I'm trying to build an intuition for the relationship of integrals to the equations of motion.
I was using this site to get an overall intuition for the relationship between integrals and derivatives. It seems clear how a cube changes in that the derivative is like adding 3 areas so that it grows by $3x^2$. And the reverse is just integrating the rate to get the cube $x^3 + c$.
Now my confusion is when we consider movement. Acceleration has units in $m/s^2$ but velocity is not cubed (at least in my understanding). So why does the integral of acceleration with respect to time produce values with cubes? I feel like I'm missing something obvious here.
If you integrate a function $f(t)$ with respect to time $t$, the result will be measured in the units of $f$, multiplied by the units of $t$; ie.
$ \begin{equation} [\int f(t)\,dt] = [f(t)]\times [t] \end{equation} $
where $[\cdot]$ means "the units of". Likewise, if you differentiate $f(t)$ with respect to $t$, the result will have the units of $f(t)$, divided by the units of $t$:
$ \begin{equation} [\frac{df}{dt}] = \frac{[f(t)]}{[t]} \end{equation} $
As you noted, in SI units acceleration has the units of $\mbox{m/s}^2$. So if you integrate $a(t)$ to find the velocity $v(t)$, you know the units of $v(t)$ will be $\mbox{m/s}$:
$ \begin{equation} [v(t)]=[\int a(t)\,dt] = [a(t)]\times [t] = \mbox{m/s}^2\times s=\mbox{m/s} \end{equation} $
However, this does not imply the acceleration must scale as some particular power $n$ of $t$, since any power of $t$ can be multiplied by some dimensional prefactor $A$ such that $At^n$ has units of $\mbox{m/s}^2$. For example, if $n=3$, and the units of $A$ are $\mbox{m/s}^5$, then $At^3$ is a valid acceleration, in the sense that its units are $\mbox{m/s}^2$.