Question

Let $f\left(x\right)=x-\left[x\right]$for $xϵR,$where$\left[x\right]$=the greatest integer$\le x$, then ${\int }_{-2}^{2}f\left(x\right)dx$ is

• A. $4$
• B. $2$
• C. $0$
• D. $1$

Answer 1

The correct option is B $2$

$I={\int }_{-2}^{2}f\left(x\right)dx={\int }_{-2}^2(x-\left[x\right])dx={\int }_{-2}^{2}xdx-{\int }_{-2}^2\left[x\right]dx={\int }_{-2}^{2}xdx-{\int }_{-2}^{-1}\left[x\right]dx-{\int }_{-1}^0\left[x\right]dx-{\int }_0^1\left[x\right]dx-{\int }_1^2\left[x\right]dx={\int }_{-2}^{2}xdx-{\int }_{-2}^{-1}(-2)dx-{\int }_{-1}^0(-1)dx-{\int }_0^1\left(0\right)dx-{\int }_1^2\left(1\right)dx={\left[\frac{x^2}{2}\right]}_{-2}^2+2{\left[x\right]}_{-2}^{-1}+{\left[x\right]}_{-1}^0+0-{\left[x\right]}_1^2=\frac{4}{2}-\frac{4}{2}+2(-1+2)+(0+1)-(2-1)=2$