JEE Main 2023 — Mathematics Calculus

Given that $\displaystyle {\int }_{0}^{2.4}[{x}^{2}]dx=\alpha +\beta \sqrt{2}+\gamma \sqrt{3}+\delta \sqrt{5}\text{, then }\alpha +\beta +\gamma +\delta$∫02.4​[x2]dx=α+β2​+γ3​+δ5​, then α+β+γ+δ

We know that for $1\leq x<2$1≤x<2, $[x]=1$[x]=1$\displaystyle \Rightarrow {\int }_{0}^{2.4}[{x}^{2}]dx={\int }_{0}^{1}0dx+{\int }_{1}^{\sqrt{2}}1dx+{\int }_{\sqrt{2}}^{\sqrt{3}}2dx+{\int }_{\sqrt{3}}^{2}3dx+{\int }_{2}^{\sqrt{5}}4dx+{\int }_{\sqrt{5}}^{2.4}5dx$⇒∫02.4​[x2]dx=∫01​0dx+∫12​​1dx+∫2​3​​2dx+∫3​2​3dx+∫25​​4dx+∫5​2.4​5dx

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

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

This is in the form of $\alpha +\beta \sqrt{2}+\gamma \sqrt{3}+\delta \sqrt{5}$α+β2​+γ3​+δ5​

$$\alpha =9,\beta =-1,\gamma =-1,\delta =-1$$α=9,β=−1,γ=−1,δ=−1

$$\Rightarrow \alpha +\beta +\gamma +\delta =9-1-1-1=6$$⇒α+β+γ+δ=9−1−1−1=6

Hence this is the required answer.