Lebesgue's criterion for Riemann integrability states that a function $f: [a, b] \longrightarrow \Bbb R$ is Riemann integrable if and only if it is bounded and continuous almost everywhere on $[a,b]$.
In the present case, $|F(x)| \leq |b| \sqrt{|b|}$ on $[0, |b|]$ and $F$ is also continuous on $[0, |b|]$, so by Lebesgue's criterion $F$ is Riemann integrable.
On the other hand, $F'$ is not bounded in any neighborhood of $x = 0$, so by Lebesgue's criterion it is not Riemann integrable.
The point of this example is that you can't always integrate a function just because that function is a derivative of some other function.
You can start by showing that if $f \colon [a,b] \rightarrow \mathbb{R}$ is a bounded function and $c \in (a,b)$ such that both $f|_{[a,c]}$ and $f|_{[c,b]}$ are Riemann integrable, then $f$ is Riemann integrable and
$$ \int_a^b f(x)\, dx = \int_a^c f(x)\, dx + \int_c^b f(x) \, dx. $$
Then, prove that for any $a < x_0 < b$ and $y_0 \in \mathbb{R}$, the function
$$ f_{x_0,y_0}(x) := \begin{cases} y_0 & x = x_0 \\ 0 & x \neq x_0 \end{cases} $$
is Riemann integrable on $[a,b]$ with integral zero. The argument will be the same argument as for your $f_1$.
Now, choose some $\frac{1}{2} < a_1 < 1$. On $[a_1,1]$, your function $f$ is just $1 - f_{1,1}(x)$ and thus, is Riemann integrable. On $[\frac{1}{2},a_1]$, your function is just $1 - f_{\frac{1}{2},1}$ and thus is also Riemann integrable. By the result above, $f|_{[\frac{1}{2},1]}$ is Riemann integrable. Continuing this way inductively, you can see that $f|_{[\frac{1}{n},1]}$ is Riemann integrable for all $n \in \mathbb{N}$. Alternatively, if you already know that a function with finitely many discontinuities is Riemann integrable, you can skip all of the above.
Finally, using the definition of the Riemann integrable directly, show that if $f \colon [a,b] \rightarrow \mathbb{R}$ is a bounded function such that $f|_{[a + \frac{1}{n},b]}$ is Riemann integrable for all sufficiently large $n$, then $f$ is Riemann integrable and
$$ \int_a^b f(x) \, dx = \lim_{n \to \infty} \int_{a + \frac{1}{n}}^b f(x)\, dx. $$
Best Answer
Your computations are correct and they show that $f'$ is bounded and it has a single discontinuity point. Therefore, it Riemann-integrable (and this would still be true if it had a countable set of discontinuity points).