JEE Main 2021 (Online) 16th March Evening ShiftQuestion 1
Consider the integral $I=\int_{0}^{10}\frac{[x]e^{[x]}}{e^{x-1}}dx$, where $[x]$ denotes the greatest integer less than or equal to $x$. Then the value of $I$ is equal to :
Definite Integration
Solution
Concept. On $[n,n+1)$, $[x]=n$, so the integrand is $ne^{n}e^{1-x}$; each piece is simple. Formulas used. (1) $\int_n^{n+1}e^{1-x}dx=e^{1-n}-e^{-n}=e^{-n}(e-1)$Step 1 - One piece. By (1), $\int_n^{n+1}ne^ne^{1-x}dx=n(e-1)$. Step 2 - Add $n=0,\dots,9$. $(0+1+\dots+9)(e-1)=45(e-1)$. Answer: (C)