`int_(0)^(1.5) [x^(2)]dx` is equal to

Options

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

Explanation

To solve the integral \( \int_{0}^{1.5} \lfloor x^2 \rfloor \, dx \), we will break it down into intervals based on the behavior of the greatest integer function, \( \lfloor x^2 \rfloor \).

### Step 1: Determine the intervals for \( \lfloor x^2 \rfloor \)

1. **From \( x = 0 \) to \( x = 1 \)**:
- For \( 0 \leq x < 1 \), \( 0 \leq x^2 < 1 \) implies \( \lfloor x^2 \rfloor = 0 \).

2. **From \( x = 1 \) to \( x = \sqrt{2} \)**:
- For \( 1 \leq x < \sqrt{2} \) (approximately \( 1.414 \)), \( 1 \leq x^2 < 2 \) implies \( \lfloor x^2 \rfloor = 1 \).

3. **From \( x = \sqrt{2} \) to \( x = 1.5 \)**:
- For \( \sqrt{2} \leq x < 1.5 \), \( 2 \leq x^2 < 2.25 \) implies \( \lfloor x^2 \rfloor = 2 \).

### Step 2: Split the integral based on these intervals

Now we can express the integral as the sum of integrals over these intervals:

\[ \int_{0}^{1.5} \lfloor x^2 \rfloor \, dx = \int_{0}^{1} 0 \, dx + \int_{1}^{\sqrt{2}} 1 \, dx + \int_{\sqrt{2}}^{1.5} 2 \, dx \]

### Step 3: Evaluate each integral

1. **First integral**:
\[ \int_{0}^{1} 0 \, dx = 0 \]

2. **Second integral**:
\[ \int_{1}^{\sqrt{2}} 1 \, dx = \left[ x \right]_{1}^{\sqrt{2}} = \sqrt{2} - 1 \]

3. **Third integral**:
\[ \int_{\sqrt{2}}^{1.5} 2 \, dx = 2 \left[ x \right]_{\sqrt{2}}^{1.5} = 2 \left( 1.5 - \sqrt{2} \right) = 3 - 2\sqrt{2} \]

### Step 4: Combine the results

Now, we combine the results of the three integrals:

\[ \int_{0}^{1.5} \lfloor x^2 \rfloor \, dx = 0 + (\sqrt{2} - 1) + (3 - 2\sqrt{2}) \]

Simplifying this gives:

\[ = \sqrt{2} - 1 + 3 - 2\sqrt{2} = 2 - \sqrt{2} \]

### Final Result

Thus, the value of the integral \( \int_{0}^{1.5} \lfloor x^2 \rfloor \, dx \) is:

\[ \boxed{2 - 2\sqrt{2}} \]