# The greatest integer function

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] $$\int_{2}^6 \lfloor 3x^2\rfloor dx$$

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