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Math Help - Riemann intergation

  1. #1
    Newbie pc31's Avatar
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    Riemann intergation

    I have read my book's chapter on this topic but somehow I still don't really get it. Can someone illustrate for me how it is done?
    Example question:
    Show whether or not the following function is Riemann integrable:
    f: [a,b] -> R
    f(x) = x + 1
    and find the Riemann integral value of f (if applicable)

    Thank you!
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  2. #2
    Eater of Worlds
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    Here are three very important facts about integrability:

    [1]: If f is continuous on [a,b], then f is integrable on [a,b]

    [2]: If f is bounded on [a,b] and has finitely many points of discontinuity on [a,b], then f is integrable on [a,b]

    [3]: If f is not bounded on [a,b], then f is not integrable on [a,b].

    Is your f continuous?. x+1 is just a line with y intercept at y=1 and slope 1.

    There are no points of discontinuity on the line.

    If f is defined on a closed interval [a,b], then it is Riemann integrable on [a,b] if the limit exists:

    \lim_{max \;\ {\Delta}x_{k}\to 0}\sum_{k=1}^{n}f(x_{k}){\Delta}x_{k}

    If it is integrable over [a,b] then we write:

    \int_{a}^{b}f(x)dx=\lim_{max \;\ {\Delta}x_{k}\to 0}\sum_{k=1}^{n}f(x_{k}){\Delta}x_{k}

    Can you integrate f?. You have:

    \int_{a}^{b}(x+1)dx=b-a+\frac{b^{2}}{2}-\frac{a^{2}}{2}=\frac{(b-a)(a+b+2)}{2}
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  3. #3
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    Quote Originally Posted by pc31 View Post
    I have read my book's chapter on this topic but somehow I still don't really get it. Example question:
    Show whether or not the following function is Riemann integrable:
    f: [a,b] -> R
    f(x) = x + 1
    and find the Riemann integral value of f (if applicable)
    There aremany variations on what textbooks use as a definition of Riemann Integral. Here is another approach.
    If your text uses the idea that an upper and a lower sum over the same partition (subdivision) can be made arbitrarily small, then the given problem is easy. Note that the function is increasing on [a,b]. Therefore for standard subdivision with n cells of [a,b] we have \left| {U(f,P) - L(f,P)} \right| = \left| {\left( {f(b) - f(a)} \right)\left( {\frac{{b - a}}{n}} \right)} \right| = \left| {\left( {b - a} \right)\left( {\frac{{b - a}}{n}} \right)} \right| = \frac{{\left( {b - a} \right)^2 }}{n}.

    For \left[ {\forall \varepsilon  > 0} \right]\left[ {\exists N:n \ge N} \right]\left( {\frac{{\left( {b - a} \right)^2 }}{n} < \varepsilon } \right).

    I hope this gives you some ideas.
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