Let $f(x) = \left| {x - 2} \right|$ and g(x) = f(f(x)), $x \in \left[ {0,4} \right]$. Then
$\int\limits_0^3 {\left( {g(x) - f(x)} \right)} dx$ is equal to:

$f(x) = |x - 2|$<br><br>$\therefore$ $f(f(x)) = \left| {|x - 2| - 2} \right| = g(x)$<br><br>$$ \Rightarrow g(x) = \left| {|x - 2| - 2} \right| = \left\{ {\matrix{ {|x - 4|} &amp; {if\,x \ge 2} \cr {| - x|} &amp; {if\,x &lt; 2} \cr } } \right.$$<br><br>$\therefore$ $\int\limits_0^3 {(g(x) - f(x))dx}$<br><br>$= \int\limits_0^3 {g(x)dx} - \int\limits_0^3 {f(x)dx}$<br><br>$$ = \int\limits_0^2 {| - x|dx} + \int\limits_2^3 { - (x - 4)dx} - \int\limits_0^2 { - (x - 2)dx} - \int\limits_2^3 {(x - 2)dx} $$<br><br>$$ = \int\limits_0^2 {x\,dx} - \int\limits_2^3 {(x - 4)dx} + \int\limits_0^2 {(x - 2)dx} - \int\limits_2^3 {(x - 2)dx} $$<br><br>$$ = \left[ {{{{x^2}} \over 2}} \right]_0^2 - \left[ {{{{x^2}} \over 2} - 4x} \right]_2^3 + \left[ {{{{x^2}} \over 2} - 2x} \right]_0^2 - \left[ {{{{x^2}} \over 2} - 2x} \right]_2^3$$<br><br>$$ = 2 - \left\{ {{9 \over 2} - 12 - 2 + 8} \right\} + \{ 2 - 4\} - \left\{ {{9 \over 2} - 6 - 2 + 4} \right\}$$<br><br>$$ = 2 - \left\{ {{9 \over 2} - 6} \right\} - 2 - \left\{ {{9 \over 2} - 4} \right\}$$<br><br>$= 10 - {9 \over 2} - {9 \over 2} = {{20 - 9 - 9} \over 2} = {2 \over 2} = 1$