The value of `int_1^3|(x-1)(x-2)(x-3)|dx` is

Options

A. `1/3`
B. `1/2`
C. `9/4`
D. `9/5`

Answer

To solve the integral \( \int_1^3 |(x-1)(x-2)(x-3)| \, dx \), we will first analyze the expression inside the absolute value to determine where it is positive and where it is negative.

### Step 1: Identify the critical points
The expression \( (x-1)(x-2)(x-3) \) has roots at \( x = 1, 2, 3 \). These points will help us determine the intervals where the expression is positive or negative.

### Step 2: Test the intervals
We will evaluate the sign of \( (x-1)(x-2)(x-3) \) in the intervals \( (-\infty, 1) \), \( (1, 2) \), \( (2, 3) \), and \( (3, \infty) \).

1. **Interval \( (-\infty, 1) \)**: Choose \( x = 0 \):
\[ (0-1)(0-2)(0-3) = (-1)(-2)(-3) = -6 \quad (\text{negative}) \]

2. **Interval \( (1, 2) \)**: Choose \( x = 1.5 \):
\[ (1.5-1)(1.5-2)(1.5-3) = (0.5)(-0.5)(-1.5) = 0.375 \quad (\text{positive}) \]

3. **Interval \( (2, 3) \)**: Choose \( x = 2.5 \):
\[ (2.5-1)(2.5-2)(2.5-3) = (1.5)(0.5)(-0.5) = -0.375 \quad (\text{negative}) \]

4. **Interval \( (3, \infty) \)**: Choose \( x = 4 \):
\[ (4-1)(4-2)(4-3) = (3)(2)(1) = 6 \quad (\text{positive}) \]

### Step 3: Rewrite the integral
Based on the sign analysis, we can rewrite the integral as:
\[ \int_1^3 |(x-1)(x-2)(x-3)| \, dx = \int_1^2 (x-1)(x-2)(x-3) \, dx - \int_2^3 (x-1)(x-2)(x-3) \, dx \]

### Step 4: Calculate the integrals
1. **For \( \int_1^2 (x-1)(x-2)(x-3) \, dx \)**:
\[ (x-1)(x-2)(x-3) = -(x-1)(x-2)(3-x) \]
We can expand it:
\[ = -(x^3 - 6x^2 + 11x - 6) \]
So,
\[ = -x^3 + 6x^2 - 11x + 6 \]

Now integrate:
\[ \int (-x^3 + 6x^2 - 11x + 6) \, dx = -\frac{x^4}{4} + 2x^3 - \frac{11x^2}{2} + 6x \Big|_1^2 \]

Evaluating at \( x = 2 \):
\[ -\frac{16}{4} + 2(8) - \frac{11(4)}{2} + 12 = -4 + 16 - 22 + 12 = 2 \]

Evaluating at \( x = 1 \):
\[ -\frac{1}{4} + 2(1) - \frac{11(1)}{2} + 6 = -\frac{1}{4} + 2 - \frac{11}{2} + 6 = -\frac{1}{4} + 8 - 5.5 = 2.25 \]

Thus,
\[ \int_1^2 (x-1)(x-2)(x-3) \, dx = 2 - 2.25 = -0.25 \]

2. **For \( \int_2^3 (x-1)(x-2)(x-3) \, dx \)**:
The expression is positive, so we can integrate directly:
\[ \int_2^3 (x-1)(x-2)(x-3) \, dx = \int_2^3 (x^3 - 6x^2 + 11x - 6) \, dx \]

Evaluating:
\[ \int (x^3 - 6x^2 + 11x - 6) \, dx = \frac{x^4}{4} - 2x^3 + \frac{11x^2}{2} - 6x \Big|_2^3 \]

Evaluating at \( x = 3 \):
\[ \frac{81}{4} - 54 + \frac{99}{2} - 18 = \frac{81}{4} - 54 + \frac{198}{4} - \frac{72}{4} = \frac{207 - 216}{4} = -\frac{9}{4} \]

Evaluating at \( x = 2 \):
\[ \frac{16}{4} - 16 + 22 - 12 = 4 - 16 + 22 - 12 = -2 \]

Thus,
\[ \int_2^3 (x-1)(x-2)(x-3) \, dx = -\frac{9}{4} + 2 = -\frac{1}{4} \]

### Step 5: Combine results
Now we combine the results:
\[ \int_1^3 |(x-1)(x-2)(x-3)| \, dx = -0.25 - (-\frac{1}{4}) = -0.25 + 0.25 = 0 \]

### Final Answer
The value of \( \int_1^3 |(x-1)(x-2)(x-3)| \, dx \) is \( 0 \).