The value of `int_(-2)^(2) |[x]| dx` is equal to

Options

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

Answer

To solve the integral \( \int_{-2}^{2} |x| \, dx \), we will break it down into manageable parts by considering the definition of the absolute value function.

### Step-by-step solution:

1. **Identify the intervals**: The function \( |x| \) can be expressed differently depending on the value of \( x \):
- For \( x < 0 \), \( |x| = -x \)
- For \( x \geq 0 \), \( |x| = x \)

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

2. **Evaluate the first integral** \( \int_{-2}^{0} |x| \, dx \):
- In this interval, \( |x| = -x \):
\[ \int_{-2}^{0} |x| \, dx = \int_{-2}^{0} -x \, dx \]
- Now, compute the integral:
\[ = -\left[ \frac{x^2}{2} \right]_{-2}^{0} = -\left( \frac{0^2}{2} - \frac{(-2)^2}{2} \right) = -\left( 0 - \frac{4}{2} \right) = -(-2) = 2 \]

3. **Evaluate the second integral** \( \int_{0}^{2} |x| \, dx \):
- In this interval, \( |x| = x \):
\[ \int_{0}^{2} |x| \, dx = \int_{0}^{2} x \, dx \]
- Now, compute the integral:
\[ = \left[ \frac{x^2}{2} \right]_{0}^{2} = \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = \left( \frac{4}{2} - 0 \right) = 2 \]

4. **Combine the results**:
\[ \int_{-2}^{2} |x| \, dx = \int_{-2}^{0} |x| \, dx + \int_{0}^{2} |x| \, dx = 2 + 2 = 4 \]

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