`int_(0)^(3//2) [x^(2)] dx`=

Options

A. `2+sqrt2`
B. `2- sqrt2`
C. `3//2`
D. 3

Answer

To solve the integral \( I = \int_{0}^{\frac{3}{2}} \lfloor x^2 \rfloor \, dx \), where \( \lfloor x^2 \rfloor \) is the greatest integer function of \( x^2 \), we will break the integral into segments based on the values of \( x^2 \).

### Step 1: Determine the range of \( x^2 \)
First, we find the range of \( x^2 \) as \( x \) varies from \( 0 \) to \( \frac{3}{2} \):
- At \( x = 0 \), \( x^2 = 0 \).
- At \( x = \frac{3}{2} \), \( x^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4} = 2.25 \).

Thus, \( 0 \leq x^2 < 2.25 \).

### Step 2: Identify the intervals for \( \lfloor x^2 \rfloor \)
The function \( \lfloor x^2 \rfloor \) will take different integer values in the following intervals:
- For \( 0 \leq x^2 < 1 \) (i.e., \( 0 \leq x < 1 \)), \( \lfloor x^2 \rfloor = 0 \).
- For \( 1 \leq x^2 < 2 \) (i.e., \( 1 \leq x < \sqrt{2} \)), \( \lfloor x^2 \rfloor = 1 \).
- For \( 2 \leq x^2 < 2.25 \) (i.e., \( \sqrt{2} \leq x < \frac{3}{2} \)), \( \lfloor x^2 \rfloor = 2 \).

### Step 3: Break the integral into parts
Now we can break the integral into three parts based on the intervals identified:
\[ I = \int_{0}^{1} \lfloor x^2 \rfloor \, dx + \int_{1}^{\sqrt{2}} \lfloor x^2 \rfloor \, dx + \int_{\sqrt{2}}^{\frac{3}{2}} \lfloor x^2 \rfloor \, dx \]

### Step 4: Evaluate each integral
1. **First Integral**:
\[ \int_{0}^{1} \lfloor x^2 \rfloor \, dx = \int_{0}^{1} 0 \, dx = 0 \]

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

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

### Step 5: Combine the results
Now, we combine all parts:
\[ I = 0 + (\sqrt{2} - 1) + (3 - 2\sqrt{2}) = 2 - \sqrt{2} \]

Thus, the final result is:
\[ I = 2 - \sqrt{2} \]