Question

Let $f\left(x\right)=\{\begin{array}{cc}2-|x|,& |x|\le 1\\ 0,& |x|>1\end{array}$ and $g\left(x\right)=f(x+1)$.
The value of $\underset{-3}{\overset{3}{\int }}g\left(x\right)dx$ is

Answer 1

$g\left(x\right)=f(x+1)\Rightarrow g\left(x\right)=\{\begin{array}{cc}2-|x+1|,& |x+1|\le 1\Rightarrow -2\le x\le 0\\ 0,& x<-2,x>0\end{array}$
$\underset{-3}{\overset{3}{\int }}g\left(x\right)dx=\underset{-3}{\overset{-2}{\int }}g\left(x\right)dx+\underset{-2}{\overset{0}{\int }}g\left(x\right)dx+\underset{0}{\overset{3}{\int }}g\left(x\right)dx$
$g\left(x\right)\text{is}0\text{for}x<-2\text{and}x>0$
$\Rightarrow \underset{-3}{\overset{3}{\int }}g\left(x\right)dx=\underset{-2}{\overset{0}{\int }}g\left(x\right)dx=\underset{-2}{\overset{0}{\int }}(2-|x+1|)dx$
$=\underset{-2}{\overset{0}{\int }}2dx-\underset{-2}{\overset{0}{\int }}|x+1|dx$
$=4+\underset{-2}{\overset{-1}{\int }}(x+1)dx-\underset{-1}{\overset{0}{\int }}(x+1)dx$
$=4+{[\frac{x^2}{2}+x]}_{-2}^{-1}-{[\frac{x^2}{2}+x]}_{-1}^0$
$=4+[\frac{1}{2}-1-(\frac{4}{2}-2)]+[\frac{1}{2}-1]$
$=3$