Let f x = x - 2 and g x = f f x ,   x ∈ 0 , 4 . Then ∫ 0 3 g x - f x   d x is equal to

$f\left(x\right)=\left|x-2\right|=\left\{\begin{array}{l}2-x; x<2\\ x-2; x\ge 2\end{array}\right.$

$g\left(x\right)=f\left(f\left(x\right)\right)=\left\{\begin{array}{l}2-f\left(x\right); f\left(x\right)<2\\ f\left(x\right)-2; f\left(x\right)\ge 2\end{array}\right.$

$=\left\{\begin{array}{l}2-\left(2-x\right); 2-x<2,x<2\\ \left(2-x\right)-2; 2-x\ge 2,x<2\\ 2-\left(x-2\right); x-2<2,x\ge 2\\ \left(x-2\right)-2; x-2\ge 2,x\ge 2\end{array}\right.$

$=\left\{\begin{array}{l}2-\left(2-x\right); \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}0<x<2\\ \left(2-x\right)-2; \hspace{0.17em}x<0\\ 2-\left(x-2\right); \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2\le x<4\\ \left(x-2\right)-2; x\ge 4\end{array}\right.$

$\underset{0}{\overset{3}{\int }}\left(g\left(x\right)-f\left(x\right)\right)dx=\underset{0}{\overset{2}{\int }}xdx+\underset{2}{\overset{3}{\int }}\left(4-x\right)dx-\underset{0}{\overset{3}{\int }}\left|x-2\right|dx$

$={\left[\frac{x^2}{2}\right]}_0^2+{\left[4x-\frac{x^2}{2}\right]}_2^3-{\left[2x-\frac{x^2}{2}\right]}_0^2-{\left[\frac{x^2}{2}-2x\right]}_2^3$

$=1$.