Let f(x) = |x - 2| and g(x) = f(f(x)),x in [0 | vedclass

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(D) Given $f(x) = |x - 2|$ and $g(x) = f(f(x)) = ||x - 2| - 2|$.We need to evaluate $I = \int_{0}^{3} (g(x) - f(x)) \, dx = \int_{0}^{3} g(x) \, dx -

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(D) Given $f(x) = |x - 2|$ and $g(x) = f(f(x)) = ||x - 2| - 2|$.We need to evaluate $I = \int_{0}^{3} (g(x) - f(x)) \, dx = \int_{0}^{3} g(x) \, dx - \int_{0}^{3} f(x) \, dx$.First,calculate $\int_{0}^{3} f(x) \, dx = \int_{0}^{3} |x - 2| \, dx$.This represents the area under the graph of $f(x)$ from $x = 0$ to $x = 3$. The graph consists of two triangles: one with base $2$ and height $2$ (from $0$ to $2$),and one with base $1$ and height $1$ (from $2$ to $3$).Area $= \frac{1}{2} 	imes 2 	imes 2 + \frac{1}{2} 	imes 1 	imes 1 = 2 + 0.5 = 2.5$.Next,calculate $\int_{0}^{3} g(x) \, dx = \int_{0}^{3} ||x - 2| - 2| \, dx$.For $x \in [0, 2]$,$g(x) = |(2 - x) - 2| = |-x| = x$. Area $= \int_{0}^{2} x \, dx = \frac{1}{2} 	imes 2 	imes 2 = 2$.For $x \in [2, 3]$,$g(x) = |(x - 2) - 2| = |x - 4| = 4 - x$. Area $= \int_{2}^{3} (4 - x) \, dx = \frac{1}{2} 	imes (2 + 1) 	imes 1 = 1.5$.Total $\int_{0}^{3} g(x) \, dx = 2 + 1.5 = 3.5$.Thus,$I = 3.5 - 2.5 = 1$.

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

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