Evaluate :
(i) `int_(0)^(2)[x]^(2) dx` (where `[*]` denotes greatest integer function)

Answer

To evaluate the integral \( \int_{0}^{2} [x^2] \, dx \), where \([x^2]\) denotes the greatest integer function (also known as the floor function), we will break the integral into segments based on the behavior of the function \([x^2]\) over the interval \([0, 2]\).

### Step 1: Determine the intervals for \([x^2]\)

1. For \(x \in [0, 1)\):
- \(x^2 \in [0, 1)\) implies \([x^2] = 0\).

2. For \(x \in [1, \sqrt{2})\):
- \(x^2 \in [1, 2)\) implies \([x^2] = 1\).

3. For \(x \in [\sqrt{2}, \sqrt{3})\):
- \(x^2 \in [2, 3)\) implies \([x^2] = 2\).

4. For \(x \in [\sqrt{3}, 2]\):
- \(x^2 \in [3, 4)\) implies \([x^2] = 3\).

### Step 2: Break the integral into segments

We can now express the integral as a sum of integrals over these intervals:

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

### Step 3: Evaluate each integral

1. **First integral**:
\[ \int_{0}^{1} [x^2] \, dx = \int_{0}^{1} 0 \, dx = 0 \]

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

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

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

### Step 4: Combine the results

Now we combine all the evaluated integrals:

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

### Step 5: Simplify the expression

Combining all terms:

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

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

\[ = 5 - \sqrt{2} - \sqrt{3} \]

### Final Answer

Thus, the value of the integral is:

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