Finding the value of the integral $\int_{1/2}^{1}[1/x^2]dx$ I came across the following problem that says:  

For any $x \in \mathbb R$,let $[x]$ denote the greatest integer smaller than or equal to $x$. Then the value of the integral  $\int_{1/2}^{1}[1/x^2]dx$ is which of the following:
  1.$\frac {1}{\sqrt 3}+\frac {1}{\sqrt 2}-\frac {1}{2}$,
  2.$\frac {1}{\sqrt 3}+\frac {1}{\sqrt 2}$,
  3.$\frac {1}{\sqrt 3}+\frac {1}{\sqrt 2}+\frac {1}{2}$,
  4.$\frac {1}{\sqrt 3}-\frac {1}{\sqrt 2}$.  

I do not know how to progress with the problem. Can someone point me in the right direction? Thanks in advance for your time.

When $x=1/2$, $[1/x^2] = 4$; when $x=1/\sqrt{3}$, it is equal to $3$, etc.  The integral is then
$$\int_{1/2}^1 dx \: [1/x^2] = -4 \left ( \frac{1}{2} - \frac{1}{\sqrt{3}} \right ) - 3 \left ( \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{2}} \right ) - 2 \left ( \frac{1}{\sqrt{2}} - 1 \right )$$
or,
$$\int_{1/2}^1 dx \: [1/x^2] =  \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{2}}   $$
