# Evaluating ∫100−100⌊x3⌋dx\int _{-100}^{100}\lfloor {x^3}\rfloor \,dx

Is there an alternative better solution?

$I=\displaystyle\int_{-100}^{100}\lfloor x^3\rfloor\,dx$
$=\displaystyle\int_{-100}^{100}\lfloor(100-100-x)^3\rfloor\,dx$ $\quad$ [$\because\int_{a}^{b}f(x)\,dx=\int_{a}^{b}f(a+b-x)\,dx$]

$=\displaystyle\int_{-100}^{100}\lfloor-x^3\rfloor\,dx$

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

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

EDIT. Is there any area interpretation of the integral?

For $$0>x\not \in \Bbb Z0>x\not \in \Bbb Z$$ we have $$[-x]=-[x]-1.[-x]=-[x]-1.$$ $$\text {So }\quad \int_{-100}^0[y^3]\;dy = \int_0^{100}(-[x^3]-1)\;dx\text {So }\quad \int_{-100}^0[y^3]\;dy = \int_0^{100}(-[x^3]-1)\;dx$$ is obtained by letting $$y=-x.y=-x.$$ $$\text { Therefore } \quad \int_{-100}^{100}[x^3]\;dx=\int_{-100}^0 [y^3]\;dy+\int_0^{100}[x^3]\;dx=\text { Therefore } \quad \int_{-100}^{100}[x^3]\;dx=\int_{-100}^0 [y^3]\;dy+\int_0^{100}[x^3]\;dx=$$ $$=\int_0^{100}(-[x^3]-1)\;dx+\int_0^{100} [x^3]\;dx==\int_0^{100}(-[x^3]-1)\;dx+\int_0^{100} [x^3]\;dx=$$ $$=\int_0^{100} (-1-[x^3]+[x^3])\;dx=\int_0^{100}(-1)\;dx=-100.=\int_0^{100} (-1-[x^3]+[x^3])\;dx=\int_0^{100}(-1)\;dx=-100.$$

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