`int_(-2)^(2) |[x]|dx=`

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the integral \( \int_{-2}^{2} |x| \, dx \), we need to break it down into manageable parts since the absolute value function \( |x| \) behaves differently on the intervals where \( x \) is negative and where \( x \) is positive.

### Step 1: Break the integral into intervals
The function \( |x| \) can be expressed as:
- \( |x| = -x \) when \( x < 0 \)
- \( |x| = x \) when \( x \geq 0 \)

Thus, we can split the integral at \( x = 0 \):
\[ \int_{-2}^{2} |x| \, dx = \int_{-2}^{0} |x| \, dx + \int_{0}^{2} |x| \, dx \]

### Step 2: Evaluate the first integral
For the interval from \(-2\) to \(0\):
\[ \int_{-2}^{0} |x| \, dx = \int_{-2}^{0} -x \, dx \]
Now, we compute this integral:
\[ = -\left[ \frac{x^2}{2} \right]_{-2}^{0} = -\left( \frac{0^2}{2} - \frac{(-2)^2}{2} \right) = -\left( 0 - \frac{4}{2} \right) = -(-2) = 2 \]

### Step 3: Evaluate the second integral
For the interval from \(0\) to \(2\):
\[ \int_{0}^{2} |x| \, dx = \int_{0}^{2} x \, dx \]
Now, we compute this integral:
\[ = \left[ \frac{x^2}{2} \right]_{0}^{2} = \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = \left( \frac{4}{2} - 0 \right) = 2 \]

### Step 4: Combine the results
Now, we add the results of the two integrals:
\[ \int_{-2}^{2} |x| \, dx = 2 + 2 = 4 \]

### Final Answer
Thus, the value of the integral \( \int_{-2}^{2} |x| \, dx \) is \( \boxed{4} \).