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

Options

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

Answer

To solve the integral \( \int_{-1}^{1} x |x| \, dx \), we need to consider the behavior of the function \( x |x| \) over the interval from -1 to 1. The absolute value function \( |x| \) can be expressed differently depending on whether \( x \) is positive or negative.

### Step 1: Split the Integral
The function \( |x| \) can be defined as:
- \( |x| = -x \) when \( x < 0 \)
- \( |x| = x \) when \( x \geq 0 \)

Thus, we can split the integral into two parts:
\[ \int_{-1}^{1} x |x| \, dx = \int_{-1}^{0} x (-x) \, dx + \int_{0}^{1} x (x) \, dx \]

### Step 2: Simplify Each Integral
Now we simplify each part:
1. For \( x \in [-1, 0] \):
\[ \int_{-1}^{0} x (-x) \, dx = \int_{-1}^{0} -x^2 \, dx \]
2. For \( x \in [0, 1] \):
\[ \int_{0}^{1} x (x) \, dx = \int_{0}^{1} x^2 \, dx \]

### Step 3: Evaluate the Integrals
Now we evaluate each integral:
1. Evaluate \( \int_{-1}^{0} -x^2 \, dx \):
\[ \int -x^2 \, dx = -\frac{x^3}{3} \Big|_{-1}^{0} = -\frac{0^3}{3} + \frac{(-1)^3}{3} = 0 + \frac{1}{3} = \frac{1}{3} \]

2. Evaluate \( \int_{0}^{1} x^2 \, dx \):
\[ \int x^2 \, dx = \frac{x^3}{3} \Big|_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} - 0 = \frac{1}{3} \]

### Step 4: Combine the Results
Now we combine the results of the two integrals:
\[ \int_{-1}^{1} x |x| \, dx = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \]

### Final Answer
Thus, the value of the integral \( \int_{-1}^{1} x |x| \, dx \) is:
\[ \frac{2}{3} \]