Answer
4512 liters
Work Step by Step
An oil storage tank ruptures at time $t=0$ and oil leaks from the tank at a rate of $r(t)=100 e^{-0.01t}$ liters per minute.
Consider $r'(t)$ amount of oil leaks out during first hour.
$ r'(t)=\int _{0}^{60}r(t)dt$
This implies
$ r'(t)=\int _{0}^{60} 100 e^{-0.01t}dt$
$=100(\frac{e^{-0.01t}}{-0.01})_{0}^{60}$
$=4512$ liters
Hence, the amount of oil leaks out during first hour will be
$4512$ liters.