Q. If $f(t) = \int_{- t}^{t}\frac{e^{- |x|}\ }{2}dx,\ then\ \lim_{t \rightarrow \infty}{f(t)}$is equal to
A) 1
B) $\frac{1}{2}$
C) 0
D) $- 1$
Explanation
(a)
$f(t) = \int_{- t}^{t}{\frac{e^{- \lbrack x\rbrack}}{2}dx = 2\int_{0}^{t}{\frac{e^{- x}}{2}dx}}$
$= - \lbrack e^{- x}\left. \ \ \right\rbrack_{0}^{t} = - e^{- t} + 1$
Now, $\lim_{t \rightarrow \infty}{f(t) = - \lim_{t \rightarrow \infty}e^{- t} + 1 = 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.