$ \int_{1}^{2} xe^{\lfloor x\rfloor +\lfloor x^3\rfloor }\,dx $

Where $\lfloor x\rfloor $ is floor function or greatest integer function

I thought since the limits are from $1$ to $2$ then I can integrate

$\int_{1}^{2} xe^{x+x^3}\,dx $

Then I tried solving via by parts by differentiating $x$ and integration the exponential function but that didn't seem to work.

How should i approach this question and other variations involving gif ?


2 Answers 2


$1^3=1$ and $2^3=8$, so the integral will need to be split into pieces for each cube root in between. And since $\lfloor x \rfloor$ is constant itself on the interval, we only have one sum in the potential double sum:

$$I = \sum_{n=1}^7 \int_\sqrt[3]{n}^\sqrt[3]{n+1}x e^{1+n}\:dx$$


Let $x$ be a real number, we always have $\lfloor x\rfloor=n$ iff $n\leqslant x<n+1$. Then, over $[1,2[$ we have $\lfloor x\rfloor =1$. Similarly, $\lfloor x^3\rfloor=n$ iff $n\leqslant x^3<n+1$, then we need break the integral up into pieces of the form $\displaystyle \int_{n^{1/3}}^{(n+1)^{1/3}}xe^{1+n}dx$. Then, the sum of these pieces over all $n$ in $[|1,7|]$. Thus, $$\int_{1}^{2}xe^{\lfloor x\rfloor+\lfloor x^3\rfloor}dx=\sum_{n\in [|1,7|]}\int_{n^{1/3}}^{(n+1)^{1/3}}xe^{1+n}dx=\sum_{n\in [|1,7|]}\left(\frac{(n+1)^{2/3}-n^{2/3}}{2}\right)e^{n+1}\approx 829.0996$$

Edit: corrected.

  • 2
    $\begingroup$ That's close, except in OP's problem the cube is inside of the floors. The answer is a little more complicated than that. $\endgroup$ Jan 30 at 16:46
  • 1
    $\begingroup$ @NinadMunshi: I have restored and corrected my answer using your observation so that the effort has not been in vain. I have added a numerical approximation of the value of the integral. $\endgroup$
    – A. P.
    Jan 30 at 17:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .