Posted by Lena on Sunday, November 14, 2010 at 4:32pm.
V is total volume leaked
first I will do the integral
dV/dt = 19e^(2t)
V = (19/2)e^2t + C
When t = 0
V = 19/2 + C = 0 I assume
so C = -19/2
V = (19/2) e^(2t) - 19/2
= (19/2)(e^2t-1)
now rectangles with left values at 0, 1, 2, 3 seconds
At t = 0, V = 0
dV/dt = 19
integral from 1 to 2 is 0
t = 1, V1 = 19 gal
dV/dt = 19e^2 = 140
so at t =2
V2 = 19 + 140 = 159
t = 2, V2 = 159
dV/dt = 19 e^4 = 1037
so at t = 3
V3 = 159 + 1037 = 1196
t = 3, V3 = 1196
dV/dt = 19 e^6 = 7665
so at t = 4
V4 = 1196 + 7665 = 8861
B)V = (19/2) e^(2t) - 19/2
at t = 4
(19/2) (e^8-1) = 28319
using left hand values vastly underestimates integral because function is increasing so quickly
are you sure it is not e^(-2t) ?
that would make more sense. If so do it the same way but with different numbers. It should give better results.
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