The integral `I= int_(0)^(2) [x^2] dx ([x]` denotes the greatest integer less than or equal to `x)` is equal to:

Options

A. `5 -2 sqrt3`
B. `5 - sqrt2 - sqrt3`
C. `6 - sqrt2 - sqrt3`
D. `3 - sqrt2`

Answer

To solve the integral \( I = \int_{0}^{2} [x^2] \, dx \), where \([x]\) denotes the greatest integer less than or equal to \(x\), we will break down the integral based on the behavior of the function \([x^2]\) over the interval \([0, 2]\).

### Step 1: Determine the range of \(x^2\) over the interval \([0, 2]\)

When \(x\) varies from \(0\) to \(2\):
- At \(x = 0\), \(x^2 = 0\)
- At \(x = 2\), \(x^2 = 4\)

Thus, \(x^2\) ranges from \(0\) to \(4\).

### Step 2: Identify the intervals where \([x^2]\) is constant

The function \([x^2]\) will take on different integer values in the following intervals:
- For \(0 \leq x < 1\): \(x^2\) ranges from \(0\) to \(1\), so \([x^2] = 0\).
- For \(1 \leq x < \sqrt{2}\): \(x^2\) ranges from \(1\) to \(2\), so \([x^2] = 1\).
- For \(\sqrt{2} \leq x < \sqrt{3}\): \(x^2\) ranges from \(2\) to \(3\), so \([x^2] = 2\).
- For \(\sqrt{3} \leq x < 2\): \(x^2\) ranges from \(3\) to \(4\), so \([x^2] = 3\).

### Step 3: Set up the integral based on these intervals

We can now express the integral \(I\) as the sum of integrals over these intervals:
\[ I = \int_{0}^{1} 0 \, dx + \int_{1}^{\sqrt{2}} 1 \, dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx + \int_{\sqrt{3}}^{2} 3 \, dx \]

### Step 4: 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 5: Combine the results

Now we combine all the results:
\[ I = 0 + (\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2}) + 3(2 - \sqrt{3}) \]
\[ I = \sqrt{2} - 1 + 2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3} \]
\[ I = (6 - 1) + (\sqrt{2} - 2\sqrt{2}) + (2\sqrt{3} - 3\sqrt{3}) \]
\[ I = 5 - \sqrt{2} - \sqrt{3} \]

### Final Answer
Thus, the value of the integral \(I\) is:
\[ I = 5 - \sqrt{2} - \sqrt{3} \]