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Fundamental Theorem

  1. Sep 25, 2004 #1
    Hello everyone, its been a while.

    It's been almost 4 months since I did anything calculus related so I forgot all of my skills. :bugeye:

    The problem is:
    Use the Fundamental Theorem of Calculus to find the derivative of the function
    [tex] h(x) = \int_{2}^{\frac{1}{x}} \arctan{t} \,dt [/tex]
  2. jcsd
  3. Sep 25, 2004 #2
    IIRC (whichI might not) the fundamental theorem of calculus says that given F(x) = S(f(x),x,a,b) F'(x) = f(b)-f(a)
  4. Sep 25, 2004 #3
    Consider the function [tex]F(x) = \int_{a}^{x} f(t) \,dt [/tex].

    The Fundamental Theorem of Calculus is given by: [tex]\frac{dF}{dx} = f(x)[/tex]. In your case the upper integration limit is [tex]1/x[/tex]. Therefore, you will have to use the chain rule. Let [tex]u=1/x \Rightarrow \frac{dh}{dx} = \frac{dh}{du}\frac{du}{dx} = -\frac{1}{x^2}arctan(\frac{1}{x})[/tex]
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