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Math Help - Shifted Riemann

  1. #1
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    Shifted Riemann

    Let  f be Riemann integrable on  [a,b] . Let  k be any real number, and let  g be the function given by  g(x) = f(x) + k . Prove that  g is Riemann integrable on  [a,b] and that  \smallint_{a}^{b} g = k(b-a) + \smallint_{a}^{b} f . What is the geometric interpretation.


    Proof. Let  \epsilon >0 . Let  P be a partition of  [a,b] with  ||P|| < \delta . Let  x_{i}^{*} be such that  x_{i-1} \leq x_{i}^{*} \leq x_i . We know that  |\mathcal{R}(f,P)-I| < \epsilon . Now  \mathcal{R}(g,P) = \sum_{i=1}^{n} [f(x_{i}^{*})+k](x_{i}-x_{i-1}) . So we just expand this out and see that  \mathcal{R}(g,P) = \mathcal{R}(f,P)+ k(b-a) ?

    And the geometric interpretation is that we are increasing the each of the rectangle's height by  k . And we are doing this for  b-a rectangles.
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  2. #2
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    Quote Originally Posted by manjohn12 View Post
    Let  f be Riemann integrable on  [a,b] . Let  k be any real number, and let  g be the function given by  g(x) = f(x) + k . Prove that  g is Riemann integrable on  [a,b] and that  \smallint_{a}^{b} g = k(b-a) + \smallint_{a}^{b} f . What is the geometric interpretation.


    Proof. Let  \epsilon >0 . Let  P be a partition of  [a,b] with  ||P|| < \delta . Let  x_{i}^{*} be such that  x_{i-1} \leq x_{i}^{*} \leq x_i . We know that  |\mathcal{R}(f,P)-I| < \epsilon . Now  \mathcal{R}(g,P) = \sum_{i=1}^{n} [f(x_{i}^{*})+k](x_{i}-x_{i-1}) . So we just expand this out and see that  \mathcal{R}(g,P) = \mathcal{R}(f,P)+ k(b-a) ?
    This is good so far. You get \mathcal{R}(g,P) = \mathcal{R}(f,P) + k(b-a).
    Therefore, |\mathcal{R}(g,P) - I - k(b-a)| = |\mathcal{R}(f,P) - I| < \epsilon. Thus, we see that \smallint_a^b g =  I + k(b-a).
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