The value of `int_(-1)^(3){|x-2|+[x]} dx`, where [.] denotes the greatest integer function, is equal to

Options

A. 5
B. 6
C. 3
D. None of these

Answer

To solve the integral \( \int_{-1}^{3} (|x-2| + [x]) \, dx \), where \([x]\) denotes the greatest integer function, we will break the integral into segments based on the behavior of the functions involved.

### Step 1: Identify the intervals
The function \( |x-2| \) changes at \( x = 2 \), and the greatest integer function \([x]\) changes at integer values. Therefore, we will break the integral into the following intervals:
1. \( [-1, 0) \)
2. \( [0, 1) \)
3. \( [1, 2) \)
4. \( [2, 3] \)

### Step 2: Evaluate the integral on each interval

#### Interval 1: \( [-1, 0) \)
On this interval:
- \( |x-2| = 2 - x \)
- \([x] = -1\)

Thus, the integral becomes:
\[ \int_{-1}^{0} (|x-2| + [x]) \, dx = \int_{-1}^{0} (2 - x - 1) \, dx = \int_{-1}^{0} (1 - x) \, dx \]
Calculating this:
\[ = \left[ x - \frac{x^2}{2} \right]_{-1}^{0} = \left( 0 - 0 \right) - \left( -1 - \frac{1}{2} \right) = \frac{3}{2} \]

#### Interval 2: \( [0, 1) \)
On this interval:
- \( |x-2| = 2 - x \)
- \([x] = 0\)

Thus, the integral becomes:
\[ \int_{0}^{1} (|x-2| + [x]) \, dx = \int_{0}^{1} (2 - x + 0) \, dx = \int_{0}^{1} (2 - x) \, dx \]
Calculating this:
\[ = \left[ 2x - \frac{x^2}{2} \right]_{0}^{1} = \left( 2 - \frac{1}{2} \right) - 0 = \frac{3}{2} \]

#### Interval 3: \( [1, 2) \)
On this interval:
- \( |x-2| = 2 - x \)
- \([x] = 1\)

Thus, the integral becomes:
\[ \int_{1}^{2} (|x-2| + [x]) \, dx = \int_{1}^{2} (2 - x + 1) \, dx = \int_{1}^{2} (3 - x) \, dx \]
Calculating this:
\[ = \left[ 3x - \frac{x^2}{2} \right]_{1}^{2} = \left( 6 - 2 \right) - \left( 3 - \frac{1}{2} \right) = 4 - 2.5 = 1.5 \]

#### Interval 4: \( [2, 3] \)
On this interval:
- \( |x-2| = x - 2 \)
- \([x] = 2\)

Thus, the integral becomes:
\[ \int_{2}^{3} (|x-2| + [x]) \, dx = \int_{2}^{3} (x - 2 + 2) \, dx = \int_{2}^{3} x \, dx \]
Calculating this:
\[ = \left[ \frac{x^2}{2} \right]_{2}^{3} = \left( \frac{9}{2} - 2 \right) = \frac{9}{2} - \frac{4}{2} = \frac{5}{2} \]

### Step 3: Combine the results
Now we combine the results from all intervals:
\[ \text{Total} = \frac{3}{2} + \frac{3}{2} + \frac{3}{2} + \frac{5}{2} = \frac{3 + 3 + 3 + 5}{2} = \frac{14}{2} = 7 \]

### Final Answer
The value of the integral \( \int_{-1}^{3} (|x-2| + [x]) \, dx \) is \( 7 \).