Let $$[t]$$ denote the greatest integer function. If $$\int_0^{2.4} [x^2] dx = \alpha + \beta\sqrt{2} + \gamma\sqrt{3} + \delta\sqrt{5}$$, then $$\alpha + \beta + \gamma + \delta$$ is equal to _____.

We need to evaluate $$\displaystyle\int_0^{2.4} [x^2]\, dx$$, where $$[t]$$ denotes the greatest integer function.

As $$x$$ ranges from $$0$$ to $$2.4$$, $$x^2$$ ranges from $$0$$ to $$5.76$$. The value $$[x^2]$$ changes at points where $$x^2$$ equals an integer, i.e., at $$x = 1, \sqrt{2}, \sqrt{3}, 2, \sqrt{5}$$.

$$\begin{array}{|c|c|c|} \hline \text{Interval for } x & \text{Range of } x^2 & [x^2] \\ \hline [0, 1) & [0, 1) & 0 \\ [1, \sqrt{2}) & [1, 2) & 1 \\ [\sqrt{2}, \sqrt{3}) & [2, 3) & 2 \\ [\sqrt{3}, 2) & [3, 4) & 3 \\ [2, \sqrt{5}) & [4, 5) & 4 \\ [\sqrt{5}, 2.4] & [5, 5.76] & 5 \\ \hline \end{array}$$

$$\int_0^{2.4} [x^2]\, dx = 0 \cdot (1 - 0) + 1 \cdot (\sqrt{2} - 1) + 2 \cdot (\sqrt{3} - \sqrt{2}) + 3 \cdot (2 - \sqrt{3}) + 4 \cdot (\sqrt{5} - 2) + 5 \cdot \left(\dfrac{12}{5} - \sqrt{5}\right)$$

$$= (\sqrt{2} - 1) + (2\sqrt{3} - 2\sqrt{2}) + (6 - 3\sqrt{3}) + (4\sqrt{5} - 8) + (12 - 5\sqrt{5})$$

Collecting constant terms: $$-1 + 6 - 8 + 12 = 9$$

Collecting $$\sqrt{2}$$ terms: $$1 - 2 = -1$$

Collecting $$\sqrt{3}$$ terms: $$2 - 3 = -1$$

Collecting $$\sqrt{5}$$ terms: $$4 - 5 = -1$$

$$\int_0^{2.4} [x^2]\, dx = 9 - \sqrt{2} - \sqrt{3} - \sqrt{5}$$

Comparing with $$\alpha + \beta\sqrt{2} + \gamma\sqrt{3} + \delta\sqrt{5}$$:

$$\alpha = 9, \quad \beta = -1, \quad \gamma = -1, \quad \delta = -1$$

$$\alpha + \beta + \gamma + \delta = 9 + (-1) + (-1) + (-1) = \boxed{6}$$