Question

If $f\left(x\right)=x-\left[x\right]$, for every real number x, where $\left[x\right]$ is integral part of x. Then, ${\int }_{-1}^{1}f\left(x\right)dx$ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is B 1

$I={\int }_{-1}^{1}f\left(x\right)dx={\int }_{-1}^{1}x-\left[x\right]dx={\int }_{-1}^{1}xdx-{\int }_{-1}^1\left[x\right]dxI={\left[\frac{x^2}{2}\right]}_{-1}^1-{\int }_{-1}^1\left[x\right]dxI=[\frac{1^2}{2}-\frac{{(-1)}^2}{2}]-{\int }_{-1}^1\left[x\right]dxI=0-{\int }_{-1}^1\left[x\right]dxI={\int }_{-1}^0\left[x\right]dx+{\int }_0^1\left[x\right]dx$

The value of $\left[x\right]$,

$\left[x\right]=0x\in (0,1)\left[x\right]=-1x\in (-1,0)$

$\therefore I=-{\int }_{-1}^0(-1)dx-{\int }_0^1\left(0\right)dxI=-{[-x]}_{-1}^0-0I=-[\left(0\right)-(-(-1))]=1$.