The value of `int_(0)^(2) x^([x^(2) +1]) dx`, where `[x]` is the greatest integer less than or equal to `x` is

Options

A. `2`
B. `8//3`
C. `4`
D. none of these

Answer

To solve the integral \( I = \int_{0}^{2} x^{[\sqrt{x^2 + 1}]} \, dx \), where \([\cdot]\) denotes the greatest integer function, we need to analyze the function \( \sqrt{x^2 + 1} \) to determine the intervals where the greatest integer value changes.

### Step-by-Step Solution

1. **Identify the Function**:
We start with the function \( \sqrt{x^2 + 1} \). We will evaluate this function at key points to determine the intervals for the greatest integer function.

2. **Evaluate the Function at Key Points**:
- At \( x = 0 \):
\[ \sqrt{0^2 + 1} = \sqrt{1} = 1 \quad \Rightarrow \quad [\sqrt{0^2 + 1}] = 1 \]
- At \( x = 1 \):
\[ \sqrt{1^2 + 1} = \sqrt{2} \approx 1.414 \quad \Rightarrow \quad [\sqrt{1^2 + 1}] = 1 \]
- At \( x = \sqrt{2} \):
\[ \sqrt{(\sqrt{2})^2 + 1} = \sqrt{2 + 1} = \sqrt{3} \approx 1.732 \quad \Rightarrow \quad [\sqrt{(\sqrt{2})^2 + 1}] = 1 \]
- At \( x = \sqrt{3} \):
\[ \sqrt{(\sqrt{3})^2 + 1} = \sqrt{3 + 1} = 2 \quad \Rightarrow \quad [\sqrt{(\sqrt{3})^2 + 1}] = 2 \]
- At \( x = 2 \):
\[ \sqrt{2^2 + 1} = \sqrt{4 + 1} = \sqrt{5} \approx 2.236 \quad \Rightarrow \quad [\sqrt{2^2 + 1}] = 2 \]

3. **Determine the Intervals**:
From the evaluations:
- For \( x \in [0, \sqrt{3}) \), \( [\sqrt{x^2 + 1}] = 1 \)
- For \( x \in [\sqrt{3}, 2] \), \( [\sqrt{x^2 + 1}] = 2 \)

4. **Break the Integral**:
We can now break the integral into two parts:
\[ I = \int_{0}^{\sqrt{3}} x^{1} \, dx + \int_{\sqrt{3}}^{2} x^{2} \, dx \]

5. **Calculate Each Integral**:
- First Integral:
\[ \int_{0}^{\sqrt{3}} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{\sqrt{3}} = \frac{(\sqrt{3})^2}{2} - 0 = \frac{3}{2} \]
- Second Integral:
\[ \int_{\sqrt{3}}^{2} x^{2} \, dx = \left[ \frac{x^3}{3} \right]_{\sqrt{3}}^{2} = \frac{2^3}{3} - \frac{(\sqrt{3})^3}{3} = \frac{8}{3} - \frac{3\sqrt{3}}{3} = \frac{8 - 3\sqrt{3}}{3} \]

6. **Combine Results**:
Now, combine the results of both integrals:
\[ I = \frac{3}{2} + \frac{8 - 3\sqrt{3}}{3} \]

7. **Find a Common Denominator**:
The common denominator for \( \frac{3}{2} \) and \( \frac{8 - 3\sqrt{3}}{3} \) is 6:
\[ I = \frac{3 \times 3}{6} + \frac{2(8 - 3\sqrt{3})}{6} = \frac{9 + 16 - 6\sqrt{3}}{6} = \frac{25 - 6\sqrt{3}}{6} \]

### Final Result:
Thus, the value of the integral is:
\[ I = \frac{25 - 6\sqrt{3}}{6} \]