`int_(0)^(3) [x]dx` is equal to

Options

A. 2
B. 4
C. 3
D. 1

Explanation

To solve the integral \( \int_{0}^{3} \lfloor x \rfloor \, dx \), where \( \lfloor x \rfloor \) is the greatest integer function (also known as the floor function), we will break the integral into intervals where \( \lfloor x \rfloor \) is constant.

### Step-by-Step Solution:

1. **Identify the intervals**:
The function \( \lfloor x \rfloor \) changes its value at integer points. Therefore, we will break the integral from 0 to 3 into three intervals:
- From \( 0 \) to \( 1 \): Here, \( \lfloor x \rfloor = 0 \).
- From \( 1 \) to \( 2 \): Here, \( \lfloor x \rfloor = 1 \).
- From \( 2 \) to \( 3 \): Here, \( \lfloor x \rfloor = 2 \).

2. **Set up the integral**:
We can express the integral as the sum of integrals over these intervals:
\[ \int_{0}^{3} \lfloor x \rfloor \, dx = \int_{0}^{1} \lfloor x \rfloor \, dx + \int_{1}^{2} \lfloor x \rfloor \, dx + \int_{2}^{3} \lfloor x \rfloor \, dx \]

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

4. **Combine the results**:
Now, we add the results of the three integrals:
\[ \int_{0}^{3} \lfloor x \rfloor \, dx = 0 + 1 + 2 = 3 \]

5. **Final answer**:
Thus, the value of the integral \( \int_{0}^{3} \lfloor x \rfloor \, dx \) is \( 3 \).