Given that ${\int }_0^{2.4}\left[x^2\right]dx=\alpha +\beta \sqrt{2}+\gamma \sqrt{3}+\delta \sqrt{5}\text{, then }\alpha +\beta +\gamma +\delta$

We know that for $1\le x<2$​, $\left[x\right]=1$​​​​​​$\Rightarrow {\int }_0^{2.4}\left[x^2\right]dx={\int }_0^{1}0dx+{\int }_1^{\sqrt{2}}1dx+{\int }_{\sqrt{2}}^{\sqrt{3}}2dx+{\int }_{\sqrt{3}}^{2}3dx+{\int }_2^{\sqrt{5}}4dx+{\int }_{\sqrt{5}}^{2.4}5dx$

$=0+(\sqrt{2}-1)+2(\sqrt{3}-\sqrt{2})+3(2-\sqrt{3})+4(\sqrt{5}-2)+5\left(2.4-\sqrt{5}\right)$

$=9-\sqrt{2}-\sqrt{3}-\sqrt{5}$

This is in the form of $\alpha +\beta \sqrt{2}+\gamma \sqrt{3}+\delta \sqrt{5}$

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

$\Rightarrow \alpha +\beta +\gamma +\delta =9-1-1-1=6$

Hence this is the required answer.