\(\int_{\sqrt{2}}^{2} f(x) dx\)  is equal to ?

Calculation:

Given,

The function is \( f(x) = \lfloor x^2 \rfloor \).

We are tasked with finding the value of \( \int_{\sqrt{2}}^{2} f(x) dx \).

We can break the integral into two parts as follows:

\( \int_{\sqrt{2}}^{2} f(x) dx = \int_{\sqrt{2}}^{\sqrt{3}} 2 dx + \int_{\sqrt{3}}^{2} 3 dx \)

For the range \( \sqrt{2} \leq x \leq \sqrt{3} \), \( \lfloor x^2 \rfloor = 2 \). So the first part of the integral is:

\( \int_{\sqrt{2}}^{\sqrt{3}} 2 dx = 2 \times (\sqrt{3} - \sqrt{2}) \)

For the range \( \sqrt{3} \leq x \leq 2 \), \( \lfloor x^2 \rfloor = 3 \). So the second part of the integral is:

\( \int_{\sqrt{3}}^{2} 3 dx = 3 \times (2 - \sqrt{3}) \)

Now, we calculate the values:

\( 2 \times (\sqrt{3} - \sqrt{2}) = 2\sqrt{3} - 2\sqrt{2} \)

\( 3 \times (2 - \sqrt{3}) = 6 - 3\sqrt{3} \)

Combining the two parts:

\( 2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3} = 6 - 2\sqrt{2} - \sqrt{3} \)

Hence, the correct answer is Option 1.