Hint: the answer might contain 2024

Here's the related video I mentioned at the beginning:
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[@maxvangulik1988]

floor(-2024pi)=-floor(2024pi)-1
I=int[-2024pi,ceil(-2024pi)](floor(x))dx (everything else cancels)
width=ceil(-2024pi)+2024pi
height=floor(-2024pi)
I=floor(-2024pi)•(ceil(-2024pi)+2024pi)
I=({2024pi}-1)•floor(-2024pi)

[@owl3math]

Here is the link to that previous video where I looked at the case with integer bounds and bounds going to infinity:
ruclips.net/video/D5kwCC_Pa9A/видео.html

[@baconboyxy]

I(a, b) = Int[a, b] floor(x)
I(a, b) = Sum[a, b-1] x
I(-c, c) = Sum[-c, c-1]
-c + 1 through c - 1 cancel each other out, leaving -c
I(-c, c) = -c
In this case I(-2024pi, 2024pi) = -2024pi

[@adandap]

I kept getting myself confused with the bounds until I drew a number line and marked various points, setting floor(2024 pi) = N. Then it was easy enough to calculate the three contributions.

[@owl3math]

Hi adandap. Yep kind of some basic book keeping here. Kind of cool how everything cancels! Not sure I’ll remember this as a formula however that you always get the lower bound as the answer.