While pondering the value of integrals and derivatives, I stumbled on to a question.

What are you left with if you take the integral of distance? In case I'm wording this wrong (it's been awhile since math, and I feel rusty) when I say distance I refer to what you get when you take the integral of velocity, given that you know C.

I see that there is no usage of the KB anymore.

Sadly, yes. It's entirely too many clicks to try to discover the same amount of information, people get to it slower, and it's being removed soon. Plus, the forum has the same outcome.

Put a question on the KB, it takes a week to get answered. Post one on the forum, the answer arrives tomorrow.

It doesn't really have a physical definition. If you have an equation for the position and you integrate, you get another equation. It is mathematically valid, but it doesn't represent anything. At least, not so far as I know.

In Dimensional Analysis (using L for Length and T for Time):

Acceleration is represented by L¹T^-2
Velocity is represented by L¹T^-1
Distance is represented by L¹T⁰

So integrating Distance would produce something represented by L¹T¹

As far as I know there are no real measurements that use this representation (or a similar Dimension, which could easily be changed)

The closest I could find to "ms" (Metre-Second) Would be a Coulomb-Metre per Amp, which breaks down to this, but would be completely fictitious:
(Including A for Amperage)

Coulomb = A¹L⁰T¹
Metre = A⁰L¹T⁰
Coulomb-Metre = A¹L¹T¹
Amp = A¹L⁰T⁰
Coulomb-Metre per Amp = A⁰L¹T¹

I have no idea how this measurement would be implemented in reality