Let f(x)=|x-2| and g(x)=f(f(x)), x[0,4].

Question

Let $f(x)=|x-2|$ and $g(x)=f(f(x))$, $x\in[0,4]$. Then $\displaystyle\int_0^3(g(x)-f(x))\,dx$ is equal to:

Accepted Answer

$f(x) = |x - 2|$

$$ \therefore $$ $$f(f(x)) = \left| {|x - 2| - 2} \right| = g(x)$$

$$ \Rightarrow g(x) = \left| {|x - 2| - 2} \right| = \left\{ {\begin{pmatrix} {|x - 4|} & {if\,x \ge 2}  \
   {| - x|} & {if\,x < 2}  \

\end{pmatrix} } \right.$$

$$ \therefore $$ $$\int\limits_0^3 {(g(x) - f(x))dx} $$

$$ = \int\limits_0^3 {g(x)dx}  - \int\limits_0^3 {f(x)dx} $$

$$ = \int\limits_0^2 {| - x|dx}  + \int\limits_2^3 { - (x - 4)dx}  - \int\limits_0^2 { - (x - 2)dx}  - \int\limits_2^3 {(x - 2)dx} $$

$$ = \int\limits_0^2 {x\,dx}  - \int\limits_2^3 {(x - 4)dx}  + \int\limits_0^2 {(x - 2)dx}  - \int\limits_2^3 {(x - 2)dx} $$

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

$$ = 2 - \left\{ {{9 \over 2} - 12 - 2 + 8} \right\} + \{ 2 - 4\}  - \left\{ {{9 \over 2} - 6 - 2 + 4} \right\}$$

$$ = 2 - \left\{ {{9 \over 2} - 6} \right\} - 2 - \left\{ {{9 \over 2} - 4} \right\}$$

$$ = 10 - {9 \over 2} - {9 \over 2} = {{20 - 9 - 9} \over 2} = {2 \over 2} = 1$$

Solutions · 1