The value of `int_1^2(x^([x^2])+[x^2]^x)dx`, where [.] denotes the greatest integer function, is equal to

Options

A. `(5)/(4)+sqrt(3)+(2^(sqrt(3))-2^(sqrt(2)))+(1)/(log 3)(9-3^(sqrt(3)))`
B. `(5)/(4)+sqrt(2)+(sqrt(2))/(3)+(1)/(log 2)(2^(sqrt(3))-2^(sqrt(2)))+(1)/(log 3)(9-3^(sqrt(3)))`
C. `(5)/(4)+(sqrt(2))/(3)+(1)/(log 2)(2^(sqrt(3))-2^(sqrt(2)))+(1)/(log 3)(9-3^(sqrt(3)))`
D. None of these

Answer

To solve the integral \( I = \int_1^2 \left( x^{[x^2]} + [x^2]^x \right) dx \), where \([.]\) denotes the greatest integer function, we will first analyze the behavior of the greatest integer function within the limits of integration.

### Step 1: Determine the intervals for the greatest integer function
The function \([x^2]\) will take different integer values in the interval from 1 to 2:
- For \( x \in [1, \sqrt{2}) \), \( x^2 \) ranges from 1 to 2, so \([x^2] = 1\).
- For \( x \in [\sqrt{2}, 2) \), \( x^2 \) ranges from 2 to 4, so \([x^2] = 2\).

### Step 2: Split the integral into two parts
We can split the integral at \( x = \sqrt{2} \):
\[ I = \int_1^{\sqrt{2}} \left( x^{[x^2]} + [x^2]^x \right) dx + \int_{\sqrt{2}}^2 \left( x^{[x^2]} + [x^2]^x \right) dx \]

### Step 3: Evaluate the first integral
For \( x \in [1, \sqrt{2}) \):
- \([x^2] = 1\)
- Thus, \( I_1 = \int_1^{\sqrt{2}} \left( x^1 + 1^x \right) dx = \int_1^{\sqrt{2}} \left( x + 1 \right) dx \)

Calculating \( I_1 \):
\[ I_1 = \int_1^{\sqrt{2}} x \, dx + \int_1^{\sqrt{2}} 1 \, dx \]
\[ = \left[ \frac{x^2}{2} \right]_1^{\sqrt{2}} + \left[ x \right]_1^{\sqrt{2}} \]
\[ = \left( \frac{(\sqrt{2})^2}{2} - \frac{1^2}{2} \right) + \left( \sqrt{2} - 1 \right) \]
\[ = \left( \frac{2}{2} - \frac{1}{2} \right) + \left( \sqrt{2} - 1 \right) \]
\[ = \frac{1}{2} + \sqrt{2} - 1 = \sqrt{2} - \frac{1}{2} \]

### Step 4: Evaluate the second integral
For \( x \in [\sqrt{2}, 2) \):
- \([x^2] = 2\)
- Thus, \( I_2 = \int_{\sqrt{2}}^2 \left( x^2 + 2^x \right) dx \)

Calculating \( I_2 \):
\[ I_2 = \int_{\sqrt{2}}^2 x^2 \, dx + \int_{\sqrt{2}}^2 2^x \, dx \]
\[ = \left[ \frac{x^3}{3} \right]_{\sqrt{2}}^2 + \left[ \frac{2^x}{\ln 2} \right]_{\sqrt{2}}^2 \]
\[ = \left( \frac{2^3}{3} - \frac{(\sqrt{2})^3}{3} \right) + \left( \frac{2^2}{\ln 2} - \frac{2^{\sqrt{2}}}{\ln 2} \right) \]
\[ = \left( \frac{8}{3} - \frac{2\sqrt{2}}{3} \right) + \left( \frac{4}{\ln 2} - \frac{2^{\sqrt{2}}}{\ln 2} \right) \]

### Step 5: Combine both integrals
Now, we combine \( I_1 \) and \( I_2 \):
\[ I = I_1 + I_2 = \left( \sqrt{2} - \frac{1}{2} \right) + \left( \frac{8}{3} - \frac{2\sqrt{2}}{3} + \frac{4}{\ln 2} - \frac{2^{\sqrt{2}}}{\ln 2} \right) \]

### Final Result
Thus, the final value of the integral \( I \) can be expressed as:
\[ I = \sqrt{2} - \frac{1}{2} + \frac{8}{3} - \frac{2\sqrt{2}}{3} + \frac{4}{\ln 2} - \frac{2^{\sqrt{2}}}{\ln 2} \]