JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Definite Integrals

$[x^2]$ changes value wherever $x^2$ crosses an integer. Since $(2.4{)}^2=5.76$, the break points in $[0,2.4]$ are $x=1,\sqrt{2},\sqrt{3},2,\sqrt{5}$.

$[x^2]=0$ on $[0,1)$; $=1$ on $[1,\sqrt{2})$; $=2$ on $[\sqrt{2},\sqrt{3})$; $=3$ on $[\sqrt{3},2)$; $=4$ on $[2,\sqrt{5})$; $=5$ on $[\sqrt{5},2.4]$.

${\int }_0^{2.4}[x^2]dx=0+1(\sqrt{2}-1)+2(\sqrt{3}-\sqrt{2})+3(2-\sqrt{3})+4(\sqrt{5}-2)+5(2.4-\sqrt{5})$

Collecting like terms:
rational part: $-1+6-8+12=9$
coefficient of $\sqrt{2}$: $1-2=-1$
coefficient of $\sqrt{3}$: $2-3=-1$
coefficient of $\sqrt{5}$: $4-5=-1$

$\therefore {\int }_0^{2.4}[x^2]dx=9-\sqrt{2}-\sqrt{3}-\sqrt{5}$

Since $1,\sqrt{2},\sqrt{3},\sqrt{5}$ are linearly independent over the rationals, comparison gives $\alpha =9$, $\beta =\gamma =\delta =-1$ uniquely.

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