`int_(0)^(sqrt(2)) [x^(2)]dx`, is

Options

A. `2-sqrt(2)`
B. `2+sqrt(2)`
C. `sqrt(2)-1`
D. `sqrt(2)-2`

Answer

To solve the integral \( \int_{0}^{\sqrt{2}} \lfloor x^2 \rfloor \, dx \), where \( \lfloor x^2 \rfloor \) is the greatest integer function (or floor function) of \( x^2 \), we will break the integral into two parts based on the behavior of \( \lfloor x^2 \rfloor \) over the interval from \( 0 \) to \( \sqrt{2} \).

### Step-by-Step Solution:

1. **Determine the intervals for \( x \)**:
- For \( x \) in the interval \( [0, 1) \):
- \( x^2 \) ranges from \( 0 \) to \( 1 \).
- Thus, \( \lfloor x^2 \rfloor = 0 \).
- For \( x \) in the interval \( [1, \sqrt{2}] \):
- \( x^2 \) ranges from \( 1 \) to \( 2 \).
- Thus, \( \lfloor x^2 \rfloor = 1 \).

2. **Break the integral into two parts**:
\[ \int_{0}^{\sqrt{2}} \lfloor x^2 \rfloor \, dx = \int_{0}^{1} \lfloor x^2 \rfloor \, dx + \int_{1}^{\sqrt{2}} \lfloor x^2 \rfloor \, dx \]

3. **Evaluate the first integral**:
\[ \int_{0}^{1} \lfloor x^2 \rfloor \, dx = \int_{0}^{1} 0 \, dx = 0 \]

4. **Evaluate the second integral**:
\[ \int_{1}^{\sqrt{2}} \lfloor x^2 \rfloor \, dx = \int_{1}^{\sqrt{2}} 1 \, dx \]
- The integral of \( 1 \) is simply \( x \):
\[ \int_{1}^{\sqrt{2}} 1 \, dx = x \bigg|_{1}^{\sqrt{2}} = \sqrt{2} - 1 \]

5. **Combine the results**:
\[ \int_{0}^{\sqrt{2}} \lfloor x^2 \rfloor \, dx = 0 + (\sqrt{2} - 1) = \sqrt{2} - 1 \]

### Final Answer:
Thus, the value of the integral \( \int_{0}^{\sqrt{2}} \lfloor x^2 \rfloor \, dx \) is \( \sqrt{2} - 1 \).