Posted by **Lena** on Sunday, November 14, 2010 at 4:32pm.

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:

- Calculus -
**Damon**, Sunday, November 14, 2010 at 5:00pm
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

- By the way -
**Damon**, Sunday, November 14, 2010 at 5:02pm
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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