The value of `12 int_0^3|x^2-3x+2|dx` is

Answer

To solve the integral \( 12 \int_0^3 |x^2 - 3x + 2| \, dx \), we will follow these steps:

### Step 1: Factor the Quadratic Expression
The expression inside the absolute value can be factored:
\[ x^2 - 3x + 2 = (x - 1)(x - 2) \]

### Step 2: Determine the Roots and Intervals
The roots of the equation \( x^2 - 3x + 2 = 0 \) are \( x = 1 \) and \( x = 2 \). These roots divide the interval \([0, 3]\) into three sub-intervals: \([0, 1]\), \([1, 2]\), and \([2, 3]\).

### Step 3: Analyze the Sign of the Expression in Each Interval
- For \( x \in [0, 1] \):
- \( (x - 1) < 0 \) and \( (x - 2) < 0 \) ⇒ \( |(x - 1)(x - 2)| = -(x - 1)(x - 2) = -(x^2 - 3x + 2) \)

- For \( x \in [1, 2] \):
- \( (x - 1) \geq 0 \) and \( (x - 2) < 0 \) ⇒ \( |(x - 1)(x - 2)| = -(x - 1)(x - 2) = -(x^2 - 3x + 2) \)

- For \( x \in [2, 3] \):
- \( (x - 1) \geq 0 \) and \( (x - 2) \geq 0 \) ⇒ \( |(x - 1)(x - 2)| = (x - 1)(x - 2) = x^2 - 3x + 2 \)

### Step 4: Set Up the Integral
Now we can express the integral as:
\[ \int_0^3 |x^2 - 3x + 2| \, dx = \int_0^1 -(x^2 - 3x + 2) \, dx + \int_1^2 -(x^2 - 3x + 2) \, dx + \int_2^3 (x^2 - 3x + 2) \, dx \]

### Step 5: Calculate Each Integral
1. **From \(0\) to \(1\)**:
\[ \int_0^1 -(x^2 - 3x + 2) \, dx = \int_0^1 (3x - x^2 - 2) \, dx \]
\[ = \left[ \frac{3x^2}{2} - \frac{x^3}{3} - 2x \right]_0^1 = \left( \frac{3}{2} - \frac{1}{3} - 2 \right) = \frac{3}{2} - \frac{1}{3} - \frac{6}{3} = \frac{3}{2} - \frac{7}{3} = \frac{9 - 14}{6} = -\frac{5}{6} \]

2. **From \(1\) to \(2\)**:
\[ \int_1^2 -(x^2 - 3x + 2) \, dx = \int_1^2 (3x - x^2 - 2) \, dx \]
\[ = \left[ \frac{3x^2}{2} - \frac{x^3}{3} - 2x \right]_1^2 = \left( \frac{3(2^2)}{2} - \frac{(2^3)}{3} - 2(2) \right) - \left( \frac{3(1^2)}{2} - \frac{(1^3)}{3} - 2(1) \right) \]
\[ = \left( 6 - \frac{8}{3} - 4 \right) - \left( \frac{3}{2} - \frac{1}{3} - 2 \right) \]
\[ = \left( 2 - \frac{8}{3} \right) - \left( \frac{3}{2} - \frac{1}{3} - 2 \right) = \left( \frac{6 - 8}{3} \right) - \left( \frac{9 - 1 - 6}{6} \right) = -\frac{2}{3} - \left( \frac{2}{6} \right) = -\frac{2}{3} - \frac{1}{3} = -1 \]

3. **From \(2\) to \(3\)**:
\[ \int_2^3 (x^2 - 3x + 2) \, dx = \left[ \frac{x^3}{3} - \frac{3x^2}{2} + 2x \right]_2^3 \]
\[ = \left( \frac{27}{3} - \frac{27}{2} + 6 \right) - \left( \frac{8}{3} - \frac{12}{2} + 4 \right) = \left( 9 - \frac{27}{2} + 6 \right) - \left( \frac{8}{3} - 6 + 4 \right) \]
\[ = \left( 15 - \frac{27}{2} \right) - \left( \frac{8}{3} - 2 \right) = \left( \frac{30 - 27}{2} \right) - \left( \frac{8 - 6}{3} \right) = \frac{3}{2} - \frac{2}{3} \]
\[ = \frac{9 - 4}{6} = \frac{5}{6} \]

### Step 6: Combine the Results
Now we combine the results from the three integrals:
\[ \int_0^3 |x^2 - 3x + 2| \, dx = -\frac{5}{6} - 1 + \frac{5}{6} = -1 \]

### Step 7: Multiply by 12
Finally, we multiply by 12:
\[ 12 \int_0^3 |x^2 - 3x + 2| \, dx = 12 \times (-1) = -12 \]

### Final Answer
Thus, the value of \( 12 \int_0^3 |x^2 - 3x + 2| \, dx \) is:
\[ \boxed{-12} \]