Solve this following


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 :


  1. $\frac{3}{2}$

  2. 0

  3. $\frac{1}{2}$


  4. 1

Correct Option: , 4


$\int_{0}^{3} g(x)-f(x)=\int_{0}^{3}|| x-2|-2| d x-\int_{0}^{3}|x-2| d x$

$=\left(\frac{1}{2} \times 2 \times 2+1+\frac{1}{2} \times 1 \times 1\right)-\left(\frac{1}{2} \times 2 \times 2+\frac{1}{2} \times 1 \times 1\right)$



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