Water flows out from the tank at a rate that is a function of time and given by the expression r(t) = 200 - 4t liters per minute.
To determine the volume of water that flows out during the first 10 minutes the function r(t) has to be integrated between the limits 0 and 10. At any time t the rate of flow of water is 200 - 4t and the volume of water that actually flows out during an infinitesimally small duration `dt` is `(200 - 4t)*dt` as the rate at which water flows out during `dt` can be considered to be constant. Adding the volume of water that flows out during each of these instants gives the total volume of water that flows out.
`int_(0)^10 200 - 4t dt`
=> `200*t - 4t^2/2` between 10 and 0
=> `200(10 - 0) - 2(10^2 - 0^2)`
=> `2000 - 200`
A volume of water equal to 1800 liters flows out during the first 10 minutes.