I figured out that
$$\int_0^x \lfloor t \rfloor \cdot dt= \frac{x\cdot(x-1)}{2} \qquad when \; x=\lfloor x \rfloor$$
But what would the equation look like when $x \neq \lfloor x \rfloor$?
I believe that there would be another positive term on the right that would look similar to the absolute value of sine.
We would have $$\int_0^x \lfloor t \rfloor dt = \int_0^{\lfloor x \rfloor} \lfloor t \rfloor dt + \int_{\lfloor x \rfloor}^x \lfloor t \rfloor dt = \frac{\lfloor x \rfloor \cdot (\lfloor x \rfloor - 1)}{2} + \lfloor x \rfloor (x - \lfloor x \rfloor) = \frac{\lfloor x \rfloor (2x - \lfloor x \rfloor - 1)}{2}$$