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First principles

  1. May 2, 2005 #1
    If f is continuous over an interval containing (a,x)find from first principles the derivative of the function f(x)=integral f(t)dt.Any help?
     
  2. jcsd
  3. May 2, 2005 #2
    dump rule

    integral from a->x of f(t)dt = f(x) * d/dx(x) = f(x)

    you can derive this easily
    let F(x) be the antiderivative of f(x)

    therefore the integral = F(x) - F(a)

    take the derivative of that... the F(a) term falls off cause its a constant

    thus you do d/dx(F(x)) = f(x)

    yay!

    if this is for AP calc, i would really reccommend looking over the fundamental theorom of calculus
     
  4. May 2, 2005 #3

    HallsofIvy

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    I don't think that's as "first principles" as kidia intended. Here is the standard proof of the fundamental theorem:
    Let [tex]F(x)= \int_a^x f(t)dt[/tex]. Then [tex]F(x+h)= \int_a^{x+h}f(t)dt[/tex]
    [tex] = \int_a^x f(t)dt+ \int_x^{x+h}f(t)dt[/tex]

    So that F(x+h)- F(x)= \int_x^{x+h}f(t)dt. Now apply the mean value theorem to the function [tex]\int_x^{x+h}f(t)dt[/tex] to argue that F(x+h)-F(x)= hf(x*) where x* is between x and x+h. Finally, divide both sides by h and take the limit as h goes to 0.
     
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