What is the value the integral `int_(-1)^(1)|x| dx` ?

Options

A. 1
B. 0
C. 2
D. -1

Explanation

To solve the integral \(\int_{-1}^{1} |x| \, dx\), we can break it down into two parts based on the definition of the absolute value function.

### Step-by-step Solution:

1. **Identify the behavior of \(|x|\)**:
- For \(x < 0\) (i.e., from \(-1\) to \(0\)), \(|x| = -x\).
- For \(x \geq 0\) (i.e., from \(0\) to \(1\)), \(|x| = x\).

2. **Break the integral into two parts**:
\[ \int_{-1}^{1} |x| \, dx = \int_{-1}^{0} |x| \, dx + \int_{0}^{1} |x| \, dx \]
This can be rewritten as:
\[ \int_{-1}^{1} |x| \, dx = \int_{-1}^{0} -x \, dx + \int_{0}^{1} x \, dx \]

3. **Evaluate the first integral**:
\[ \int_{-1}^{0} -x \, dx \]
- The antiderivative of \(-x\) is \(-\frac{x^2}{2}\).
- Evaluate from \(-1\) to \(0\):
\[ = \left[-\frac{x^2}{2}\right]_{-1}^{0} = \left[-\frac{0^2}{2}\right] - \left[-\frac{(-1)^2}{2}\right] = 0 - \left[-\frac{1}{2}\right] = \frac{1}{2} \]

4. **Evaluate the second integral**:
\[ \int_{0}^{1} x \, dx \]
- The antiderivative of \(x\) is \(\frac{x^2}{2}\).
- Evaluate from \(0\) to \(1\):
\[ = \left[\frac{x^2}{2}\right]_{0}^{1} = \left[\frac{1^2}{2}\right] - \left[\frac{0^2}{2}\right] = \frac{1}{2} - 0 = \frac{1}{2} \]

5. **Combine the results**:
\[ \int_{-1}^{1} |x| \, dx = \frac{1}{2} + \frac{1}{2} = 1 \]

### Final Answer:
The value of the integral \(\int_{-1}^{1} |x| \, dx\) is \(1\).