Question

The value of integral $\underset{-1}{\overset{2}{\int }}\left[\frac{\left[x\right]}{1+x^2}\right]dx,$ where $[\cdot ]$denotes the greatest integer function, is equal to

• A. $-2$
• B. $-1$
• C. $0$
• D. $1$

Answer 1

The correct option is B $-1$

We know, $\left[x\right]=0,\forall x\in [0,1)$
For $x\in [1,2),\left[x\right]=1$
$\Rightarrow \frac{\left[x\right]}{1+x^2}=\frac{1}{1+x^2}$
Clearly, $\frac{1}{1+x^2}<1,\forall x\in [1,2)$
$\Rightarrow \left[\frac{\left[x\right]}{1+x^2}\right]=0$
For $x\in [-1,0),\left[x\right]=-1$
$\Rightarrow \left[\frac{\left[x\right]}{1+x^2}\right]=\left[\frac{-1}{1+x^2}\right]$
In the interval $x\in [-1,0)$
$\Rightarrow \frac{1}{2}\le \frac{1}{1+x^2}<1$$\Rightarrow -\frac{1}{2}\ge \frac{-1}{1+x^2}>-1$
$\Rightarrow \left[\frac{\left[x\right]}{1+x^2}\right]=-1\forall x\in [-1,0)$
Thus, the given integral is
$I=\underset{-1}{\overset{0}{\int }}(-1)dx+\underset{0}{\overset{1}{\int }}\left(0\right)dx+\underset{1}{\overset{2}{\int }}\left(0\right)dx\Rightarrow I=-1$