14
$\begingroup$

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?

$\endgroup$
2
  • 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

8
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.