Does $\int_a^bf(x)\,\mathrm dx$ always exist and equal $F(b)-F(a)$? Consider, you have $F'(x) = f(x)$ for all $x$ from $[a,b]$, where $F(x)$ is antiderivative. Does $\int_a^bf(x)\,\mathrm dx$ necessarily exist and is equal to $F(b)-F(a)$?
From condition we could conclude that $f(x)$ is integrable and continuous. It seems that definite integral exists. And I have no example, when definite integral isn't equal to $F(b) - F(a)$. Please, can you provide an example for such situation and list the properties of a function $f(x)$. I know correct answer for question(my second sentence) is "no", but I want to understand it. Also, I've read that if $f(x)$ is integrable this function can have no antiderivative. May be it's key for my question, but again, I don't understand how it possible and I don't know any examples. Any help is appreciated
 A: The correct statement is: if $F'(x)=f(x)$ for all $x\in[a,b]$, and $f$ is Riemann integrable on $[a,b]$, then
$$
\int_{a}^{b}f(x) \, dx=F(b)-F(a) \, .
$$
To understand why the condition "$f$ is Riemann integrable on $[a,b]$" is important, consider the function
$$
F(x)=\begin{cases}
x^2\sin\left(\dfrac{1}{x^2}\right) &\text{ if $x\neq0$} \\
0 &\text{ if $x=0$} \, .
\end{cases}
$$
Its derivative is
$$
f(x)=\begin{cases}
\displaystyle{2x\sin\left(\frac{1}{x^2}\right)-\frac{2\cos\left(\frac{1}{x^2}\right)}{x}} &\text{if $x\neq0$} \\
0 &\text{if $x=0$} \, .
\end{cases}
$$
We might be tempted to think that
$$
\int_{-1}^{1}f(x) \, dx = F(1)-F(-1) \, ,
$$
but $f$ is unbounded on $[-1,1]$, and so is not Riemann integrable on $[-1,1]$. So the fact that $f$ is the derivative of another function $F$ on $[a,b]$ is by no means a guarantee that $\int_{a}^{b}f(x) \, dx$ exists. There are even 'worse' examples: consider Volterra's function.
The good news is that if $f$ is continuous on $[a,b]$, then $f$ is Riemann integrable on $[a,b]$. We tend to integrate elementary functions, and every elementary function is continuous on its domain. So in practice these problems rarely crop up.
A: if:
$$F'(x)=f(x)$$
then:
$$\int_a^bdF=\int_a^bf(x)\,dx$$
$$\implies F(b)-F(a)=\int_a^bf(x)\,dx$$
however, there is a requirement that both $F,F'$ exist and $F'$ is integrable on $[a,b]$
