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Math Help - Integral Order

  1. #1
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    Integral Order

    Theorem. Let  a,b,M and  m be real numbers with  a<b and  m \leq M . Let  f and  g be real-valued functions that are Riemann integrable on  [a,b] . Then the following hold:


    1. If  f(x) \geq 0 for all  x \in [a,b] then  \int_{a}^{b} f \geq 0 .



    2. If  f(x) \leq g(x) for all  x \in [a,b] then  \int_{a}^{b} f \leq \int_{a}^{b} g .



    3. If  m \leq f(x) \leq M for all  x \in [a,b] , then  m(b-a) \leq \int_{a}^{b} f \leq M(b-a) .



    Outline of Proof. Basically we consider the values  f and  g can take, and look at  \mathcal{R}(f,P) and  \mathcal{R}(g,P) ? For instance, in #2 we look at  g-f (which is Riemann integrable) and show that  \int_{a}^{b} g-f \geq 0 ?
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  2. #2
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    Quote Originally Posted by manjohn12 View Post
    Theorem. Let  a,b,M and  m be real numbers with  a<b and  m \leq M . Let  f and  g be real-valued functions that are Riemann integrable on  [a,b] . Then the following hold:


    1. If  f(x) \geq 0 for all  x \in [a,b] then  \int_{a}^{b} f \geq 0 .



    2. If  f(x) \leq g(x) for all  x \in [a,b] then  \int_{a}^{b} f \leq \int_{a}^{b} g .



    3. If  m \leq f(x) \leq M for all  x \in [a,b] , then  m(b-a) \leq \int_{a}^{b} f \leq M(b-a) .
    If you prove #1 then #2 follows immediately by considering f-g which in consequence #3 follows immediately.

    For #1 assume that I < 0 we know that |\mathcal{R}(f,P) - I|  < \epsilon for ||P||<\delta.
    But, |\mathcal{R}(f,P) - I| = \mathcal{R}(f,P) - I since I < 0.
    Thus, |\mathcal{R}(f,P) - 0| = \mathcal{R}(f,P)  < \mathcal{R}(f,P) - I < \epsilon.

    We have a contradition because we have shown the integral of f is 0 and I but the integral is well-defined and so this forces I = 0 which is where we get a contradiction.
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