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Problem With fundamental theorum of Calculus

  1. Dec 17, 2008 #1
    1. The problem statement, all variables and given/known data

    The question: F'([tex]\pi[/tex]/2) if F(X)= [tex]\int^{cosx}_{0} e^{t^{2}}[/tex]

    3. The attempt at a solution

    I thought I thought F'(X) = f(t) = e^{t^{2}} replacing t with cos^{2}x

    But my book writes:

    F'(x) = [tex](-sinx)e^{cos^{2}x}[/tex]
  2. jcsd
  3. Dec 17, 2008 #2
    the deal is that you need to apply chain rule here, since your upper limit bound of integration is not an independent variable, but rather a function as well.
  4. Dec 17, 2008 #3


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    I assume your question is, "Find the value of [itex]F'(\pi/2)[/itex], given

    [tex]F(x)=\int_0^{\cos x}e^{t^2}\,dt\;.[/tex]

    The fundamental theorem of calculus states that if

    [tex]F(x)=\int_a^x f(t)dt\;,[/tex]



    Can you see how this differs from your problem?
  5. Dec 17, 2008 #4
    hmm... I'm sorry but I still don't see it...

    The book then substitutes pi/2 into the F'(x) that I wrote above and ends it there...

    the final answer is -1....
  6. Dec 17, 2008 #5

    What does the chain rule say?

    if you have[tex] [F(g(x))]'=F'(g(x))*g'(x)=e^{cos^2x}(cosx)'=?[/tex]

    In your problem you have g(x)=cosx and

    [tex]F(g(x))=\int_0^{\ g(x)}e^{t^2}\,dt=\int_0^{\cos x}e^{t^2}\,dt[/tex]

    Now just look above and you are done.
    Last edited: Dec 17, 2008
  7. Dec 17, 2008 #6
    Ah, I see whats going on...

    I was differentiating [tex]e^{cos^{2}x}[/tex] which was giving me something else...
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