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Find the derivative of the integral

  1. May 7, 2006 #1
    Find the interval on which the curve y = [tex]\int_0^{x}[/tex](1/(1 + t + t2))dt
    is concave upward.

    I know that I need to find the derivative of the integral.
    But I can't figure out how.

    I tried to use substitution, letting u = (1 + t + t 2).
    Then du = (1 + 2t)dt. But that doesn't work.
     
  2. jcsd
  3. May 7, 2006 #2
    You need to find the derivative, and the second derivative of the integral, you don't actually need to find the integral, maybe the second fundamental theorem of calculus would be useful here..
     
  4. May 7, 2006 #3
    Ok, the Fundamental Theorem of Calculus Part 2 says to take the antiderivative of the function evaluated at b and a...
    i.e.: F(b) -F(a)

    I don't know how to find the antiderivative any other way than to evaluate the integral...(because the integral is the antiderivative of the function, right?).

    Or are you saying to take the derivative of the integral to get the function back?

    So, would y' = 1/(1 + t + t2)?
     
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