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Calculus

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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 -

    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 -

    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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