`int_(-1)^(1)x|x|dx` is equal to

Options

A. `0`
B. `2/3`
C. `2`
D. `-2`

Explanation

To solve the integral \( \int_{-1}^{1} x |x| \, dx \), we need to consider the definition of the absolute value function \( |x| \).

### Step-by-Step Solution:

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

2. **Split the integral**:
We can split the integral at \( x = 0 \):
\[ \int_{-1}^{1} x |x| \, dx = \int_{-1}^{0} x |x| \, dx + \int_{0}^{1} x |x| \, dx \]

3. **Evaluate the integral from -1 to 0**:
For \( x \) in the interval \([-1, 0]\):
\[ |x| = -x \implies x |x| = x(-x) = -x^2 \]
Thus,
\[ \int_{-1}^{0} x |x| \, dx = \int_{-1}^{0} -x^2 \, dx \]

4. **Integrate**:
\[ \int -x^2 \, dx = -\frac{x^3}{3} \]
Evaluating from -1 to 0:
\[ \left[-\frac{x^3}{3}\right]_{-1}^{0} = \left[-\frac{0^3}{3}\right] - \left[-\frac{(-1)^3}{3}\right] = 0 - \left[\frac{1}{3}\right] = -\frac{1}{3} \]

5. **Evaluate the integral from 0 to 1**:
For \( x \) in the interval \([0, 1]\):
\[ |x| = x \implies x |x| = x \cdot x = x^2 \]
Thus,
\[ \int_{0}^{1} x |x| \, dx = \int_{0}^{1} x^2 \, dx \]

6. **Integrate**:
\[ \int x^2 \, dx = \frac{x^3}{3} \]
Evaluating from 0 to 1:
\[ \left[\frac{x^3}{3}\right]_{0}^{1} = \left[\frac{1^3}{3}\right] - \left[\frac{0^3}{3}\right] = \frac{1}{3} - 0 = \frac{1}{3} \]

7. **Combine the results**:
Now we combine the two results:
\[ \int_{-1}^{1} x |x| \, dx = \left(-\frac{1}{3}\right) + \left(\frac{1}{3}\right) = 0 \]

### Final Answer:
\[ \int_{-1}^{1} x |x| \, dx = 0 \]