Let [t] denote the greatest integer function.

Question

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 \_\_\_\_\_\_\_\_\_\_.

Answer: 6

Accepted Answer

$\begin{gathered} \int _{0}^{2.4}[{x}^{2}]dx =\int _{0}^{1}[{x}^{2}]dx+\int _{1}^{\sqrt{2}}[{x}^{2}]dx \ \ +\int _{\sqrt{2}}^{\sqrt{3}}[{x}^{2}]dx+\int _{\sqrt{3}}^{2}[{x}^{2}]dx+\int _{2}^{\sqrt{5}}[{x}^{2}]dx+\int _{\sqrt{5}}^{2.4}[{x}^{2}]dx \end{gathered}$

$\begin{gathered} ={\int }_{0}0.dx+{\int }_{\sqrt{2}}^{1}1\cdot dx+{\int }_{\sqrt{3}}^{\sqrt{2}}2dx+{\int }_{2}^{\sqrt{3}}3dx+{\int }_{\sqrt{5}}^{2}4dx+{\int }_{2.4}^{\sqrt{5}}5dx \ =0+[x{]}_{\sqrt{2}}^{1}+2[x{]}_{\sqrt{3}}^{\sqrt{2}}+3[x{]}_{2}^{\sqrt{3}}+4[x{]}_{\sqrt{5}}^{2}+5[x{]}_{2.4}^{\sqrt{5}} \ =\sqrt{2}-1+2\sqrt{3}-2\sqrt{2}+6-3\sqrt{3}+4\sqrt{5}-8+12-5\sqrt{5} \ =-\sqrt{2}-\sqrt{3}-\sqrt{5}+9 \ 	herefore \alpha =9,\beta =-1,\gamma =-1,\delta =-1 \ \alpha +\beta +\gamma +\delta =9-1-1-1=6 \end{gathered}$

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