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Math Help - Fundamental Theorem of Calculus

  1. #1
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    Fundamental Theorem of Calculus

    Theorem. Let  a \in \mathbb{R} , and let  I be an interval containing  a . Let  I \subseteq K \subseteq \mathbb{R}. Suppose that the function  f: K \to \mathbb{R} is integrable on  [a,x] for all  x \in I with  x > a and on  [x,a] for all  x \in I with  x < a . Let  F_a be the area accumulation function for  f on  I based at  a . Then  F_a is uniformly continuous on  I . If  f happens to be continuous on  I , then  F_a is differentiable on  I and  F_{a}'(x) = f(x) for all  x \in I .

    Proof. Suppose  a \in \mathbb{R} ,  I an interval containing  a , and let  f: I \to \mathbb{R} be a function. The area accumulation function  F_a: I \to \mathbb{R} is defined by  F_{a}(x) = \int_{a}^{x} f(t) \ dt . (1) We want to show that  F_a is uniformly continuous on  I . To do this, would we use a sequence approach? Because  F_a is not Lipschitz. (2) Suppose  f is continuous on  I . We want to show that  F_a is differentiable at some point  c . So we look at the following:

     \frac{F_{a}(x)-F_{a}(c)}{x-c} = \frac{\int_{a}^{x} f(t) \ dt- \int_{a}^{c} f(t) \ dt}{x-c} = \frac{1}{x-c} \int_{x}^{c} f(t) \ dt

    We want to show that  \lim\limits_{x \to c} \frac{1}{x-c} \int_{c}^{x} f(t) \ dt = f(c) . And so  f(c) = \frac{1}{x-c} \int_{c}^{x} f(c) \ dt . The result follows.  \diamond

    Is the second part correct? In the first part, would the easiest way be to use a sequence approach? Or just the regular  \epsilon-\delta definition?
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  2. #2
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    Quote Originally Posted by manjohn12 View Post
    Theorem. Let  a \in \mathbb{R} , and let  I be an interval containing  a . Let  I \subseteq K \subseteq \mathbb{R}. Suppose that the function  f: K \to \mathbb{R} is integrable on  [a,x] for all  x \in I with  x > a and on  [x,a] for all  x \in I with  x < a . Let  F_a be the area accumulation function for  f on  I based at  a . Then  F_a is uniformly continuous on  I . If  f happens to be continuous on  I , then  F_a is differentiable on  I and  F_{a}'(x) = f(x) for all  x \in I .

    Proof. Suppose  a \in \mathbb{R} ,  I an interval containing  a , and let  f: I \to \mathbb{R} be a function. The area accumulation function  F_a: I \to \mathbb{R} is defined by  F_{a}(x) = \int_{a}^{x} f(t) \ dt . (1) We want to show that  F_a is uniformly continuous on  I . To do this, would we use a sequence approach? Because  F_a is not Lipschitz. (2) Suppose  f is continuous on  I . We want to show that  F_a is differentiable at some point  c . So we look at the following:

     \frac{F_{a}(x)-F_{a}(c)}{x-c} = \frac{\int_{a}^{x} f(t) \ dt- \int_{a}^{c} f(t) \ dt}{x-c} = \frac{1}{x-c} \int_{x}^{c} f(t) \ dt

    We want to show that  \lim\limits_{x \to c} \frac{1}{x-c} \int_{c}^{x} f(t) \ dt = f(c) . And so  f(c) = \frac{1}{x-c} \int_{c}^{x} f(c) \ dt . The result follows.  \diamond

    Is the second part correct? In the first part, would the easiest way be to use a sequence approach? Or just the regular  \epsilon-\delta definition?

    So for (1), would a sequence approach be the best?
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