Evaluate: `int_(0.5)^(3.5) [x] dx`

Answer

To evaluate the integral \( \int_{0.5}^{3.5} [x] \, dx \), where \([x]\) is the greatest integer function (also known as the floor function), we will break the integral into segments based on the behavior of the greatest integer function over the interval from \(0.5\) to \(3.5\).

### Step-by-Step Solution:

1. **Identify the intervals**:
The greatest integer function \([x]\) takes constant integer values over specific intervals. We can identify the intervals within \(0.5\) to \(3.5\):
- From \(0.5\) to \(1\), \([x] = 0\)
- From \(1\) to \(2\), \([x] = 1\)
- From \(2\) to \(3\), \([x] = 2\)
- From \(3\) to \(3.5\), \([x] = 3\)

2. **Break the integral into segments**:
We can express the integral as the sum of integrals over these intervals:
\[ \int_{0.5}^{3.5} [x] \, dx = \int_{0.5}^{1} [x] \, dx + \int_{1}^{2} [x] \, dx + \int_{2}^{3} [x] \, dx + \int_{3}^{3.5} [x] \, dx \]

3. **Evaluate each integral**:
- For \( \int_{0.5}^{1} [x] \, dx = \int_{0.5}^{1} 0 \, dx = 0 \)
- For \( \int_{1}^{2} [x] \, dx = \int_{1}^{2} 1 \, dx = 1 \cdot (2 - 1) = 1 \)
- For \( \int_{2}^{3} [x] \, dx = \int_{2}^{3} 2 \, dx = 2 \cdot (3 - 2) = 2 \)
- For \( \int_{3}^{3.5} [x] \, dx = \int_{3}^{3.5} 3 \, dx = 3 \cdot (3.5 - 3) = 3 \cdot 0.5 = 1.5 \)

4. **Combine the results**:
Now, we sum all the evaluated integrals:
\[ \int_{0.5}^{3.5} [x] \, dx = 0 + 1 + 2 + 1.5 = 4.5 \]

### Final Answer:
\[ \int_{0.5}^{3.5} [x] \, dx = 4.5 \]