The value of `int_(1)^(3) (|x-2|+[x])dx` is `([x]` stands for greatest integer less than or equal to x)

Answer

To solve the integral \( I = \int_{1}^{3} (|x-2| + [x]) \, dx \), where \([x]\) denotes the greatest integer less than or equal to \(x\), we will break the integral into two parts based on the behavior of the absolute value and the greatest integer function.

### Step 1: Break the Integral into Intervals
The absolute value function \( |x-2| \) changes at \( x = 2 \). Therefore, we can split the integral into two parts:
\[ I = \int_{1}^{2} (|x-2| + [x]) \, dx + \int_{2}^{3} (|x-2| + [x]) \, dx \]

### Step 2: Evaluate Each Interval
1. **For \( x \in [1, 2] \)**:
- Here, \( |x-2| = 2 - x \) (since \( x < 2 \)).
- The greatest integer function \([x] = 1\) (since \( 1 \leq x < 2 \)).
- Thus, the integral becomes:
\[ \int_{1}^{2} ((2 - x) + 1) \, dx = \int_{1}^{2} (3 - x) \, dx \]

2. **For \( x \in [2, 3] \)**:
- Here, \( |x-2| = x - 2 \) (since \( x \geq 2 \)).
- The greatest integer function \([x] = 2\) (since \( 2 \leq x < 3 \)).
- Thus, the integral becomes:
\[ \int_{2}^{3} ((x - 2) + 2) \, dx = \int_{2}^{3} x \, dx \]

### Step 3: Calculate Each Integral
1. **Calculate \( \int_{1}^{2} (3 - x) \, dx \)**:
\[ \int (3 - x) \, dx = 3x - \frac{x^2}{2} \]
Evaluating from 1 to 2:
\[ \left[ 3(2) - \frac{(2)^2}{2} \right] - \left[ 3(1) - \frac{(1)^2}{2} \right] = \left[ 6 - 2 \right] - \left[ 3 - 0.5 \right] = 4 - 2.5 = 1.5 \]

2. **Calculate \( \int_{2}^{3} x \, dx \)**:
\[ \int x \, dx = \frac{x^2}{2} \]
Evaluating from 2 to 3:
\[ \left[ \frac{(3)^2}{2} \right] - \left[ \frac{(2)^2}{2} \right] = \left[ \frac{9}{2} \right] - \left[ \frac{4}{2} \right] = \frac{9}{2} - 2 = \frac{9}{2} - \frac{4}{2} = \frac{5}{2} \]

### Step 4: Combine the Results
Now, we combine the results from both intervals:
\[ I = 1.5 + \frac{5}{2} = \frac{3}{2} + \frac{5}{2} = \frac{8}{2} = 4 \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{4} \]