I am having hard time understanding this. What is meant by "greatest integer?" Can anyone refer me to any visual/graphical explanation for $\lfloor x\rfloor$? I am trying with this question but could not do it. [find the greatest integer function][1] $$\int_{2}^6 \lfloor 3x^2\rfloor dx$$

  • $\begingroup$ $[x]$ (most commonly written $\lfloor x\rfloor$) is the greatest integer which is smaller or equal to $x$. For example, $[4]=4$, $[4.56]=4$, $[3.9999999]=3$, $[-3.2]=-4$, and so on. $\endgroup$ – Luiz Cordeiro Sep 19 '16 at 9:38
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    $\begingroup$ have a look at [Floor_and_ceiling_functions][1] [1]: en.wikipedia.org/wiki/Floor_and_ceiling_functions $\endgroup$ – G Cab Sep 19 '16 at 9:39
  • $\begingroup$ Another question with some useful answers about integrating the greatest integer function is math.stackexchange.com/questions/408953/… $\endgroup$ – David K Apr 22 '17 at 11:06

For $x\in [2,6)$, $x\to 3x^2$ is an increasing function and it attains the values in $[3\cdot 2^2,3\cdot 6^2)=[12,108)$. For any integer $k\in [12,107]$, let $$I_k=\{x\in [2,6): k\leq 3x^2<k+1\}=[\sqrt{k/3},\sqrt{(k+1)/3}).$$ Note that if $x\in I_k$ then $\lfloor 3x^2\rfloor=k$ which means that $\lfloor 3x^2\rfloor$ is constant on each interval $I_k$.

Moreover $|I_k|=\sqrt{(k+1)/3}-\sqrt{k/3}$ where $|I_k|$ is the length of $I_k$.

Hence by the definition of integral, $$\int_{2}^6 \lfloor 3x^2\rfloor dx=\sum_{k=12}^{107} k|I_k| =\frac{1}{\sqrt{3}}\sum_{k=12}^{107} k(\sqrt{k+1}-\sqrt{k}) \approx 206.005$$ where $|I_k|$ is the length of the interval $I_k$.

P.S. For general information about the floor function take a look HERE.

  • $\begingroup$ thank you so much Mr. Robert. can you please explain more or show some steps that get you to the answer. I really appreciate it. $\endgroup$ – Sarah Sep 21 '16 at 13:33
  • $\begingroup$ @Sarah I edited my answer with more details. Further doubts? $\endgroup$ – Robert Z Sep 21 '16 at 13:43

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