Question

Evaluate ${\int }_{-1}^{2}f\left(x\right)dx,wheref\left(x\right)=|x+1|+|x|+|x+1|$

Answer 1

We can redefine $f$ as $f\left(x\right)=2-x,if-1<x\le 0$
$=x+2,if0<x\le 1$

$=3x,if1<x\le 2$

Therefore ${\int }_{-1}^{2}f\left(x\right)dx={\int }_{-1}^0(2-x)dx+{\int }_0^1(x+2)dx+{\int }_1^{2}3xdx$ (by $P_2$)
$={(2x-\frac{x^2}{2})}_{-1}^0+{(\frac{x^2}{2}+2x)}_0^1{\left(\frac{3x^2}{2}\right)}_1^1$
$=0-(-2-\frac{1}{2})+(\frac{1}{2}+2)+3(\frac{4}{2}-\frac{1}{2})=\frac{5}{2}+\frac{5}{2}+\frac{9}{2}=\frac{19}{2}$