Define \( h(x) \) in Terms of \( f(x) \):
Since \( h(x) = f(x) + |f(x)| \), we can evaluate \( h(x) \) separately on the intervals where \( f(x) \) takes different forms:
For \(-2 \leq x < 0\), \( f(x) = -x \), so \( |f(x)| = x \). Therefore:
\[ h(x) = f(x) + |f(x)| = -x + x = 0 \]
For \( 0 < x \leq 2 \), \( f(x) = x - 2 \), so \( |f(x)| = 2 - x \) (since \( x - 2 < 0 \)). Thus:
\[ h(x) = f(x) + |f(x)| = (x - 2) + (2 - x) = 0 \]
This implies \( h(x) = 0 \) on both intervals.
Calculate \( \int_{-2}^2 h(x) dx \): Since \( h(x) = 0 \) on the entire interval \(-2 \leq x \leq 2\), we have: \[ \int_{-2}^2 h(x) dx = \int_{-2}^2 0 \, dx = 0 \]
Conclusion: \( \int_{-2}^0 h(x) dx = 0 \) and \( \int_0^2 h(x) dx = 2 \).
Therefore, the answer is \(2\).
Match List-I with List-II: List-I