Your expression, with a $(dx)^2$ in the integral, makes me uncomfortable. My comfort is not of overwhelming importance. But your expression is also highly non-standard, and may contribute to confusion later.
It is not how I do it when thinking informally.
Imagine the vertical strip from $x$ to $x+h$ where $h$ is very very small. If you like, you can write $dx$ instead of $h$, and even think of $dx$ as being "infinitesimal." I will use $h$, but there are good arguments for $dx$, including a centuries-old tradition that still survives in some applied areas.
We find an approximate expression for the moment of our strip about the $y$-axis. If you had been using $h$, your expression would have been $hf(x)(x+h/2)$. That's a somewhat better approximation than $(hf(x))(x)$. The intuition you had is that it is more accurate to think of the mass of the strip as concentrated at a distance $x+h/2$ from the $y$-axis. For reasonably-behaved functions, this is true.
Note that your expression differs from the simpler expression $(hf(x))(x)$ by $(h^2/2)(xf(x))$. If $h$ is very small, and $f$ is at all respectable, this difference is tiny in comparison with $(hf(x))(x)$.
Actually, your small change, from $x$ to $x+h/2$, could have been improved on. For the area of the strip is better approximated by $hf(x+h/2)$. If our function is differentiable, then by the tangent line approximation, $f(x+h/2)\approx f(x)+(h/2)f'(x)$. If you take this correction into account, the estimate for the moment becomes
So it would have been more accurate to use $hxf(x)[1+h/2)(f(x)+xf'(x)]$. But still, here we also have a main term $hxf(x)$, plus a (partial) correction term that is very small compared to the main term.
However, we are getting too fancy. A standard way to think informally is that the moment of the strip is $hxf(x)$ plus a term that approaches $0$ faster than any constant multiple of $h$, so that the error in our approximation is negligible compared to $hxf(x)$. For nice functions $f$, the error in our approximation behaves like a constant times $h^2$, and for $h$ small enough, $h^2$ is negligible compared to $h$.
Even more informally, we throw away terms in $h^2$ or higher powers of $h$!
So we have divided our interval $[a,b]$ into strips of tiny width. We want to "add up" the moments of these strips. There is a mnemonic advantage to using $dx$ instead of $h$. We are "adding up" moments $xf(x)dx$. The "sum" is $\int_a^b xf(x)dx$. Recall that the integral symbol $\int$ was originally an elongated $S$, for sum (in Latin).
We have discarded the terms in powers of $h$ that are $2$ or higher before writing down the integral. They leave no trace in the integral.