if you look at the graph below, the "ceiling" function, is f(x), on that interval of [-4,0], whilst the "floor" function, is g(x)
so, taking that into account, then [tex]\bf \begin{cases}
f(x)=x^3+x^2-6x\\\\
g(x)=6x
\end{cases}\qquad \displaystyle \int\limits_{-4}^{0}\
\begin{array}{llll}
[ceiling]&-&[floor]\\
f(x)&&g(x)
\end{array}
\\\\\\
thus
\\\\\\
\displaystyle \int\limits_{-4}^{0}\ [[x^3+x^2-6x]-[6x]]dx\implies
\displaystyle \int\limits_{-4}^{0}\ x^3+x^2-12x
\\\\\\
\left. \cfrac{x^4}{4}+\cfrac{x^3}{3}-6x^2 \right]_{0}^{-4}[/tex]
