Let $[t]$ denote the greatest integer function. If $$\int_\limits{0}^{2.4}\left[x^{2}\right] d x=\alpha+\beta \sqrt{2}+\gamma \sqrt{3}+\delta \sqrt{5}$$, then $\alpha+\beta+\gamma+\delta$ is equal to __________.

$$ \begin{aligned} \int\limits_0^{2.4}\left[x^2\right] d x & =\int\limits_0^1\left[x^2\right] d x+\int\limits_1^{\sqrt{2}}\left[x^2\right] d x \\\\ & +\int\limits_{\sqrt{2}}^{\sqrt{3}}\left[x^2\right] d x+\int\limits_{\sqrt{3}}^2\left[x^2\right] d x+\int\limits_2^{\sqrt{5}}\left[x^2\right] d x+\int\limits_{\sqrt{5}}^{2.4}\left[x^2\right] d x \end{aligned} $$ <br/><br/>$$ \begin{aligned} & =\int_0 0 . d x+\int_1^{\sqrt{2}} 1 \cdot d x+\int_{\sqrt{2}}^{\sqrt{3}} 2 d x+\int_{\sqrt{3}}^2 3 d x+\int_2^{\sqrt{5}} 4 d x+\int_{\sqrt{5}}^{2.4} 5 d x \\\\ & = 0+[x]_1^{\sqrt{2}}+2[x]_{\sqrt{2}}^{\sqrt{3}}+3[x]_{\sqrt{3}}^2+4[x]_2^{\sqrt{5}}+5[x]_{\sqrt{5}}^{2.4} \\\\ & =\sqrt{2}-1+2 \sqrt{3}-2 \sqrt{2}+6-3 \sqrt{3}+4 \sqrt{5}-8+12-5 \sqrt{5} \\\\ & =-\sqrt{2}-\sqrt{3}-\sqrt{5}+9 \\\\ & \therefore \alpha=9, \beta=-1, \gamma=-1, \delta=-1 \\\\ & \text { So, } \alpha+\beta+\gamma+\delta=9-1-1-1=6 \end{aligned} $$