Let `f(x) ={(-2,-2 le x le 0), (x-2, 0 lt x le 2):}`
`h(x) = f(|x|) + |f(x)|`. Then `int_(-2)^2 h(x) dx` is equal to

Options

A. 4
B. 1
C. 2
D. 6

Answer

To solve the problem, we need to evaluate the integral \( \int_{-2}^{2} h(x) \, dx \), where \( h(x) = f(|x|) + |f(x)| \) and the function \( f(x) \) is defined as follows:

\[ f(x) = \begin{cases} -2 & \text{for } -2 \leq x \leq 0 \\ x - 2 & \text{for } 0 < x \leq 2 \end{cases} \]

### Step 1: Determine \( f(|x|) \)

- For \( x \in [-2, 0] \):
- \( |x| = -x \) (which is non-negative), hence \( f(|x|) = f(-x) = -2 \).

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

### Step 2: Determine \( |f(x)| \)

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

- For \( x \in (0, 2] \):
- \( f(x) = x - 2 \). Since \( x - 2 \) is non-positive in this interval, \( |f(x)| = 2 - x \).

### Step 3: Combine the results to find \( h(x) \)

Now we can express \( h(x) \) for the two intervals:

- For \( x \in [-2, 0] \):
\[ h(x) = f(|x|) + |f(x)| = -2 + 2 = 0 \]

- For \( x \in (0, 2] \):
\[ h(x) = f(|x|) + |f(x)| = (x - 2) + (2 - x) = 0 \]

### Step 4: Evaluate the integral

Now we can evaluate the integral:

\[ \int_{-2}^{2} h(x) \, dx = \int_{-2}^{0} 0 \, dx + \int_{0}^{2} 0 \, dx = 0 + 0 = 0 \]

### Final Answer

Thus, the value of the integral \( \int_{-2}^{2} h(x) \, dx \) is:

\[ \boxed{0} \]