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: , 4
Solution:
$\int_{0}^{3} g(x)-f(x)=\int_{0}^{3}|| x-2|-2| d x-\int_{0}^{3}|x-2| d x$
$=\left(\frac{1}{2} \times 2 \times 2+1+\frac{1}{2} \times 1 \times 1\right)-\left(\frac{1}{2} \times 2 \times 2+\frac{1}{2} \times 1 \times 1\right)$
$=\left(2+1+\frac{1}{2}\right)-\left(2+\frac{1}{2}\right)=1$
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