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Riemann sum

  1. Apr 11, 2008 #1
    [SOLVED] Riemann sum

    Important stuff:

    [tex]\sum i^2 = \frac{n(n+1)(2n+1)}{6}[/tex]

    [tex]\sum i = \frac{n(n+1)}{2}[/tex]

    And the solution: (Where I write "lim" I mean limit as n-->infinity. Where I write the summation sign I mean from i=1 to n.)

    [tex]lim \sum t^2 + 6t - 4 \Delta t[/tex]

    [tex]\Delta t = \frac{5 - (-2)}{n}[/tex]

    [tex]t = i \Delta t = \frac{7i}{n}[/tex]

    etc, etc.

    [tex]= \frac{343}{3} + 147 + 28[/tex]

    Which is about 233.3...

    When I did the same thing by just doing the integral, I got -54.66....
     
  2. jcsd
  3. Apr 11, 2008 #2

    HallsofIvy

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    The difficulty is that you haven't shown HOW you got either of those values. Your [itex]\Delta x[/itex] should be - since you are integrating from 5 down to -2. You have also neglected the "-" on "- 4" in the integrand.
     
  4. Apr 11, 2008 #3
    Sorry, I messed up the thing I meant for it to be from -2 to 5 but I put the numbers in the wrong places when I was typing it up :P

    Ok :) This is the "etc, etc." Where did I miss a "-" on the -4 ?

    [tex]lim \sum t^2 + 6t - 4 \Delta t[/tex]

    [tex]lim \sum ( ( \frac{7i}{n})^2 + 6 \frac{7i}{n} - 4 ) \frac{7}{n}[/tex]

    [tex]lim \sum ( \frac{7i}{n})^2 * \frac{7}{n} + 6 \frac{7i}{n} * \frac{7}{n} - 4 * \frac{7}{n}[/tex]

    [tex]lim \sum \frac{7^2}{n^2} * i^2 * \frac{7}{n} + 6 \frac{7}{n} * i * \frac{7}{n} - 4 * \frac{7}{n}[/tex]

    [tex]lim \frac{7^2}{n^2} * \frac{n(n+1)(2n+1)}{6} * \frac{7}{n} + 6 \frac{7}{n} * \frac{n(n+1)}{2} * \frac{7}{n} - 4n * \frac{7}{n}[/tex]

    [tex]lim \frac{7^3}{6} * \frac{2n^2 + 3n + 1}{n^2} + 21 * 7 * \frac{n+1}{n} - 4 * 7[/tex]

    [tex]\frac{7^3}{3} + 21 * 7 - 4 * 7[/tex]

    If you can follow the algebra there :)

    ------

    Sorry all for not realizing this part of the forum wasn't for homework. If anyone has the ability to move it to the right part of the forum, it would be shiny.
     
    Last edited: Apr 11, 2008
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