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Thread: Integral Order

  1. #1
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    Integral Order

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


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



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



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



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


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



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



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

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

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