The value of `int _(1)^(2)(x^([x^(2)])+[x^(2)]^(x)) d x ` is equal to where `[.]` denotes the greatest integer function

Options

A. `(5)/(4)sqrt3+(2^(sqrt3)-2^(sqrt2)) +(1)/( log 3)(9-3^(sqrt3))`
B. `(5)/(4)+sqrt3+(sqrt2)/(3)+(1)/( log 2) (2^(sqrt3)-2^(sqrt2))+(1)/( log 3)(9-3^(sqrt3))`
C. `(5)/(4)+(sqrt2)/(2)+(1)/(log 2)(2^(sqrt3)-2^(sqrt2))+(1)/(log 3)(9-3^(sqrt3))`
D. None of the above

Answer

To solve the integral

\[ I = \int_{1}^{2} \left( x^{\lfloor x^2 \rfloor} + \lfloor x^2 \rfloor^x \right) dx \]

where \(\lfloor . \rfloor\) denotes the greatest integer function, we will break down the integral into segments based on the value of \(x\).

### Step 1: Determine the ranges for \(\lfloor x^2 \rfloor\)

For \(x\) in the interval \([1, 2]\):
- When \(1 \leq x < \sqrt{2}\), \(x^2\) is in the interval \([1, 2)\), so \(\lfloor x^2 \rfloor = 1\).
- When \(\sqrt{2} \leq x < \sqrt{3}\), \(x^2\) is in the interval \([2, 3)\), so \(\lfloor x^2 \rfloor = 2\).
- When \(\sqrt{3} \leq x \leq 2\), \(x^2\) is in the interval \([3, 4]\), so \(\lfloor x^2 \rfloor = 3\).

### Step 2: Break the integral into parts

We can now express \(I\) as:

\[ I = \int_{1}^{\sqrt{2}} \left( x^{1} + 1^{x} \right) dx + \int_{\sqrt{2}}^{\sqrt{3}} \left( x^{2} + 2^{x} \right) dx + \int_{\sqrt{3}}^{2} \left( x^{3} + 3^{x} \right) dx \]

### Step 3: Evaluate each integral separately

1. **First Integral:**

\[ \int_{1}^{\sqrt{2}} \left( x + 1 \right) dx = \int_{1}^{\sqrt{2}} x \, dx + \int_{1}^{\sqrt{2}} 1 \, dx \]

Calculating these:

\[ \int x \, dx = \frac{x^2}{2} \Big|_{1}^{\sqrt{2}} = \frac{(\sqrt{2})^2}{2} - \frac{1^2}{2} = \frac{2}{2} - \frac{1}{2} = \frac{1}{2} \]

\[ \int 1 \, dx = x \Big|_{1}^{\sqrt{2}} = \sqrt{2} - 1 \]

Thus,

\[ \int_{1}^{\sqrt{2}} \left( x + 1 \right) dx = \frac{1}{2} + (\sqrt{2} - 1) = \sqrt{2} - \frac{1}{2} \]

2. **Second Integral:**

\[ \int_{\sqrt{2}}^{\sqrt{3}} \left( x^{2} + 2^{x} \right) dx = \int_{\sqrt{2}}^{\sqrt{3}} x^{2} \, dx + \int_{\sqrt{2}}^{\sqrt{3}} 2^{x} \, dx \]

Calculating these:

\[ \int x^{2} \, dx = \frac{x^3}{3} \Big|_{\sqrt{2}}^{\sqrt{3}} = \frac{(\sqrt{3})^3}{3} - \frac{(\sqrt{2})^3}{3} = \frac{3\sqrt{3}}{3} - \frac{2\sqrt{2}}{3} = \sqrt{3} - \frac{2\sqrt{2}}{3} \]

\[ \int 2^{x} \, dx = \frac{2^{x}}{\ln 2} \Big|_{\sqrt{2}}^{\sqrt{3}} = \frac{2^{\sqrt{3}}}{\ln 2} - \frac{2^{\sqrt{2}}}{\ln 2} \]

Thus,

\[ \int_{\sqrt{2}}^{\sqrt{3}} \left( x^{2} + 2^{x} \right) dx = \left( \sqrt{3} - \frac{2\sqrt{2}}{3} \right) + \left( \frac{2^{\sqrt{3}} - 2^{\sqrt{2}}}{\ln 2} \right) \]

3. **Third Integral:**

\[ \int_{\sqrt{3}}^{2} \left( x^{3} + 3^{x} \right) dx = \int_{\sqrt{3}}^{2} x^{3} \, dx + \int_{\sqrt{3}}^{2} 3^{x} \, dx \]

Calculating these:

\[ \int x^{3} \, dx = \frac{x^4}{4} \Big|_{\sqrt{3}}^{2} = \frac{(2)^4}{4} - \frac{(\sqrt{3})^4}{4} = \frac{16}{4} - \frac{9}{4} = \frac{7}{4} \]

\[ \int 3^{x} \, dx = \frac{3^{x}}{\ln 3} \Big|_{\sqrt{3}}^{2} = \frac{3^{2}}{\ln 3} - \frac{3^{\sqrt{3}}}{\ln 3} = \frac{9 - 3^{\sqrt{3}}}{\ln 3} \]

Thus,

\[ \int_{\sqrt{3}}^{2} \left( x^{3} + 3^{x} \right) dx = \frac{7}{4} + \frac{9 - 3^{\sqrt{3}}}{\ln 3} \]

### Step 4: Combine all parts

Now, we can combine all the parts to find the total value of the integral \(I\):

\[ I = \left( \sqrt{2} - \frac{1}{2} \right) + \left( \sqrt{3} - \frac{2\sqrt{2}}{3} + \frac{2^{\sqrt{3}} - 2^{\sqrt{2}}}{\ln 2} \right) + \left( \frac{7}{4} + \frac{9 - 3^{\sqrt{3}}}{\ln 3} \right) \]

### Final Result

This gives the final value of the integral \(I\).

---