JEE Main 2021 (Online) 3rd August Morning ShiftQuestion 1
$\int_{1}^{2}\left([x^{2}-1]+[x^{2}+1]\right)dx$ is equal to ($[t]$ denotes the greatest integer $\le t$)
Definite Integration
Solution
Concept. $[t\pm1]=[t]\pm1$, so the integrand is $2[x^2]$; split at $\sqrt2,\sqrt3$. Formulas used. (1) $[x^2+1]=[x^2]+1$, $[x^2-1]=[x^2]-1$(2) On $[1,2)$: $[x^2]=1,2,3$ on $[1,\sqrt2)$, $[\sqrt2,\sqrt3)$, $[\sqrt3,2)$Step 1 - Simplify. By (1), the integrand is $2[x^2]$. Step 2 - Integrate. By (2), $$\int_1^2[x^2]dx=(\sqrt2-1)+2(\sqrt3-\sqrt2)+3(2-\sqrt3)=5-\sqrt2-\sqrt3.$$ Step 3 - Value. $2(5-\sqrt2-\sqrt3)=10-2\sqrt2-2\sqrt3$. Answer: (D)