Calculus - Integral - Draining tank of water?

Question

Suppose that a tank initially contains 2000 gal of water and the rate of change of its volume after the tank drains for t min is `(t) = (0.5)t -30 (in gallons per minute). How much water does the tank contain after it has been draining for 25 minutes?

Answer 1

R(t) = .5 t - 30 --- This is your rate equation

S(t) = .25t^2 - 30t + C --- Integrate, this is your "how much water" equation

you know when time is 0, that there is 2000 gal of water so plug that in

S(0) = 0 + 0 + C ---- C = 2000 --- constant of integration

S(t) = .25t^2 - 30t + 2000 --- Adjusted general equation

Now just plug in 25 into t (since the units match ) and you get oyur answer

1406.25 Gal of water remains in the tank

Answer 2

The answer would be the integral of ((0.5)t-30)dt over the period, ie. from 0 to 25.

1411.25 gallons