Question

Let $f\left(x\right)=$ maximum $\{x+\left|x\right|,x-\left[x\right]\},$ where $\left[x\right]=$ greatest integer $\le x.$ Then, ${\int }_{-2}^{2}f\left(x\right)dx$ is equal to

• A. $\frac{1}{2}$
• B. $5$
• C. $\frac{3}{2}$
• D. $2$

Answer 1

The correct option is B $5$

$f\left(x\right)=max\{x+\left|x\right|,x-\left[x\right]\}$
$I={\int }_{-2}^{2}f\left(x\right)dx={\int }_{-2}^{-1}f\left(x\right)dx+{\int }_{-1}^{0}f\left(x\right)dx+{\int }_0^{1}f\left(x\right)dx+{\int }_1^{2}f\left(x\right)dx$
$f\left(x\right)=\{\begin{array}{c}x+2;-2<x<-1\\ x+1;-1<x<0\\ 2x;0<x<1\\ 2x;1<x<2\end{array}$
We get
$I={\int }_{-2}^{-1}(x+2)dx+{\int }_{-1}^0(x+1)dx+{\int }_0^1\left(2x\right)dx+{\int }_1^2\left(2x\right)dx$
$={[\frac{x^2}{2}+2x]}_{-2}^{-1}+{[\frac{x^2}{2}+x]}_{-1}^0+{\left[x^2\right]}_0^1+{\left[x^2\right]}_1^2$
$=\frac{1}{2}-2-\frac{4}{2}+4+0-\frac{1}{2}+1+1+4-1=5$