Let [.]denote the greatest integer function, then the value of `int_(0)^(1.5) x[x^(2)]dx`, is

Options

A. `(3)/(2)`
B. `(3)/(4)`
C. `(5)/(4)`
D. 0

Explanation

To solve the integral \( \int_{0}^{1.5} x \lfloor x^2 \rfloor \, dx \), we need to analyze the behavior of the greatest integer function \( \lfloor x^2 \rfloor \) over the interval from 0 to 1.5.

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

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

2. **For \( 1 \leq x < \sqrt{2} \)**:
- \( x^2 \) ranges from \( 1 \) to \( 2 \).
- Therefore, \( \lfloor x^2 \rfloor = 1 \).

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

### Step 2: Break the integral into parts

We can break the integral into three parts based on the intervals determined above:

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

### Step 3: Calculate each integral

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

2. **Second integral**:
\[ \int_{1}^{\sqrt{2}} x \cdot 1 \, dx = \int_{1}^{\sqrt{2}} x \, dx = \left[ \frac{x^2}{2} \right]_{1}^{\sqrt{2}} = \frac{(\sqrt{2})^2}{2} - \frac{1^2}{2} = \frac{2}{2} - \frac{1}{2} = 1 - 0.5 = 0.5 \]

3. **Third integral**:
\[ \int_{\sqrt{2}}^{1.5} x \cdot 2 \, dx = 2 \int_{\sqrt{2}}^{1.5} x \, dx = 2 \left[ \frac{x^2}{2} \right]_{\sqrt{2}}^{1.5} = \left[ x^2 \right]_{\sqrt{2}}^{1.5} = (1.5)^2 - (\sqrt{2})^2 = 2.25 - 2 = 0.25 \]

### Step 4: Combine the results

Now, we can combine the results of the three integrals:

\[ \int_{0}^{1.5} x \lfloor x^2 \rfloor \, dx = 0 + 0.5 + 0.25 = 0.75 \]

### Final Answer

Thus, the value of the integral \( \int_{0}^{1.5} x \lfloor x^2 \rfloor \, dx \) is \( \frac{3}{4} \) or \( 0.75 \).