The value of `int_(0)^(2)[x^(2)-1]dx`, where [x] denotes the greatest integer function, is given by:

Options

A. `3-sqrt(3)-sqrt(2)`
B. 2
C. 1
D. None of these

Answer

To solve the integral \( I = \int_{0}^{2} [x^2 - 1] \, dx \), where \([x]\) denotes the greatest integer function, we first need to analyze the expression inside the integral.

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

The expression \( x^2 - 1 \) can be rewritten as:
- At \( x = 0 \): \( 0^2 - 1 = -1 \)
- At \( x = 1 \): \( 1^2 - 1 = 0 \)
- At \( x = \sqrt{2} \): \( (\sqrt{2})^2 - 1 = 1 \)
- At \( x = \sqrt{3} \): \( (\sqrt{3})^2 - 1 = 2 \)
- At \( x = 2 \): \( 2^2 - 1 = 3 \)

Thus, we can break the integral into intervals based on the values of \( x^2 - 1 \):
- From \( 0 \) to \( 1 \): \( x^2 - 1 \) goes from \(-1\) to \(0\)
- From \( 1 \) to \( \sqrt{2} \): \( x^2 - 1 \) goes from \(0\) to \(1\)
- From \( \sqrt{2} \) to \( \sqrt{3} \): \( x^2 - 1 \) goes from \(1\) to \(2\)
- From \( \sqrt{3} \) to \( 2 \): \( x^2 - 1 \) goes from \(2\) to \(3\)

### Step 2: Set up the integral

We can express the integral as:
\[ I = \int_{0}^{1} [x^2 - 1] \, dx + \int_{1}^{\sqrt{2}} [x^2 - 1] \, dx + \int_{\sqrt{2}}^{\sqrt{3}} [x^2 - 1] \, dx + \int_{\sqrt{3}}^{2} [x^2 - 1] \, dx \]

### Step 3: Evaluate each integral

1. **From \( 0 \) to \( 1 \)**:
\[ [x^2 - 1] = -1 \quad \text{(since } -1 \leq x^2 - 1 < 0\text{)} \]
\[ \int_{0}^{1} -1 \, dx = -x \bigg|_{0}^{1} = -1 \]

2. **From \( 1 \) to \( \sqrt{2} \)**:
\[ [x^2 - 1] = 0 \quad \text{(since } 0 \leq x^2 - 1 < 1\text{)} \]
\[ \int_{1}^{\sqrt{2}} 0 \, dx = 0 \]

3. **From \( \sqrt{2} \) to \( \sqrt{3} \)**:
\[ [x^2 - 1] = 1 \quad \text{(since } 1 \leq x^2 - 1 < 2\text{)} \]
\[ \int_{\sqrt{2}}^{\sqrt{3}} 1 \, dx = x \bigg|_{\sqrt{2}}^{\sqrt{3}} = \sqrt{3} - \sqrt{2} \]

4. **From \( \sqrt{3} \) to \( 2 \)**:
\[ [x^2 - 1] = 2 \quad \text{(since } 2 \leq x^2 - 1 < 3\text{)} \]
\[ \int_{\sqrt{3}}^{2} 2 \, dx = 2x \bigg|_{\sqrt{3}}^{2} = 2(2) - 2\sqrt{3} = 4 - 2\sqrt{3} \]

### Step 4: Combine the results

Now we combine all the results:
\[ I = (-1) + 0 + (\sqrt{3} - \sqrt{2}) + (4 - 2\sqrt{3}) \]
\[ I = -1 + \sqrt{3} - \sqrt{2} + 4 - 2\sqrt{3} \]
\[ I = 3 - \sqrt{3} - \sqrt{2} \]

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