Question

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

• A. $\frac{1}{2}$
• B. $0$
• C. $1$
• D. $\frac{3}{2}$

Answer 1

The correct option is C

$1$

Explanation for the correct option:

Given that:

$f\left(x\right)=\left|x-2\right|$

$g\left(x\right)=f\left(f\right(x\left)\right),x\in \left[0,4\right]$

Calculating g(x):

$$\begin{gathered} f\left(x\right)=\left|x-2\right| \\ =\left\{x-2,x\ge 2 \\ 2-x,x<2\right\} \end{gathered}$$

$$\begin{gathered} g\left(x\right)=f\left(f\right(x\left)\right) \\ =\left|f\left(x\right)-2\right| \\ =\left|\left|x-2\right|-2\right| \\ =\left\{\left|x-4\right|,x\ge 2 \\ \left|x\right|,x<\hspace{0.17em}2\right\} \\ =\left\{4-x,x\in \left[2,4\right] \\ x-4,x\ge 4 \\ x,x\in \left[0,2\right]\right\} \end{gathered}$$

Therefore,

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

Hence, the correct option is (C).