Results 1 to 3 of 3

Math Help - Integral Theorem

  1. #1
    Member
    Joined
    Oct 2008
    Posts
    156

    Integral Theorem

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

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


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


    3. The function  f-g is Riemann integrable on  [a,b] and  \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  \mathcal{R}(f+g,P), \ \mathcal{R}(kf, P), and  \mathcal{R}(f-g, P) and show that they are in fact equal to those right hand sides?
    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
    1. The function  f+g is Riemann integrable on  [a,b] and  \int_{a}^{b} (f+g) = \int_{a}^{b} f + \int_{a}^{b} g .
    Let I_1 = \smallint_a^b f \text{ and }I_2 = \smallint_a^b g. For \epsilon > 0 there exists \delta > 0 so that | \mathcal{R}(f,P) - I_1| < \frac{\epsilon}{2} and |\mathcal{R}(g,P) - I_2| < \frac{\epsilon}{2} as long as ||P|| < \delta.

    Now, \mathcal{R}(f+g,P) = \mathcal{R}(f,P) + \mathcal{R}(g,P).
    Thus, |\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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2008
    Posts
    156
    Quote Originally Posted by ThePerfectHacker View Post
    Let I_1 = \smallint_a^b f \text{ and }I_2 = \smallint_a^b g. For \epsilon > 0 there exists \delta > 0 so that | \mathcal{R}(f,P) - I_1| < \frac{\epsilon}{2} and |\mathcal{R}(g,P) - I_2| < \frac{\epsilon}{2} as long as ||P|| < \delta.

    Now, \mathcal{R}(f+g,P) = \mathcal{R}(f,P) + \mathcal{R}(f,P).
    Thus, |\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.
    \mathcal{R}(f+g,P) = \mathcal{R}(f,P) + \mathcal{R}(g,P).

    Thus, |\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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Volume integral to a surface integral ... divergence theorem
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: October 7th 2010, 07:11 PM
  2. Replies: 2
    Last Post: August 31st 2010, 08:38 AM
  3. Integral using Stokes Theorem
    Posted in the Calculus Forum
    Replies: 0
    Last Post: July 18th 2010, 03:13 PM
  4. Cauchy's Integral theorem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 18th 2010, 07:01 AM
  5. Integral Mean Value Theorem
    Posted in the Calculus Forum
    Replies: 5
    Last Post: February 15th 2009, 06:56 PM

Search Tags


/mathhelpforum @mathhelpforum