`int_(0)^(10)|x(x-1)(x-2)|dx` is equal to

Options

A. `160.05`
B. `1600.5`
C. `16.005`
D. none of these

Explanation

To solve the integral \( I = \int_{0}^{10} |x(x-1)(x-2)| \, dx \), we will first analyze the expression inside the absolute value to determine where it is positive and negative.

### Step 1: Identify the roots of the function
The expression \( x(x-1)(x-2) \) has roots at \( x = 0, 1, 2 \). We will evaluate the sign of the expression in the intervals determined by these roots: \( (-\infty, 0) \), \( (0, 1) \), \( (1, 2) \), and \( (2, \infty) \).

### Step 2: Determine the sign of \( x(x-1)(x-2) \)
- For \( x < 0 \): The expression is negative.
- For \( 0 < x < 1 \): The expression is positive.
- For \( 1 < x < 2 \): The expression is negative.
- For \( x > 2 \): The expression is positive.

### Step 3: Set up the integral with absolute values
Now we can express the integral as:
\[ I = \int_{0}^{1} x(x-1)(x-2) \, dx - \int_{1}^{2} x(x-1)(x-2) \, dx + \int_{2}^{10} x(x-1)(x-2) \, dx \]

### Step 4: Calculate each integral
1. **From \( 0 \) to \( 1 \)**:
\[ \int_{0}^{1} x(x-1)(x-2) \, dx = \int_{0}^{1} (x^3 - 3x^2 + 2x) \, dx \]
Evaluating this integral:
\[ = \left[ \frac{x^4}{4} - x^3 + x^2 \right]_{0}^{1} = \left( \frac{1}{4} - 1 + 1 \right) - 0 = \frac{1}{4} \]

2. **From \( 1 \) to \( 2 \)**:
\[ -\int_{1}^{2} x(x-1)(x-2) \, dx = -\int_{1}^{2} (x^3 - 3x^2 + 2x) \, dx \]
Evaluating this integral:
\[ = -\left[ \frac{x^4}{4} - x^3 + x^2 \right]_{1}^{2} = -\left( \left( \frac{16}{4} - 8 + 4 \right) - \left( \frac{1}{4} - 1 + 1 \right) \right) \]
\[ = -\left( 4 - 8 + 4 - \frac{1}{4} + 1 - 1 \right) = -\left( 0 - \frac{1}{4} \right) = \frac{1}{4} \]

3. **From \( 2 \) to \( 10 \)**:
\[ \int_{2}^{10} x(x-1)(x-2) \, dx = \int_{2}^{10} (x^3 - 3x^2 + 2x) \, dx \]
Evaluating this integral:
\[ = \left[ \frac{x^4}{4} - x^3 + x^2 \right]_{2}^{10} = \left( \frac{10000}{4} - 1000 + 100 \right) - \left( \frac{16}{4} - 8 + 4 \right) \]
\[ = \left( 2500 - 1000 + 100 \right) - (4 - 8 + 4) = 1600 - 0 = 1600 \]

### Step 5: Combine the results
Now we combine the results of the integrals:
\[ I = \frac{1}{4} + \frac{1}{4} + 1600 = \frac{2}{4} + 1600 = \frac{1}{2} + 1600 = 1600.5 \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{1600.5} \]