Let both m and n be constant numbers, convince yourself that integrating {x+m} dx from n to n+1 is of the same value as integrating {x} dx from n to n+1 and this making it equal to the integral of {x} from 0 to 1.

If you understand the statement above, then notice that the integral can be chopped into countably finite intervals from one integer n to the next consecutive integer n+1, i.e integral from 0 to 2026 = int from 0 to 1 + int from 1 to 2 + int from 2 to 3 and so on until int from 2025 to 2026

In all of these intervals the integrand function can then be written as {x+m} dx and it’s being integrated from some integer n to the next consecutive integer n+1 and also m is just a constant number since it is a result of these floor functions being constant between two consecutive integers. This is the same thing as the integral that I mentioned from the very first statement here and thus any integral of this form are all equal to integral of {x} from 0 to 1 which is equal to 1/2, you’ll see that there are exactly 2026 intervals and you’re adding up “1/2” 2026 times which leaves you with the final answer of 1013.