Is there an alternative better solution?

I=∫100−100⌊x3⌋dx

=∫100−100⌊(100−100−x)3⌋dx [∵]=\displaystyle\int_{-100}^{100}\lfloor-x^3\rfloor\,dx

=\displaystyle\int_{-100}^{100}(-\lfloor x^3\rfloor-1)\,dx \quad [\because \lfloor x\rfloor+\lfloor-x\rfloor=-1 when x\notin \mathbb{Z}]

\Rightarrow I=-I-200 \quad \Rightarrow I=-100

EDIT.Is there any area interpretation of the integral?

**Answer**

For 0>x\not \in \Bbb Z we have [-x]=-[x]-1. \text {So }\quad \int_{-100}^0[y^3]\;dy = \int_0^{100}(-[x^3]-1)\;dx is obtained by letting y=-x. \text { Therefore } \quad \int_{-100}^{100}[x^3]\;dx=\int_{-100}^0 [y^3]\;dy+\int_0^{100}[x^3]\;dx= =\int_0^{100}(-[x^3]-1)\;dx+\int_0^{100} [x^3]\;dx= =\int_0^{100} (-1-[x^3]+[x^3])\;dx=\int_0^{100}(-1)\;dx=-100.

Remark: (Added July 2019). We have [-x] \ne -[x]-1 if 0\ge x\in \Bbb Z. But the integrals in the 1st displayed line are still equal because \{y\in [-100,0]: [y^3]\ne -[y^3]-1\} is finite.

**Attribution***Source : Link , Question Author : StubbornAtom , Answer Author : DanielWainfleet*