Three floors

We note that:

$\left \lfloor x +\left \lfloor x+\lfloor x \rfloor \right \rfloor \right \rfloor = \begin{cases} 0 \text{ for } 0\le x < 1\\ 3 \text{ for } 1\le x<2\end{cases}$

Therefore,

$\begin{aligned} \int_0^2 \left \lfloor x +\left \lfloor x+\lfloor x \rfloor \right \rfloor \right \rfloor & =\int_0^10 d x+\int_1^23 dx \\ & =0+3=\boxed{3} \end{aligned}$

As we are integrating between non-negative numbers, $\lfloor x \rfloor = x$ .

Thus we can say $\lfloor x +\lfloor x+\lfloor x \rfloor \rfloor \rfloor = \lfloor 3x \rfloor = 3 \lfloor x \rfloor$ for $x \ge 0$ .

Therefore, $\displaystyle \int^{2}_{0} \lfloor x + \lfloor x+\lfloor x \rfloor \rfloor \rfloor dx= \int^{2}_{0} 3\lfloor x \rfloor dx = 3 \int^{2}_{0}\lfloor x \rfloor dx$ .

Observing the graph for $y= \lfloor x \rfloor$ , we see that $\displaystyle \int^{2}_{0} \lfloor x \rfloor dx = 1$ .

$\Rightarrow \displaystyle \int^{2}_{0} \left \lfloor x +\left \lfloor x+\lfloor x \rfloor \right \rfloor \right \rfloor dx = \large\color{#20A900}{\boxed{3}}$