If `a lt 0 lt b`, then `int_(a)^(b) x|x| dx`=

Options

A. `(1)/(2) (a^(2) +b^(2))`
B. `(1)/(3) (b^(2)-a^(2))`
C. `(1)/(3) (a^(3) +b^(3))`
D. None

Answer

To solve the integral \( \int_{a}^{b} x |x| \, dx \) where \( a < 0 < b \), we will break the integral into two parts based on the behavior of the absolute value function \( |x| \).

### Step-by-Step Solution:

1. **Break the Integral**:
Since \( a < 0 < b \), we can split the integral at 0:
\[ \int_{a}^{b} x |x| \, dx = \int_{a}^{0} x |x| \, dx + \int_{0}^{b} x |x| \, dx \]

2. **Evaluate the First Integral**:
For \( x < 0 \) (which applies for \( x \) in the interval \( [a, 0] \)), we have \( |x| = -x \). Thus, the first integral becomes:
\[ \int_{a}^{0} x |x| \, dx = \int_{a}^{0} x (-x) \, dx = -\int_{a}^{0} x^2 \, dx \]

3. **Calculate the First Integral**:
Now, we compute \( -\int_{a}^{0} x^2 \, dx \):
\[ -\int_{a}^{0} x^2 \, dx = -\left[ \frac{x^3}{3} \right]_{a}^{0} = -\left( \frac{0^3}{3} - \frac{a^3}{3} \right) = -\left( 0 + \frac{a^3}{3} \right) = \frac{-a^3}{3} \]

4. **Evaluate the Second Integral**:
For \( x > 0 \) (which applies for \( x \) in the interval \( [0, b] \)), we have \( |x| = x \). Thus, the second integral becomes:
\[ \int_{0}^{b} x |x| \, dx = \int_{0}^{b} x^2 \, dx \]

5. **Calculate the Second Integral**:
Now, we compute \( \int_{0}^{b} x^2 \, dx \):
\[ \int_{0}^{b} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{b} = \frac{b^3}{3} - 0 = \frac{b^3}{3} \]

6. **Combine the Results**:
Now, we combine the results of both integrals:
\[ \int_{a}^{b} x |x| \, dx = \frac{-a^3}{3} + \frac{b^3}{3} = \frac{-a^3 + b^3}{3} \]

### Final Answer:
Thus, the final result of the integral is:
\[ \int_{a}^{b} x |x| \, dx = \frac{-a^3 + b^3}{3} \]