Let f(x) = begin{cases} -2, & -2 leq x leq 0 | vedclass

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(A) Given $f(x) = \begin{cases} -2, & -2 \leq x \leq 0 \ x-2, & 0 < x \leq 2 \end{cases}$.We need to find $h(x) = f(|x|) + |f(x)|$.For $x \in [0, 2]$

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$2$

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(A) Given $f(x) = \begin{cases} -2, & -2 \leq x \leq 0 \ x-2, & 0 We need to find $h(x) = f(|x|) + |f(x)|$.For $x \in [0, 2]$,$f(|x|) = f(x) = x-2$ and $|f(x)| = |x-2| = 2-x$. Thus,$h(x) = (x-2) + (2-x) = 0$.For $x \in [-2, 0)$,$f(|x|) = f(-x) = -2$ (since $-x \in (0, 2]$ is not true,actually for $x \in [-2, 0)$,$|x| \in (0, 2]$,so $f(|x|) = |x|-2 = -x-2$).Wait,let's re-evaluate: $f(|x|) = \begin{cases} -2, & |x| \leq 0 	ext{ (only } x=0) \ |x|-2, & 0 So $f(|x|) = |x|-2$.$|f(x)| = \begin{cases} |-2| = 2, & -2 \leq x \leq 0 \ |x-2| = 2-x, & 0 Thus,$h(x) = f(|x|) + |f(x)| = \begin{cases} (-x-2) + 2 = -x, & -2 \leq x Now,$\int_{-2}^2 h(x) dx = \int_{-2}^0 (-x) dx + \int_0^2 0 dx = \left[ -\frac{x^2}{2} ight]_{-2}^0 = 0 - (-\frac{(-2)^2}{2}) = 0 - (-2) = 2$.

Let $f(x) = \begin{cases} -2, & -2 \leq x \leq 0 \\ x-2, & 0 < x \leq 2 \end{cases}$ and $h(x) = f(|x|) + |f(x)|$. Then $\int_{-2}^2 h(x) dx$ is equal to:

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