Question

Let $f\left(x\right)=|x-2|$ and $g\left(x\right)=f\left(f\right(x\left)\right),x\in [0,4].$ Then ${\int }_0^3\left(g\right(x)-f(x\left)\right)dx$ equal to

Answer 1

$f\left(x\right)=|x-2|$$=\{\begin{array}{cc}x-2,& x\ge 2\\ 2-x,& x<2\end{array}$
$g\left(x\right)=||x-2|-2|$
$=\{\begin{array}{cc}|x-4|,& x\ge 2\\ |x|,& x<2\end{array}$
$=\{\begin{array}{cc}x,& 0\le x<2\\ 4-x,& 2\le x<4\\ x-4,& x\ge 4\end{array}$


$\therefore {\int }_0^3\left(g\right(x)-f(x\left)\right)dx$
$={\int }_0^{3}g\left(x\right)-{\int }_0^{3}f\left(x\right)dx$
$=(\frac{1}{2}\times 2\times 2+1+\frac{1}{2}\times 1\times 1)-(\frac{1}{2}\times 2\times 2+\frac{1}{2}\times 1\times 1)$
$=\frac{7}{2}-\frac{5}{2}$
$\therefore {\int }_0^3\left(g\right(x)-f(x\left)\right)dx=1$