Results 1 to 2 of 2

Thread: Shifted Riemann

  1. #1
    Member
    Joined
    Oct 2008
    Posts
    156

    Shifted Riemann

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


    Proof. Let $\displaystyle \epsilon >0 $. Let $\displaystyle P $ be a partition of $\displaystyle [a,b] $ with $\displaystyle ||P|| < \delta $. Let $\displaystyle x_{i}^{*} $ be such that $\displaystyle x_{i-1} \leq x_{i}^{*} \leq x_i $. We know that $\displaystyle |\mathcal{R}(f,P)-I| < \epsilon $. Now $\displaystyle \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 $\displaystyle \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 $\displaystyle k $. And we are doing this for $\displaystyle b-a $ rectangles.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by manjohn12 View Post
    Let $\displaystyle f $ be Riemann integrable on $\displaystyle [a,b] $. Let $\displaystyle k $ be any real number, and let $\displaystyle g $ be the function given by $\displaystyle g(x) = f(x) + k $. Prove that $\displaystyle g $ is Riemann integrable on $\displaystyle [a,b] $ and that $\displaystyle \smallint_{a}^{b} g = k(b-a) + \smallint_{a}^{b} f $. What is the geometric interpretation.


    Proof. Let $\displaystyle \epsilon >0 $. Let $\displaystyle P $ be a partition of $\displaystyle [a,b] $ with $\displaystyle ||P|| < \delta $. Let $\displaystyle x_{i}^{*} $ be such that $\displaystyle x_{i-1} \leq x_{i}^{*} \leq x_i $. We know that $\displaystyle |\mathcal{R}(f,P)-I| < \epsilon $. Now $\displaystyle \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 $\displaystyle \mathcal{R}(g,P) = \mathcal{R}(f,P)+ k(b-a) $?
    This is good so far. You get $\displaystyle \mathcal{R}(g,P) = \mathcal{R}(f,P) + k(b-a)$.
    Therefore, $\displaystyle |\mathcal{R}(g,P) - I - k(b-a)| = |\mathcal{R}(f,P) - I| < \epsilon$. Thus, we see that $\displaystyle \smallint_a^b g = I + k(b-a)$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Shifted Legendre Polynomial
    Posted in the Advanced Math Topics Forum
    Replies: 3
    Last Post: Oct 24th 2010, 07:32 AM
  2. Shifted fcn...
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Jun 30th 2010, 10:15 PM
  3. Obtaining the maximum of a shifted function
    Posted in the Calculus Forum
    Replies: 7
    Last Post: Jun 19th 2010, 11:42 AM
  4. Shifted exponential
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: Feb 18th 2009, 05:19 PM
  5. Riemann Sums and Riemann Integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 1st 2006, 01:08 PM

Search Tags


/mathhelpforum @mathhelpforum