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FTC integrate multiple functions.

  1. Jul 28, 2012 #1
    1. The problem statement, all variables and given/known data
    If F(x)=∫from 0 to g(x) of 1/(√(1+t^3)) dt and g(x)= ∫from 0 to cos(x) of 1+sin(t^2) dt, then find f'(pi/2)


    2. Relevant equations



    3. The attempt at a solution
    Tried FTC parts 1 and 2
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 28, 2012 #2

    SammyS

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    Hello tropic9393. Welcome to PF !

    Please, show us explicitly what you have tried so that we may help you.
     
  4. Jul 28, 2012 #3
    I know I need to solve for g(x). I did the integral with respect to t of and got realized that sin(t^2) does not have an elementary integral, and I havent learned how to do solve for that yet.
    Then i tried applying FTC directly and but i thought that only applied to the derivative of an integral.
    I have a major conceptual block, so i dont really have a lot of work to show :frown:
     
  5. Jul 28, 2012 #4

    SammyS

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    You are correct regarding the anti-derivative of sin(t2).

    Neither does [itex]\displaystyle 1/\sqrt{1+t^3}\ [/itex] have an anti-derivative which can be expressed in terms of elementary functions.

    Approach the solution in two parts, similar to the way the problem is set-up.

    According to the Fundamental Theorem of Calculus, what is f '(x), if [itex]\displaystyle f(x)=\int_{0}^{g(x)}\frac{1}{\sqrt{1+t^3}}\,dt\ ?[/itex] (Without using the explicit form of g(x). )
     
  6. Jul 28, 2012 #5
    f'(x)=1/(√(1+g(x)^3)

    Then do solve for f'(pi/2) by plugging g(pi/2) in for g(x)?
     
  7. Jul 28, 2012 #6

    SammyS

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    Not quite right. You forgot to use the chain rule.



    Suppose that the anti-derivative of [itex]1/\sqrt{1+t^3}[/itex] is [itex]H(t)[/itex].

    That says that [itex]\displaystyle \int_{0}^{g(x)}\frac{1}{\sqrt{1+t^3}}\,dt=H(g(x))-H(0)\ .[/itex]

    Use the chain rule to find the derivative, [itex]\displaystyle \frac{d}{dx}H(g(x))\ .[/itex]
     
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