Evaluate :
`int_(0)^(2)[x^(2)]dx`

Answer

To evaluate the integral \( \int_{0}^{2} \lfloor x^2 \rfloor \, dx \), where \( \lfloor x^2 \rfloor \) is the greatest integer function, we will first determine the intervals where \( \lfloor x^2 \rfloor \) takes constant values as \( x \) varies from 0 to 2.

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

1. **For \( x \) in the interval [0, 1]:**
- \( x^2 \) varies from \( 0^2 = 0 \) to \( 1^2 = 1 \).
- Thus, \( \lfloor x^2 \rfloor = 0 \) in this interval.

2. **For \( x \) in the interval [1, \( \sqrt{2} \)]:**
- \( x^2 \) varies from \( 1^2 = 1 \) to \( (\sqrt{2})^2 = 2 \).
- Thus, \( \lfloor x^2 \rfloor = 1 \) in this interval.

3. **For \( x \) in the interval [\( \sqrt{2} \), \( \sqrt{3} \)]:**
- \( x^2 \) varies from \( (\sqrt{2})^2 = 2 \) to \( (\sqrt{3})^2 = 3 \).
- Thus, \( \lfloor x^2 \rfloor = 2 \) in this interval.

4. **For \( x \) in the interval [\( \sqrt{3} \), 2]:**
- \( x^2 \) varies from \( (\sqrt{3})^2 = 3 \) to \( 2^2 = 4 \).
- Thus, \( \lfloor x^2 \rfloor = 3 \) in this interval.

### Step 2: Set up the integral with the determined intervals

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

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

### Step 3: Evaluate each integral

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

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

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

4. **Fourth integral:**
\[ \int_{\sqrt{3}}^{2} 3 \, dx = 3[x]_{\sqrt{3}}^{2} = 3(2 - \sqrt{3}) \]

### Step 4: Combine the results

Now we can combine all the results:

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

Simplifying this expression:

\[ = \sqrt{2} - 1 + 2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3} \]
\[ = (6 - 1) + (\sqrt{2} - 2\sqrt{2}) + (2\sqrt{3} - 3\sqrt{3}) \]
\[ = 5 - \sqrt{2} - \sqrt{3} \]

### Final Answer

Thus, the value of the integral is:

\[ \int_{0}^{2} \lfloor x^2 \rfloor \, dx = 5 - \sqrt{2} - \sqrt{3} \]