Check dimensions of the integral of a function

It will be the latter case, $m^2/s^3m$ which is just $m^3/s^3$.

Remember that the integral is the sum of all the products $f(x)\;\text{times}\; dx$. $dx$ is a tiny piece of the path from $0$ to $x$, so it is in units of $m$ as well. Each of the products $f(x)dx$ have units $m^3/s^3$, and the sum of all these products keeps those units.

The dimensions of the integral are simply those of $f(x)dx$, so in this case they would be $m^2/s^3 \times m = m^3/s^3$.