JEE Main 2021 (Online) 24th February Evening ShiftQuestion 1
The value of the integral $\int_{1}^{3}\left[x^{2}-2x-2\right]dx$, where $[x]$ denotes the greatest integer less than or equal to $x$, is :
Definite Integration
Solution
Concept. Complete the square: $x^2-2x-2=(x-1)^2-3$. With $y=x-1$ the integrand is $[y^2-3]$ on $[0,2]$. Formulas used. (1) $[y^2-3]=[y^2]-3$, and $[y^2]=0,1,2,3$ on $[0,1)$, $[1,\sqrt2)$, $[\sqrt2,\sqrt3)$, $[\sqrt3,2)$Step 1 - Shift. $y=x-1$: $\int_1^3[x^2-2x-2]\,dx=\int_0^2[y^2-3]\,dy$. Step 2 - Steps. By (1):$$\int_0^2[y^2-3]\,dy=-3\cdot1-2(\sqrt2-1)-1\cdot(\sqrt3-\sqrt2)+0.$$ Step 3 - Simplify. $-3-2\sqrt2+2-\sqrt3+\sqrt2=-1-\sqrt2-\sqrt3$. Answer: (C)