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

Options

A. 2
B. 1
C. `1/4`
D. 0

Answer

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

### Step 1: Break the integral into two parts
The absolute value function \( |x| \) can be expressed differently depending on the sign of \( x \):
- For \( x < 0 \), \( |x| = -x \)
- For \( x \geq 0 \), \( |x| = x \)

Thus, we can split the integral as follows:
\[ \int_{-1}^{1} x |x| \, dx = \int_{-1}^{0} x |x| \, dx + \int_{0}^{1} x |x| \, dx \]

### Step 2: Evaluate the integral from -1 to 0
For \( x \) in the interval \([-1, 0]\), we have \( |x| = -x \). Therefore, the integral becomes:
\[ \int_{-1}^{0} x |x| \, dx = \int_{-1}^{0} x (-x) \, dx = \int_{-1}^{0} -x^2 \, dx \]

Now, we can calculate this integral:
\[ \int -x^2 \, dx = -\frac{x^3}{3} + C \]
Evaluating from -1 to 0:
\[ \left[-\frac{x^3}{3}\right]_{-1}^{0} = \left[-\frac{0^3}{3}\right] - \left[-\frac{(-1)^3}{3}\right] = 0 - \left[-\frac{-1}{3}\right] = 0 + \frac{1}{3} = \frac{1}{3} \]

### Step 3: Evaluate the integral from 0 to 1
For \( x \) in the interval \([0, 1]\), we have \( |x| = x \). Therefore, the integral becomes:
\[ \int_{0}^{1} x |x| \, dx = \int_{0}^{1} x \cdot x \, dx = \int_{0}^{1} x^2 \, dx \]

Now, we can calculate this integral:
\[ \int x^2 \, dx = \frac{x^3}{3} + C \]
Evaluating from 0 to 1:
\[ \left[\frac{x^3}{3}\right]_{0}^{1} = \left[\frac{1^3}{3}\right] - \left[\frac{0^3}{3}\right] = \frac{1}{3} - 0 = \frac{1}{3} \]

### Step 4: Combine the results
Now, we can combine the results of both integrals:
\[ \int_{-1}^{1} x |x| \, dx = \int_{-1}^{0} x |x| \, dx + \int_{0}^{1} x |x| \, dx = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \]

### Final Answer
Thus, the value of the integral \( \int_{-1}^{1} x |x| \, dx \) is:
\[ \boxed{0} \]