Question

If$f\left(x\right)=\{\begin{array}{cc}1-|x|,|x|\le 1& \\ |x|-1,|x|>1& \end{array}$, and $g\left(x\right)=f(x-1)+f(x+1)$. The value of $\underset{-3}{\overset{5}{\int }}g\left(x\right)dx$

Answer 1

$f\left(x\right)=\{\begin{array}{cc}-x-1,& x<-1\\ 1+x& -1\le x<0\\ 1-x,& 0\le x\le 1\\ x-1& x>1\end{array}$
;
$f(x-1)=\{\begin{array}{ccc}-x,& x-1<-1& \Rightarrow x<0\\ x,& -1\le x-1<0& \Rightarrow 0\le x<1\\ 2-x,& 0\le x-1\le 1& \Rightarrow 1\le x\le 2\\ x-2& x-1>1& \Rightarrow x>2\end{array}$
Similarly,
$f(x+1)=\{\begin{array}{ccc}-x-2,& x+1<1-1& \Rightarrow x<-2\\ x+2& -1\le x+1<0& \Rightarrow -2\le x<-1\\ -x,& 0\le x+1\le 1& \Rightarrow -1\le x\le 0\\ x-1& x+1>1& \Rightarrow x>0\end{array}$
$\Rightarrow g\left(x\right)=f(x-1)+f(x+1)=\{\begin{array}{cc}-2x-2& x<-2,\\ 2,& -2\le x<-1\\ -2x& -1\le x\le 0\\ 2x& 0<x<1\\ 2& 1<x\le 2\\ 2x-2,& 2<x\end{array}$
clearly $g\left(x\right)$ is even,
$\text{Now}\underset{-3}{\overset{5}{\int }}g\left(x\right)dx=2{\int }_0^{3}g\left(x\right)dx+\underset{3}{\overset{5}{\int }}g\left(x\right)dx=2\left(\underset{0}{\overset{1}{\int }}2xdx+\underset{1}{\overset{2}{\int }}2dx+\underset{2}{\overset{3}{\int }}(2x-2)dx\right)+\underset{3}{\overset{5}{\int }}(2x-2)dx=2(1+2+3)+12=24$