The value of `int _(2)^(4) { |x-2|+|x-3|} dx` is

Options

A. 1
B. 2
C. 3
D. 5

Explanation

To solve the integral \( \int_{2}^{4} (|x-2| + |x-3|) \, dx \), we will break it down into intervals based on the points where the expressions inside the absolute values change sign, which are \( x = 2 \) and \( x = 3 \).

### Step 1: Identify the intervals
The integral can be split into two parts:
1. From \( 2 \) to \( 3 \)
2. From \( 3 \) to \( 4 \)

### Step 2: Evaluate \( |x-2| \) and \( |x-3| \) in each interval
- For \( 2 \leq x < 3 \):
- \( |x-2| = x - 2 \)
- \( |x-3| = -(x - 3) = 3 - x \)

Thus, in this interval:
\[ |x-2| + |x-3| = (x - 2) + (3 - x) = 1 \]

- For \( 3 \leq x \leq 4 \):
- \( |x-2| = x - 2 \)
- \( |x-3| = x - 3 \)

Thus, in this interval:
\[ |x-2| + |x-3| = (x - 2) + (x - 3) = 2x - 5 \]

### Step 3: Set up the integral
Now we can set up the integral as follows:
\[ \int_{2}^{4} (|x-2| + |x-3|) \, dx = \int_{2}^{3} 1 \, dx + \int_{3}^{4} (2x - 5) \, dx \]

### Step 4: Calculate the first integral
\[ \int_{2}^{3} 1 \, dx = [x]_{2}^{3} = 3 - 2 = 1 \]

### Step 5: Calculate the second integral
\[ \int_{3}^{4} (2x - 5) \, dx = \left[ x^2 - 5x \right]_{3}^{4} \]

Calculating the limits:
- At \( x = 4 \):
\[ 4^2 - 5 \cdot 4 = 16 - 20 = -4 \]
- At \( x = 3 \):
\[ 3^2 - 5 \cdot 3 = 9 - 15 = -6 \]

Thus,
\[ \int_{3}^{4} (2x - 5) \, dx = -4 - (-6) = -4 + 6 = 2 \]

### Step 6: Combine the results
Now, we combine the results of both integrals:
\[ \int_{2}^{4} (|x-2| + |x-3|) \, dx = 1 + 2 = 3 \]

### Final Answer
The value of the integral is \( \boxed{3} \).