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Fundamental Theorem of Calculus Part II

  1. Jul 9, 2012 #1
    1. Find the derivative of:
    ∫cos3(t)


    where a = 1/x and b = ∏/3

    This was a part of a question on my first calc exam and I just wanted to know if I did it correctly.



    We can solve this using the Fundamental Theorem of Calculus, Part II



    The solution would be to simply plug in the values for a and b, which should give a final answer of -1 - cos3(1/x)
     
  2. jcsd
  3. Jul 9, 2012 #2

    SammyS

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    What was the problem exactly ?

    Was it, find [itex]\displaystyle \frac{d}{dx}\int_{1/x}^{\pi/3}{\cos^3(t)}\,dt\ ?[/itex]
     
  4. Jul 9, 2012 #3
    Yes. That was it
     
  5. Jul 9, 2012 #4
    It said find the derivative of: and then the integral you gave
     
  6. Jul 9, 2012 #5

    HallsofIvy

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    No, what you have is incorrect. You do NOT "just plug in values for a and b"- the problem is not quite that simple. In particular, the "Fundamental Theorem of Calculus" says that
    [tex]\frac{d}{dx}\int_a^x f(t)dt= f(x)[/tex]
    so you have to do something about the facts that
    1) "x" is in the lower limit, not the upper.
    2) The limit is 1/x, not x.

    [tex]\frac{d}{dx}\int_{1/x}^{\pi/3} cos^3(t)dt= -\frac{d}{dx}\int_{\pi/3}^{1/x}cos^3(t)dt[/tex]
    If you let u= 1/x, that is
    [tex]-\frac{d}{dx}\int_{\pi/3}^u cos^3(t)dt= -\frac{du}{dx}\frac{d}{du}\int_{\pi/3}^u cos^3(t)dt[/tex]

    The derivative of dF/dx is (du/dx)(dF/du) so you need to find du/dx from the definition of u and dF/du from the "Fundamental Theorem of Calculus".
     
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