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

Break down $$h(x)$$:

1. For $$x \in [-2, 0]$$: $$|x| \in [0, 2]$$. So $$f(|x|) = |x|-2 = -x-2$$.

$$|f(x)| = |-2| = 2$$.

$$h(x) = (-x-2) + 2 = -x$$.

2. For $$x \in (0, 2]$$: $$|x| = x$$. So $$f(|x|) = x-2$$.

$$|f(x)| = |x-2| = -(x-2) = 2-x$$ (since $$x \le 2$$).

$$h(x) = (x-2) + (2-x) = 0$$.

• Integral:

$$\int_{-2}^{2} h(x) \, dx = \int_{-2}^{0} (-x) \, dx + \int_{0}^{2} 0 \, dx = \left[ -\frac{x^2}{2} \right]_{-2}^{0} = 0 - (-2) = \mathbf{2}$$