You can take, for instance$$\begin{array}{rccc}f\colon&[0,1]&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}x&\text{ if }x\in\Bbb Q\\0&\text{ if }x\notin\Bbb Q.\end{cases}\end{array}$$It is bounded. And it is not Riemann integrable since it is discontinuous at every point of $(0,1]$, and this is is uncountable. Another possibility would be$$\begin{array}{rccc}g\colon&[0,1]&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}x&\text{ if }x\in\Bbb Q\\-x&\text{ if }x\notin\Bbb Q.\end{cases}\end{array}$$
Things become clearer if you look at Barrow's original proof, which can be found in The Geometrical Lectures of Isaac Barrow (Proposition 11 of Lecture X).
According to the picture below, Barrow (= Newton's advisor) proved that, given two curves $ZGE$ and $AIF$ (with $AZ$, $PG$, $DE$, etc. increasing), if the area of the region $ADEZ$ is $DF\cdot R$ (for any $D$ and a given constant $R$) then $TF$ is tangent to $AIF$ at $F$, where $T$ is defined by the relation $\frac{DE}{DF}=\frac{R}{DT}$.
This is the Barrow's Fundamental Theorem of Calculus. In modern notation, if we take $R=1$, $ZGE$ is the graph of $g(x)$, $f(x)=-g(x)$, $AIF$ is the graph of $F(x)$, $x$ is the abscissa of $D$ and $a$ is abscissa of $A$, then Barrow's result implies that
$$\begin{aligned}
\frac{d}{dx}\left[\int_a^x f(s)\,ds\right]&=\frac{d}{dx}\left[\text{area}(ADEZ)\right]&&\text{(modern meaning of integral)}\\
&=\frac{d}{dx}\left[F(x)\right]&&\text{(Barrow's hypothesis)}\\
&=\frac{DF}{DT}&&\text{(modern meaning of derivative)}\\
&=DE&&\text{(Barrow's hypothesis)}\\
&=f(x),&&
\end{aligned}$$
which is one part of the modern Fundamental Theorem of Calculus.
The other part can be found in Proposition 19 of Lecture XI: if $FT$ is tangent, then $\text{area}(APGZ)=PI\cdot R$. In modern notation, taking $R=1$ and using the fact that $F(a)=0$ in the considered case and labeling $p$ the abscissa of $P$, we obtain the usual result
$$\int_a^p f(x)\,dx=F(p)-F(a).$$
In a broad sense:
- From the presented statements, we can see that Barrow's result is not about integrals and derivative, but about areas and tangents.
- From the proofs, that I've omitted but can be found in the said book as well as here (with more details), we can see that the Barrow's integral (at least in this particular context of the FTC) is "sum of infinitesimal rectangles".
- Leibniz (and at least some of his contemporaries, as the Bernoullis) used a similar interpretation.
Best Answer
If a function has more integrals, all of them are the same.
For example, the function signum on a limited interval can be integrated by Riemann but not by Newton. It's because signum doesn't have any primitive function.