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Simple Riemann Sums question with answers not lining up

  1. Oct 11, 2004 #1
    Please tell me if I am doing the summation of rectangular areas wrongly.

    Using summation of rectangles, find the area enclosed between the curve y = 3x^2 and the x-axis from x = 1 to x = 4.

    Now, before I answer the way it asks, I want to use antidifferentiation first to see what I should expect.
    A'(x) = F'(x) = 3x^2
    Therefore F(x) = x^3

    Area = F(4) - F(1)
    = 64 - 1
    = 63 square units, correct?

    Now, using the summation of rectangles I do this:

    I start by finding the width of the rectangles: (4-1)/n = 3/n
    [tex]\sum_{k=1}^n\ f(x) * (3/n)[/tex]
    [tex]= \sum_{k=1}^n\ [3(k * 3/n)^2] * (3/n)[/tex]
    [tex]= \sum_{k=1}^n\ [3 * k^2 * 9/n^2] * (3/n)[/tex]
    [tex]= 81/n^3\sum_{k=1}^n\ k^2[/tex]
    [tex]= 81/n^3 * n(n+1)(2n+1)/6[/tex]
    [tex]= 81/6 * n(n+1)(2n+1)/n^3[/tex]
    [tex]= 27/2 * (n+1)(2n+1)/n^2[/tex]

    Then I do the limit and find 27/2 * (n+1)/n * (2n+1)/n
    = 27/2 * (1 + 1/n)(2 + 1/n)
    As n approaches infinity, 1/n = 0

    27/2 * (1 + 0)(2 + 0)
    Thus I end up with 27 square units.

    Two very different answers. I'm not sure what I'm doing wrong with the sums, as far as my book is concerned I'm following all "legal" rules. I've read and re-read this a few times and I don't notice any glaring mistake (on my sore eyes).
     
  2. jcsd
  3. Oct 11, 2004 #2

    HallsofIvy

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    You have [tex]= \sum_{k=1}^n\ [3(k * 3/n)^2] * (3/n)[/tex]

    so, I take it, x= (3k)/n. what is that when k= 1? Your integral is supposed to start at x= 1 isn't it?

    What is x when k= n? Wasn't your integral supposed to end at x= 4?
     
  4. Oct 11, 2004 #3
    yes, x = 3k/n (the iteration of rectangle times the width).

    when k = 1, x = 3/n :bugeye:

    When k = n... x = 3 :uhh:

    Back to the drawing board :tongue2: (sorry for something so stupid!)

    And thanks for making me think :biggrin:

    I'm thinking it should be x = 1 + k * 3/n
    I'll punch it in and see how that works out. It would seem to satisfy both k = 1 and when k = n.
    (Off by one...shudder...reminds me of my C programming :biggrin:)
     
    Last edited: Oct 11, 2004
  5. Oct 11, 2004 #4
    Thanks again, HallsofIvy

    I gave it a work through and it turned out 63 square units like the other method does.

    I would list my results for others to see, only it takes me forever to edit tex by hand :(
     
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