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If f'(x) = 10t / ∛(t – 2) and f(8) = –20, calculate f(x).

  1. Jul 22, 2014 #1

    s3a

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    1. The problem statement, all variables and given/known data
    Problem:
    If f'(x) = 10t / ∛(t – 2) and f(8) = –20, calculate f(x).

    Solution:
    Let u = t – 2 ⇒ dx = du. Then f(x) = –20 + ∫_8^x [10t / ∛(t – 2)] dt = –20 + ∫_6^(x – 2) [10(u + 2) / ∛(u)] du = –20 + 10 ∫_6^(x – 2) [u^(2/3) + 2u^(–1/3)] du = 30 ∛[(x – 2)^2] + 6(x – 2)^(5/3) – 66 ∛(3) ∛(12) – 20

    Additionally, the problem is attached as TheProblem.png, and the solution is attached as TheSolution.png.

    2. Relevant equations
    I'm not sure, but I think this has to do with the Fundamental Theorem of Calculus.

    3. The attempt at a solution
    I understand all the algebraic manipulations done; I'm just confused as to how the author went from the problem to the expression f(x) = –20 + ∫_8^x [10t / ∛(t – 2)] dt. Also, is it okay/valid that f'(x) (which is a function of x) = 10t / ∛(t – 2) (which is a function of t)?

    Any help in clearing my confusions would be greatly appreciated!
     

    Attached Files:

  2. jcsd
  3. Jul 22, 2014 #2
    Remember that when you integrate a function you have find the constant of integration. He just combined the two steps of finding the constant and integrating into one step, essentially.
     
  4. Jul 22, 2014 #3

    HallsofIvy

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    The integral [itex]\int_a^a h(t)dt= 0[/itex] for any integrable function h. So that [itex]\int_8^x h(t)dt[/itex] gives a function that is 0 when x= 8. Knowing that f(8)= -20 means that [itex]f(x)= -20+ \int_8^x h(t)dt[/itex]
     
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