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Homework Help: Antiderivative of a function problem

  1. Jan 8, 2010 #1
    1. The problem statement, all variables and given/known data
    [tex]$g(x)=\int _{2 }^{\sin x}\sqrt{1- t^2}dt$[/tex]
    whats g'(x)...

    2. Relevant equations

    3. The attempt at a solution
    how to find the antiderivative of sqrt(1-t^2)?
  2. jcsd
  3. Jan 9, 2010 #2


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    Try t=sin(u) or t=cos(u)
  4. Jan 9, 2010 #3


    Staff: Mentor

    I have edited your formula slightly so that it now shows g(x), not just (x). You had extra $ characters that shouldn't have been there.
    Use the Fundamental Theorem of Calculus to find g'(x). You will also need the chain rule since your integral isn't strictly a function of just x, but is a function of sin(x).
  5. Jan 9, 2010 #4


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    Hi Slimsta! :smile:
    Just to add to what Mark44 :smile: says …

    the beauty of using the Fundamental Theorem of Calculus is that you don't need to know the antiderivative. :wink:
  6. Jan 9, 2010 #5


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    The problem does not ask you to find the anti-derivative nor is it necessary.
    Letting y= sin(x), this is
    [tex]g(y)=\int_2^y \sqrt{1- t^2} dt[/itex]
    You can find dg/dy directly from the "Fundamental Theorem of Calculus" and then use the chain rule to find dg/dx.
  7. Jan 10, 2010 #6
    okay, that make sense but what if i have a function like this:
    [tex]$\int _{3\pi /4}^{\pi }(3 \sec ^2x -\frac{6 }{\pi })dx$ [/tex]

    this is confusing me :S
  8. Jan 10, 2010 #7
    That's a definite integral, so that gives you the signed area between the limits of integration, or just a number. What's the derivative of a constant?

    A variable (other than x since x would be the dummy variable) in either of the limits of integration would make it a function.
  9. Jan 11, 2010 #8
    hh i already figured that out.. i just took out the 3 and then it becomes tanx - 2/pi x and then its easy..
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