What is the greatest integer function, and how do you integrate it? $[x]$ denotes the greatest integer $\leq x$. Let $f(x)=[x]$ and let $g(x)=[2x].$  
I am having hard time understanding this. What is meant by "greatest integer?" Can anyone refer me to any visual/graphical explanation for $[x]?$ 
For example, how would we draw the graph for $f(x)=[\sqrt{x}]$? 
Also I am a little confused by the including properties such as $[2x]=[x]+[x+1/2]$ and with $[3x],[4x]...$ 

Lastly, how would you integrate it?

For example, what would be the evaluation of the integral of $\int_1^3{[x]dx}$ and integral of $\int_0^9[\sqrt t]dt$? 
In general, I am completely confused by $[x]$ and its meaning and its functional ability. So any intuitive explanation would be much appreciated. 

The more usual notation is nowadays $\lfloor x\rfloor$. "The greatest integer that is $\le x$" means what it says. Everything will be clear once we do some examples.
The number $\lfloor 3.6\rfloor$ is the biggest integer which is $\le 3.6$. This is $3$.  So $\lfloor 3.6\rfloor=3$. Similarly, $\lfloor \pi\rfloor=3$. Also, $\rfloor 17\rfloor =17$. Informally, if $x$ is an integer, then $\lfloor x\rfloor=x$. If $x$ is not an integer, we go down until we meet the first integer. For positive $x$, that means knocking off the part after the decimal point. 
But it is easy to get a little confused when we apply the greatest integer function to negative numbers. Note that $\lfloor -4.7\rfloor=-5$, since the biggest integer which is $\le -4.7$ is $-5$. Luckily, one meets $\lfloor x\rfloor$ mainly with positive $x$. 
Now let us do an integrations, such as 
$$\int_0^9 \left\lfloor \sqrt{x}\right\rfloor \,dx.$$
Let $f(x)=\left\lfloor \sqrt{x}\right\rfloor$.  First we will see what $f(x)$ looks like. 
For $0\le x\lt 1$, we have $0\le \sqrt{x}\lt 1$, and therefore $\left\lfloor \sqrt{x}\right\rfloor=0$. 
For $1\le x\lt 4$, we have $1\le \sqrt{x}\lt 2$, and therefore $\left\lfloor \sqrt{x}\right\rfloor =1$. 
For $4\le x\lt 9$, similarly we have $\left\lfloor \sqrt{x}\right\rfloor=2$. And finally for $x=9$ we have $\left\lfloor \sqrt{x}\right\rfloor=3$. 
So $f(x)$ is defined by different formulas over different parts of the world. It is $0$ in the interval $[0,1)$, and $1$ in the interval $[1,4)$, and $2$ in the interval $[4,9)$, and $0$ at $9$. You should graph $y=f(x)$ to see clearly what comes next.
To integrate, we integrate separately over our intervals, and add up. So we want
$$\int_0^1 (0)\,dx+\int_1^4 1\,dx+\int_4^9 2\,dx+\int_9^9 3\,dx.$$
The first integral and the last are both $0$. The middle two are $3$ and $10$, so our integral is $13$.  The individual integrations are easy: geometrically, we are just finding the areas of rectangles. 
