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
Oil is leaking out of a ruptured tanker at a rate of r = 19e^(2t) gallons per hour. Write a definite integral that represents the total quantity of oil which leaks out of the tanker in the first 3 hours. Using your formula answer the following questions:
A. Estimate the integral using a left hand sum with 3 subdivisions of equal length. Round your answer to one decimal place.
Answer:
B. Use your calculator to estimate the 'exact' amount. (Your calculator does not give the exact value, therefore your answer to this question is also an approximation, but much more precise than the one you gave in part A. In this case round your answer to two decimal places.)
Answer:
2 answers
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.
that would make more sense. If so do it the same way but with different numbers. It should give better results.