Water is pumped out of a holding tank at a rate of 6-6e^-0.13t liters/minute, where t is in minutes since the pump is started. If the holding tank contains 1000 liters of water when the pump is started, how much water does it hold one hour later?
Sorry I'm posting so many questions, I really just need help!Please:)

Question

Water is pumped out of a holding tank at a rate of 6-6e^-0.13t liters/minute, where t is in minutes since the pump is started. If the holding tank contains 1000 liters of water when the pump is started, how much water does it hold one hour later?
 Sorry I'm posting so many questions, I really just need help!Please:)

anonymous · Answer

Okay... so you can integrate for how much water is pumped out in total, right?

anonymous · Answer

t goes from 0 to 60, since t is in minutes and water is pumped for one hour.

anonymous · Answer

So the total amount pumped would be $$\int\limits_{0}^{60}6-6*e^(-0.13t) dt$$

anonymous · Answer

Since the rate of pumping is changing, you would need to take the integral from 0 to 60 of your function.$$6\int\limits_{0}^{60}1-e^{-0.13t}dt$$

anonymous · Answer

Yes. What he said is correct, since he factored out a 6.

anonymous · Answer

Once you complete that definite integral to get the amount of water pumped out, subtract that from the original 1000 liters to get how much is left.

anonymous · Answer

Thank you so much! I just wrote and did it all out& got it rightttttttt:)

anonymous · Answer

Awwww yeeeeeah. High five.