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Definite Integral w/ Second Fundamental Theorem of Calculus

  1. Dec 9, 2007 #1
    1. The problem statement, all variables and given/known data
    Find F'(x) if

    [tex]F(x)=\int_{0}^{x^3}(\sin (t^2))dt[/tex]



    3. The attempt at a solution

    Here's what I did:

    [tex] F(x)= -\cos (t^2)\biggr]^{x^3}_{0}[/tex]

    and I get: [tex] F(x)= -\cos (x^6) +1[/tex]

    [tex] F'(x)= sin (x^6)(6x^5) [/tex]

    However, the book's answer is [tex]F'(x)= 3x^2 \sin(x^6) [/tex]

    How did they come up with this answer if the derivative of x^6 is 6x^5 ??

    Thanks.
     
    Last edited: Dec 9, 2007
  2. jcsd
  3. Dec 9, 2007 #2

    arildno

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    Eeh, wherever did your cosines come from???
     
  4. Dec 9, 2007 #3
    Sorry, original post edited. I was looking at the wrong problem.
     
  5. Dec 9, 2007 #4

    arildno

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    Allright!
    Your anti-derivative is totally wrong.
    The derivative of -cos(t^{2})=2tsin(t^{2}), which does not agree with your original integrand.
     
  6. Dec 9, 2007 #5
    Oh ok. How do you solve it then?

    I tired the u-substitution method:

    [tex]F(x)=\frac{1}{2}\int^{x^3}_{0}\sin u(u^{-1/2}) du [/tex]

    Am I on the right track?
     
  7. Dec 9, 2007 #6

    arildno

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    Not at all, you can't write an anti-derivative to that integrand in terms of elementary functions.

    Use the fundamental theorem of calculus instead.
     
  8. Dec 10, 2007 #7

    HallsofIvy

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    You don't find the antiderivative- you don't need to. Use the Fundamental Theorem of Calculus:
    [tex]\frac{d}{dx}\int_a^x f(t)dt= f(x)[/tex]
    Handle the fact that the upper limit is a function of x rather than x itself with the chain rule:
    [tex]\frac{d}{dx}\int_a^{g(x)}f(t)dt= f(x)\frac{dg}{dx}[/itex]
     
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