Let [] denote the greatest integet function then the value `int_0^1.5 x[x^2].dx`

Options

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

Answer

To solve the integral \( \int_0^{1.5} x \lfloor x^2 \rfloor \, dx \), where \( \lfloor x \rfloor \) denotes the greatest integer function, we will break the integral into segments based on the behavior of the greatest integer function.

### Step-by-Step Solution:

1. **Identify the intervals**:
The function \( \lfloor x^2 \rfloor \) changes its value at certain points. We need to determine where \( x^2 \) crosses integer values in the range \( [0, 1.5] \).
- For \( x \in [0, 1) \), \( x^2 \) ranges from \( 0 \) to \( 1 \), so \( \lfloor x^2 \rfloor = 0 \).
- For \( x \in [1, \sqrt{2}) \), \( x^2 \) ranges from \( 1 \) to \( 2 \), so \( \lfloor x^2 \rfloor = 1 \).
- For \( x \in [\sqrt{2}, 1.5] \), \( x^2 \) ranges from \( 2 \) to \( 2.25 \), so \( \lfloor x^2 \rfloor = 2 \).

2. **Split the integral**:
We can split the integral into three parts based on the intervals identified:
\[ \int_0^{1.5} x \lfloor x^2 \rfloor \, dx = \int_0^1 x \lfloor x^2 \rfloor \, dx + \int_1^{\sqrt{2}} x \lfloor x^2 \rfloor \, dx + \int_{\sqrt{2}}^{1.5} x \lfloor x^2 \rfloor \, dx \]

3. **Evaluate each integral**:
- **First integral**:
\[ \int_0^1 x \lfloor x^2 \rfloor \, dx = \int_0^1 x \cdot 0 \, dx = 0 \]
- **Second integral**:
\[ \int_1^{\sqrt{2}} x \lfloor x^2 \rfloor \, dx = \int_1^{\sqrt{2}} x \cdot 1 \, dx = \int_1^{\sqrt{2}} x \, dx \]
This integral evaluates to:
\[ \left[ \frac{x^2}{2} \right]_1^{\sqrt{2}} = \frac{(\sqrt{2})^2}{2} - \frac{1^2}{2} = \frac{2}{2} - \frac{1}{2} = \frac{1}{2} \]
- **Third integral**:
\[ \int_{\sqrt{2}}^{1.5} x \lfloor x^2 \rfloor \, dx = \int_{\sqrt{2}}^{1.5} x \cdot 2 \, dx = 2 \int_{\sqrt{2}}^{1.5} x \, dx \]
This integral evaluates to:
\[ 2 \left[ \frac{x^2}{2} \right]_{\sqrt{2}}^{1.5} = \left[ x^2 \right]_{\sqrt{2}}^{1.5} = (1.5)^2 - (\sqrt{2})^2 = 2.25 - 2 = 0.25 \]

4. **Combine the results**:
Now, we combine the results of the three integrals:
\[ \int_0^{1.5} x \lfloor x^2 \rfloor \, dx = 0 + \frac{1}{2} + 0.25 = \frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4} \]

### Final Answer:
The value of the integral \( \int_0^{1.5} x \lfloor x^2 \rfloor \, dx \) is \( \frac{3}{4} \).