Is there an alternative better solution?

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


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

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

EDIT. Is there any area interpretation of the integral?

  • 1
    $\begingroup$ No, but it's a good approach. $\endgroup$
    – Arthur
    Jul 2, 2016 at 12:10
  • $\begingroup$ Nothing wrong with that but I posted an A that's a little bit different. $\endgroup$ Jul 2, 2016 at 16:35

1 Answer 1


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.

  • $\begingroup$ Do you mean to say that x€Z, constitute only isolated points on the graph and hence doesn't affect the integration? $\endgroup$
    – user600016
    Jul 21, 2019 at 12:26
  • $\begingroup$ @thewitness .Exactly.... I received an upvote yesterday so I looked at this and was surprised to see that after 3 years and 8 upvotes and an Acceptance, my 1st sentence, which said "For $0>x...$", not "For $0>x\not \in \Bbb Z$..." was false, but nobody else noticed. $\endgroup$ Jul 21, 2019 at 17:09

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