Let [ t ] denote the greatest integer function. If ∫ _ 0 ^ 2 . 4 [...

Given that ${∫}_{0}^{2 . 4}\left[{x}^{2}\right]dx = \alpha + \beta\sqrt{2} + \gamma\sqrt{3} + \delta\sqrt{5}\text{, then}\alpha + \beta + \gamma + \delta$

We know that for $1 \le x < 2$, \left[x\right] = 1$$⇒ {∫}_{0}^{2 . 4}\left[{x}^{2}\right]dx = {∫}_{0}^{1}0dx + {∫}_{1}^{\sqrt{2}}1dx + {∫}_{\sqrt{2}}^{\sqrt{3}}2dx + {∫}_{\sqrt{3}}^{2}3dx + {∫}_{2}^{\sqrt{5}}4dx + {∫}_{\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 $\alpha + \beta\sqrt{2} + \gamma\sqrt{3} + \delta\sqrt{5}$

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

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

Hence this is the required answer.