Question

Given $f(x)=\left\{\begin{array}{ll}x|x| & \text { for } x \leq-1 \\ {[x+1]+[1-x]} & \text { for }-1

Answer 1

$f\left(x\right)=-x^{2} \, for \, x\leq -1, \, f\left(x\right)=1 \, for \, -1 < x < 0,$
$f\left(x\right)=2 \, for \, x=0,f\left(x\right)=1 \, for \, 0 < x < 1 \, andf\left(x\right)=-x^{2} \, for \, x\geq 1$
$\Rightarrow f\left(x\right)$ is even
$\therefore I=2\displaystyle \int _{0}^{2} f \left(x\right) \, dx=2 \, \displaystyle \int _{0}^{1} f \left(x\right) \, dx+2\displaystyle \int _{1}^{2} f \left(x\right) \, dx=2\displaystyle \int _{0}^{1} \left(1\right) \, dx+2\displaystyle \int _{1}^{2} - x^{2}dx=\frac{- 8}{3}$
$\therefore \, \left|\right.3I\left|\right. \, = \, \left|\right.-8\left|\right. \, =8$