Floor it up

$\cot^{-1} x$ is strictly decreasing on the interval $[0, 2018\pi]$ and converges to 0, so $\lfloor \cot^{-1} x \rfloor = 0$ for $x > \cot 1 \approx 0.642$ . Therefore, we can ignore this part and focus on the interval $[0, \cot 1]$ .

$\cot^{-1} 0 = \frac {\pi}2 \approx 1.57 < 2 \Rightarrow \lfloor \cot^{-1} x \rfloor = 1$ (Again, because $\cot^{-1} x$ is strictly decreasing). This means that the area under the graph is simply made up of a rectangle with height $1$ and width $\cot 1 \approx 0.642$ , so its area is $\cot 1 \approx \boxed{0.642}$ .