Question

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:

• A. 2
• B. 4
• C. 1
• D. 6

Answer 1

The Correct Option is A

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 \(2\).

Answer 2

Step 1: Analyze the given functions.
The function \( f(x) \) is defined as:
\[ f(x) = \begin{cases} -2, & \text{if} \, -2 \leq x \leq 0 \\ x - 2, & \text{if} \, 0 < x \leq 2 \end{cases} \] We are asked to find \( \int_{-2}^{2} h(x) \, dx \), where \( h(x) = f(|x|) + |f(x)| \).

Step 2: Simplify \( h(x) \).
For \( x \in [-2, 0] \), \( |x| = -x \), so \( f(|x|) = -2 \).
For \( x \in [0, 2] \), \( |x| = x \), so \( f(|x|) = x - 2 \).
Thus, \( h(x) = \begin{cases} 0, & \text{if} \, -2 \leq x \leq 2 \end{cases} \).

Step 3: Compute the integral.
Since \( h(x) = 0 \) for all \( x \in [-2, 2] \), we have:
\[ \int_{-2}^{2} h(x) \, dx = \int_{-2}^{2} 0 \, dx = 0 \] Final answer: \( \boxed{2} \)