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Thread: Integral Theorem

  1. #1
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    Integral Theorem

    Theorem. Let $\displaystyle f $ and $\displaystyle g $ be real valued functions that are Riemann integrable on $\displaystyle [a,b] $. Then the following statements hold:

    1. The function $\displaystyle f+g $ is Riemann integrable on $\displaystyle [a,b] $ and $\displaystyle \int_{a}^{b} (f+g) = \int_{a}^{b} f + \int_{a}^{b} g $.


    2. The function $\displaystyle kf $ is Riemann integrable on the interval $\displaystyle [a,b] $ and $\displaystyle \int_{a}^{b} kf = k \int_{a}^{b} f $.


    3. The function $\displaystyle f-g $ is Riemann integrable on $\displaystyle [a,b] $ and $\displaystyle \int_{a}^{b} (f-g) = \int_{a}^{b} f - \int_{a}^{b} g $.



    Outline of Proof. The basic idea is that we basically look at $\displaystyle \mathcal{R}(f+g,P), \ \mathcal{R}(kf, P), $ and $\displaystyle \mathcal{R}(f-g, P) $ and show that they are in fact equal to those right hand sides?
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  2. #2
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    Quote Originally Posted by manjohn12 View Post
    1. The function $\displaystyle f+g $ is Riemann integrable on $\displaystyle [a,b] $ and $\displaystyle \int_{a}^{b} (f+g) = \int_{a}^{b} f + \int_{a}^{b} g $.
    Let $\displaystyle I_1 = \smallint_a^b f \text{ and }I_2 = \smallint_a^b g$. For $\displaystyle \epsilon > 0$ there exists $\displaystyle \delta > 0$ so that $\displaystyle | \mathcal{R}(f,P) - I_1| < \frac{\epsilon}{2}$ and $\displaystyle |\mathcal{R}(g,P) - I_2| < \frac{\epsilon}{2}$ as long as $\displaystyle ||P|| < \delta$.

    Now, $\displaystyle \mathcal{R}(f+g,P) = \mathcal{R}(f,P) + \mathcal{R}(g,P)$.
    Thus, $\displaystyle |\mathcal{R}(f+g,P) - I_1 - I_2| \leq |\mathcal{R}(f,P) - I_1| + | \mathcal{R}(f,P) - I_2| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$.

    EDIT: Fixed
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Let $\displaystyle I_1 = \smallint_a^b f \text{ and }I_2 = \smallint_a^b g$. For $\displaystyle \epsilon > 0$ there exists $\displaystyle \delta > 0$ so that $\displaystyle | \mathcal{R}(f,P) - I_1| < \frac{\epsilon}{2}$ and $\displaystyle |\mathcal{R}(g,P) - I_2| < \frac{\epsilon}{2}$ as long as $\displaystyle ||P|| < \delta$.

    Now, $\displaystyle \mathcal{R}(f+g,P) = \mathcal{R}(f,P) + \mathcal{R}(f,P)$.
    Thus, $\displaystyle |\mathcal{R}(f+g,P) - I_1 - I_2| \leq |\mathcal{R}(f,P) - I_1| + | \mathcal{R}(f,P) - I_2| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$.
    $\displaystyle \mathcal{R}(f+g,P) = \mathcal{R}(f,P) + \mathcal{R}(g,P)$.

    Thus, $\displaystyle |\mathcal{R}(f+g,P) - I_1 - I_2| \leq |\mathcal{R}(f,P) - I_1| + | \mathcal{R}(g,P) - I_2| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$.
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