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 \( \int_{0}^{2} [x^2 - 1] \, dx \), where \([x]\) denotes the greatest integer function, we will break the integral into segments based on the behavior of the function \(x^2 - 1\) and the greatest integer function.

### Step 1: Identify the intervals for the greatest integer function
The function \(x^2 - 1\) will change its value at certain points. We need to find where \(x^2 - 1\) takes on integer values within the interval \([0, 2]\).

- At \(x = 0\): \(x^2 - 1 = -1\)
- At \(x = 1\): \(x^2 - 1 = 0\)
- At \(x = \sqrt{2}\): \(x^2 - 1 = 1\)
- At \(x = 2\): \(x^2 - 1 = 3\)

Thus, we can break the integral into the following intervals:
1. From \(0\) to \(1\): \(x^2 - 1\) ranges from \(-1\) to \(0\)
2. From \(1\) to \(\sqrt{2}\): \(x^2 - 1\) ranges from \(0\) to \(1\)
3. From \(\sqrt{2}\) to \(2\): \(x^2 - 1\) ranges from \(1\) to \(3\)

### Step 2: Evaluate the integral over each interval
1. **Interval \([0, 1]\)**:
- Here, \(x^2 - 1 < 0\), so \([x^2 - 1] = -1\).
\[ \int_{0}^{1} [x^2 - 1] \, dx = \int_{0}^{1} -1 \, dx = -1 \cdot (1 - 0) = -1 \]

2. **Interval \([1, \sqrt{2}]\)**:
- Here, \(0 \leq x^2 - 1 < 1\), so \([x^2 - 1] = 0\).
\[ \int_{1}^{\sqrt{2}} [x^2 - 1] \, dx = \int_{1}^{\sqrt{2}} 0 \, dx = 0 \]

3. **Interval \([\sqrt{2}, 2]\)**:
- Here, \(1 \leq x^2 - 1 < 3\), so \([x^2 - 1] = 1\).
\[ \int_{\sqrt{2}}^{2} [x^2 - 1] \, dx = \int_{\sqrt{2}}^{2} 1 \, dx = 1 \cdot (2 - \sqrt{2}) = 2 - \sqrt{2} \]

### Step 3: Combine the results
Now we can combine the results from each interval:
\[ \int_{0}^{2} [x^2 - 1] \, dx = \left(-1\right) + 0 + \left(2 - \sqrt{2}\right) = 1 - \sqrt{2} \]

### Final Answer
Thus, the value of the integral \( \int_{0}^{2} [x^2 - 1] \, dx \) is:
\[ \boxed{1 - \sqrt{2}} \]