Answer
$$8190{\text{ Liters}}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( t \right){\text{be the liters of water flowing from the tank}}{\text{. By the}} \cr
& {\text{concept of the derivative }}f'\left( t \right){\text{ is the rate of the flow}} \cr
& f'\left( t \right) = \left( {500 - 5t} \right){\text{ in Lit per minute, then for 18 minutes}} \cr
& f'\left( t \right) = \int_0^{18} {\left( {500 - 5t} \right)dt} \cr
& f'\left( t \right) = \left[ {500t - \frac{{5{t^2}}}{2}} \right]_0^{18} \cr
& {\text{Evaluating}} \cr
& f'\left( t \right) = \left[ {500\left( {18} \right) - \frac{{5{{\left( {18} \right)}^2}}}{2}} \right] - \left[ {500\left( 0 \right) - \frac{{5{{\left( 0 \right)}^2}}}{2}} \right] \cr
& f'\left( t \right) = 8190 - 0 \cr
& f'\left( t \right) = 8190{\text{ Liters}} \cr} $$
