If `f(x)={(-2,,,-2,le,x, <,0),(x-2,,,0,le,x, le,2):}` and `h(x) = f(|x|) |(x)|` then `int_0^k h(x) dx` is equal to (where `0

Options

A. k
B. `k/2`
C. `(2k)/3`
D. 0

Answer

To solve the problem, we need to evaluate the integral \( \int_0^k h(x) \, dx \) where \( h(x) = f(|x|) + |f(x)| \) and \( f(x) \) is defined as follows:

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

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

Since we are interested in \( h(x) = f(|x|) + |f(x)| \), we first need to evaluate \( f(|x|) \):

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

- For \( x < 0 \), \( |x| = -x \). Since \( -x \) is positive, we have:
\[ f(|x|) = f(-x) = -2 \]

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

Next, we need to evaluate \( |f(x)| \):

- For \( x \) in the interval \( [0, 2] \):
\[ f(x) = x - 2 \implies |f(x)| = |x - 2| = 2 - x \quad \text{(since \( x - 2 \) is negative in this interval)} \]

- For \( x < 0 \):
\[ f(x) = -2 \implies |f(x)| = 2 \]

### Step 3: Define \( h(x) \)

Now we can define \( h(x) \):

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

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

Thus, we conclude that:
\[ h(x) = 0 \quad \text{for all } x \]

### Step 4: Evaluate the Integral

Now we need to evaluate the integral:
\[ \int_0^k h(x) \, dx = \int_0^k 0 \, dx = 0 \]

### Final Answer

The value of the integral \( \int_0^k h(x) \, dx \) is:
\[ \boxed{0} \]