What  \(\int_{\sqrt{2}}^{\sqrt{3}} f(x) dx\) equal to?

Calculation:

Given,

The function is \( f(x) = \left\lfloor x^2 \right\rfloor \), where \( \left\lfloor x^2 \right\rfloor \) is the greatest integer function.

We are tasked with finding:

\( \int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}} \left\lfloor x^2 \right\rfloor \, dx \)

Decomposing the integral based on the function:

For \\(\frac{\sqrt{3}}{2} \leq x < 1 \), x² lies between \(\frac{3}{4} \) and 1, so \(\left\lfloor x^2 \right\rfloor = 0 \). Therefore,

\( \int_{\frac{\sqrt{3}}{2}}^1 0 \, dx = 0 \)

For\( 1 \leq x < \sqrt{2} \), x²  lies between 1 and 2, so \(\left\lfloor x^2 \right\rfloor = 1 .\) Therefore,

\( \int_1^{\sqrt{2}} 1 \, dx = \sqrt{2} - 1 \)

For \(\sqrt{2} \leq x < \sqrt{3} \), x²  lies between 2 and 3, so \(\left\lfloor x^2 \right\rfloor = 2\) . Therefore,

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

Summing up all the results:

\( \int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}} \left\lfloor x^2 \right\rfloor \, dx = 0 + (\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2}) \)

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

Hence, the correct answer is Option 2.