Question

Let f(x) = x - [x], for every real number x, where [x] is the greatest integer less than or equal to x. Then, the value of ${\int }_{-1}^{1}f\left(x\right)\mathrm{dx}$ is

• A. 1
• B. \N
• C. $\frac{1}{2}$
• D. $\frac{3}{2}$

Answer 1

The correct option is A 1

Clearly, $x-\left(x\right]=(\begin{array}{c}x-(-1)=x+1,\mathrm{if}-1\le x<0\\ x-0=x,\mathrm{if}0\le x<1\end{array}\therefore {\int }_{-1}^0(x-\left[x\right])\mathrm{dx}+{\int }_0^1(x-\left[x\right])\mathrm{dx}={\int }_{-1}^0(x+1)\mathrm{dx}+{\int }_0^1(x-0)\mathrm{dx}={\left(\frac{{\left(x+1\right)}^2}{2}\right]}_{-1}^0+{\left(\frac{x^2}{2}\right]}_0^1=\frac{1}{2}+\frac{1}{2}=1$