,y)\in A$".
$$(*)\quad F|_{[-n,n]}=F_n$$ $(*)$ looks pretty obvious (I didn't checked yet..) If $(*)$ were true, $F$ is differentiable on $\mathbb R$ and its derivative is $f$. Because for any given $x$, we have $n$ $(>x)$. $$(F|_{[-n,n]})'=F_n'=f|_{[-n,n]
}'=f'|_{[-n,n]}$$
Therefore, $F$ is primitive function which is contradiction.
Will this work? I'm not sure…
Best Answer
The fundamental theorem of calculus works also for functions defined on $\mathbb{R}$. You can either check the proof and see that it goes through (note there is no need to consider improper Riemann integrals, the integral is always done over a finite interval) or just use the theorem for arbitrary large intervals.
Namely, choose some $a \in \mathbb{R}$ and define $F(x) = \int_a^x f(x) \, dx$. This is well-defined because $f$ is continuous on each interval $[a,x]$. Assume for simplicity $x > a$. Choose some $a < x < b$ and apply the fundamental theorem for $f|_{[a,b]}$ to deduce that $F$ is differentiable at $x$ with derivative $F'(x) = f(x)$.