Q. $\int_{0}^{1000}{e^{x - \lbrack x\rbrack}dx\ is}$
A) $e^{1000} - 1$
B) $\frac{e^{1000} - 1}{e - 1}$
C) $1000(e - 1)$
D) $\frac{e - 1}{1000}$
Explanation
(c)
Since, $e^{x - \lbrack x\rbrack}\$is a periodic function with period 1.
$\therefore\int_{0}^{1000}{e^{x - \lbrack x\rbrack}dx = 1000\int_{0}^{1}{e^{x}dx = 1000\left\lbrack e^{x} \right\rbrack_{0}^{1}}}$
$= 1000(e - 1)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.