Boxing Integral

We note that $\left \lfloor t + \left \lfloor t + \left \lfloor t \right \rfloor \right \rfloor \right \rfloor = 3 \left \lfloor t \right \rfloor$ , therefore:

$\begin{aligned} \int_{-4}^4 \left \lfloor t + \left \lfloor t + \left \lfloor t \right \rfloor \right \rfloor \right \rfloor dt & = \int_{-4}^4 3 \left \lfloor t \right \rfloor dt \\ & = 3\left( \int_{-4}^{-3} \left \lfloor t \right \rfloor dt + \int_{-3}^{-2} \left \lfloor t \right \rfloor dt + \int_{-2}^{-1} \left \lfloor t \right \rfloor dt + \int_{-1}^{0} \left \lfloor t \right \rfloor dt + \int_{0}^{1} \left \lfloor t \right \rfloor dt + \int_{1}^{2} \left \lfloor t \right \rfloor dt + \int_{2}^{3} \left \lfloor t \right \rfloor dt + \int_{3}^{4} \left \lfloor t \right \rfloor dt \right) \\ & = 3\left( \int_{-4}^{-3} -4 \space dt + \int_{-3}^{-2} -3 \space dt + \int_{-2}^{-1} -2 \space dt + \int_{-1}^{0} -1 \space dt + \int_{0}^{1} 0 \space dt + \int_{1}^{2} 1 \space dt + \int_{2}^{3} 2 \space dt + \int_{3}^{4} 3 \space dt \right) \\ & = 3(-4-3-2-1+0+1+2+3) \\ & = \boxed{-12} \end{aligned}$

$\lfloor t + \lfloor t + \lfloor t \rfloor \rfloor \rfloor = 3\lfloor t \rfloor$

$\displaystyle \int_{-4}^{4}\lfloor t + \lfloor t + \lfloor t \rfloor \rfloor \rfloor dt = \displaystyle 3\int_{-4}^{4} \lfloor t \rfloor dt$
Using the property,

$\displaystyle \int_{-a}^{a} f(t)dt = \int_{0}^{a} f(t) + f(-t) dt$
$I = \displaystyle 3\int_{-4}^{4} \lfloor t \rfloor dt = 3\int_{0}^{4} \lfloor t \rfloor + \lfloor -t \rfloor dt = 3\int_{0}^{4} \lfloor t \rfloor - \lfloor t \rfloor - 1 dt = -3\int_{0}^{4} dt$
I used the property , for t > 0 $\lfloor -t \rfloor = -\lfloor t \rfloor - 1$
$\therefore I = \displaystyle -3\int_{0}^{4}dt = -3 \cdot 4 = -12$