Consider the following for the two (02) items that follow: Let f(x)...

The given integral is $\int_{\sqrt{2}}^{2} [x^2] \, dx$.

As $x$ varies from $\sqrt{2} to 2$, $x^2$ varies from $2 to 4$. The greatest integer function $[x^2]$ changes its value when $x^2$ is an integer. The integer values of $x^2$ in this interval are $2, 3, and 4$, which correspond to $x = \sqrt{2}, \sqrt{3}, and 2$.

We split the integral at $x = \sqrt{3}$:

For $x \in (\sqrt{2}, \sqrt{3})$, $2 < x^2 < 3 \Rightarrow [x^2] = 2$

For $x \in (\sqrt{3}, 2)$, $3 < x^2 < 4 \Rightarrow [x^2] = 3$

The integral can be written as:

\$int_{\sqrt{2}}^{2} [$x^2] \, dx = \$int_{\sqrt{2}}^{\sqrt{3}}$2 \, dx + \$int_{\sqrt{3}}^{2}$3 \, dx

Evaluating the $integrals:<br><br>\int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx = 2[x]_{\sqrt{2}}^{\sqrt{3}} = 2(\sqrt{3} - \sqrt{2}) = 2\sqrt{3} - 2\sqrt{2}$

$\int_{\sqrt{3}}^{2} 3 \, dx = 3[x]_{\sqrt{3}}^{2} = 3(2 - \sqrt{3}) = 6 - 3\sqrt{3}$

Adding the two values:

$\int_{\sqrt{2}}^{2} [x^2] \, dx = (2\sqrt{3} - 2\sqrt{2}) + (6 - 3\sqrt{3})$

$\int_{\sqrt{2}}^{2} [x^2] \, dx = 6 - \sqrt{3} - 2\sqrt{2}$

Answer: $6 - \sqrt{3} - 2\sqrt{2}