Question

Given $f(x)=\{\begin{array}{ll}x|x|& \text{ for }x\le -1\\ [x+1]+[1-x]& \text{ for }-1<x<1,\text{ where }[\\ -x|x|& \text{ for }x\ge 1\end{array}$ denotes the greatest integer function. If $I={\int }_{-2}^{2}f(x)dx,$ then $|3I|=$

Answer 1

Answer: 8

$f\left(x\right)=-x^{2}forx\le -1,f\left(x\right)=1for-1<x<0,$
$f\left(x\right)=2forx=0,f\left(x\right)=1for0<x<1andf\left(x\right)=-x^{2}forx\ge 1$
$\Rightarrow f\left(x\right)$ is even
$\therefore I=2{\int }_0^{2}f\left(x\right)dx=2{\int }_0^{1}f\left(x\right)dx+2{\int }_1^{2}f\left(x\right)dx=2{\int }_0^1\left(1\right)dx+2{\int }_1^2-x^{2}dx=\frac{-8}{3}$
$\therefore |3I|=|-8|=8$