Question:
Let $f(x)=|x-2|$ and $g(x)=f(f(x)), x \in[0,4]$.
Then $\int_{0}^{3}(g(x)-f(x)) d x$ is equal to :
Correct Option: 1
Solution:
$f(x)=|x-2|= \begin{cases}2-x, & x<2 \\ x-2, & x \geq 2\end{cases}$
$g(x)=f(f(x))= \begin{cases}2-f(x), & f(x)<2 \\ f(x)-2, & f(x) \geq 2\end{cases}$
$=\left\{\begin{array}{lll}2-(2-x), & 2-x<2, & x<2 \\ (2-x)-2, & 2-x \geq 2, & x<2 \\ 2-(x-2), & x-2<2, & x \geq 2 \\ (x-2)-2, & x-2 \geq 2, & x \geq 2\end{array}\right.$
$=\left\{\begin{array}{cc}-x & 0 $\therefore \int_{0}^{3}[g(x)-f(x)] d x$ $=\int_{0}^{2} x d x+\int_{2}^{3}(4-x) d x-\int_{0}^{3}|x-2| d x=1$
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