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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...

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    $\begingroup$ Primitive is a fairly out-dated term. In the US, at least, we almost always use antiderivative. $\endgroup$
    – pancini
    Sep 6, 2017 at 6:42
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    $\begingroup$ @ElliotG I've only ever seen the word "antiderivative" in elementary calculus textbooks. I wonder how many trees could be saved by changing all the antiderivatives back to primitives. $\endgroup$
    – bof
    Sep 6, 2017 at 6:58
  • $\begingroup$ In the line beginning "Define the set $A_n$" the function $F_n(x)$ pops up out of nowhere. You need to tell us what it is. $\endgroup$
    – bof
    Sep 6, 2017 at 7:00
  • $\begingroup$ I really don't see the necessity of this overly complicated argument when FTC does the job. If $f$ is continuous on $\mathbb{R} $ then the symbol $\int_{0}^{x}f(t)\,dt$ defines a function of $x$ over $\mathbb{R}$. If we denote this function by $F$ then via FTC $F'(x) =f(x) $ for all $x\in\mathbb{R} $. $\endgroup$
    – Paramanand Singh
    Sep 6, 2017 at 7:12

2 Answers 2

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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 $F'(x) = f(x)$.

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One can be more concise; in particular one need not phrase it as a contradiction proof. For each $n$, there is a unique primitive $F_n$ defined on $[-n,n]$ with $F_n(0)=0$. Then $F_{n+1}$ restricted to $[-n,n]$ is $F_n$. As in your solution you can then prove the $F_n$ fit together to make a function $F$ on $\Bbb R$, and whose restriction to $[-n,n]$ is $F_n$.

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