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Integration and Computing Area

  1. Jul 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Evaluate the sum:
    [tex] \sum_{k=100}^{250} k^3 [/tex]

    2. Relevant equations
    [tex] \sum_{j=1}^{N} j^3 = \frac{n^2(n+1)^2}{4} [/tex]

    3. The attempt at a solution
    Little confused as how to integrate from 100 to 250. I have no examples to go off of and what I try is incorrect. But heres what I tried:
    [tex] \sum_{k=1}^{250} k^3 - \sum_{k=1}^{100} k^3 = \sum_{k=100}^{250} k^3 [/tex]

    984390625-25502500=958888135 (incorrect)

    Any suggestions? I'm sure this is relatively easy but I just haven't seen it before and thus don't know how to go about solving it. Thanks in advance!
  2. jcsd
  3. Jul 16, 2008 #2
    You know in sigma notation, [tex] \sum_{k= m}^{n} f(k) [/tex]

    n is the ending value of k & m is the starting value of k. So;

    Why can't you subtract the solutions of the definite integrals, [tex]\int^{250}_{100} x^3 dx[/tex] and [tex]\int^{100}_{1} x^3 dx[/tex] ?
    Last edited: Jul 16, 2008
  4. Jul 16, 2008 #3
    Beeorz, you're method will work out if you correct the bounds.
    \sum_{k=1}^{100} k^3 = k_1^3 + k_2^3 + ... + k_99^3 + k_100^3

    If you subtract that from [tex] \sum_{k=1}^{250} k^3 [/tex] , then you are left with

    \sum_{k=101}^{250} k^3

    You should subtract [tex] \sum_{k=1}^{99} k^3 [/tex] from the sum from 1 to 250 in order to get the answer.
  5. Jul 17, 2008 #4


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    What in the world does this have to do with either "integration" or "Computing area"?
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