The calculation of the integral will give us the answer.
Amount = Int (2000e^(-0,2t))dt
With t = 0 to t = 10
A = 2000.In(e^(-0,2t))dt
Making U = -0,2t
U = -0,2t
dU/dt = d(-0,2t)/dt
dU/dt = -0,2
dU = -0,2dt
dt = dU/-0,2
_______________
Then,
A = 2000.Int (e^U).dU/-0,2
A = -10.000.Int(e^U)dU
A = -10.000.e^U
A = -10.000.e^(-0,2t) | (0,10)
A = -10.000.(e^(-2)) - [ -10.000.e^(0)]
A = -10.000.e^(-2) + 10.000
A ~ 8,646 Gallons
Total volume of water=12*10*11
Let final height of water be h
Total volume of water=h*(12*10+7*10+6*5)
