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...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

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`

=> `1800`

**A volume of water equal to 1800 liters flows out during the first 10 minutes.**

**Further Reading**