Another way to think about it: floor rounds down, so there are intervals of numbers which all have the same floor, so you can break the problem up into interval

Another way to think about it: floor rounds down, so there are intervals of numbers which all have the same floor, so you can break the problem up into intervals where floor(x/2) is constant. On [0, 2), floor(x/2) = 0, on [2, 4), floor(x/2) = 1, on [4, 6), floor(x/2) = 2, etc. So in each interval you are integrating e^(x/2) multiplied by e^0 or e^-1 or e^-2 etc. Effectively the same integral 1000 times, and the different limits and constants balance out in the end (try it to see that).