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Riemann sum limit

  1. Jul 9, 2011 #1
    1. The problem statement, all variables and given/known data
    a.) Use definition 2 to find an expression for the area under the curve y=x^3 from 0 to 1 as a limit.
    b.)Evaluate the (above) limit using the sum of the cubes of the n integers.

    2. Relevant equations
    [tex](\frac{n(n+1)}{2})^{2}[/tex]

    3. The attempt at a solution
    For part a.) I wrote my limit like this:
    [tex]\lim_{n \to \infty} \Sigma_{i=1}^{n}(\frac{i}{n})^{3}\frac{1}{n}[/tex]


    The "Definition 2" they have listed just says:
    [tex]A=\lim_{n \to \infty} R_{n} = \lim_{n \to \infty}(f(x_{1})\Delta x + f(x_{2})\Delta x + . . . + f(x_{n})\Delta x)[/tex]

    Now for part b, I understand that the formula is the sum of all cubes and so on. So I am thinking that the limit should look like this?

    [tex]\lim_{n \to \infty}(\frac{n(n+1)}{2})^{2}[/tex]
    That should handle the limit and the sum of cubes, now I need each one to multiply by delta x right? So that it comes out to:
    [tex]\lim_{n \to \infty}(\frac{n(n+1)}{2})^{2}\frac{1}{n}[/tex]
    ..because the integral is from 0 to 1, so [itex]\Delta x = \frac{1-0}{n}[/itex]

    But I am not sure how to write it in this manner and take the limit from here yet. Is this correct so far? I should just simplify the expression after the limit and then take the limit?
     
    Last edited: Jul 9, 2011
  2. jcsd
  3. Jul 9, 2011 #2

    Dick

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    No, no. The sum of i^3 for i=0 to n is (n(n+1)/2))^2. That's only a PART of the sum of (i/n)^3*(1/n) for i=0 to n. Do some algebra in that sum to separate out the i^3 part and then substitute your summation formula and take the limit.
     
  4. Jul 10, 2011 #3
    I understand now. Sorry, this is the first time I have ever actually used a summation, other than just knowing what it is.
    73p1z5.jpg

    I checked with the definite integral and this is correct as far as I can tell.
    I'm going to go learn the rules of summation and find some problems to do. Thanks, I didn't realize that this was how it worked.
     
  5. Jul 10, 2011 #4

    Dick

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    Yes, the integral from x=0 to x=1 is 1/4. And I think you did it correctly through the limit.
     
  6. Jul 11, 2011 #5
    Thanks for the help!
     
  7. Mar 15, 2012 #6
    Hey can someone explain to me how did he jumped from lim n-> Ʃ i^3/n * 1/n to lim n->1/n^4 Ʃ i^3??
     
  8. Mar 15, 2012 #7

    Dick

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    He didn't jump anywhere. He used a formula the sum of i^3 for i=1 to n.
     
  9. Mar 15, 2012 #8
    I see... but he committed a mistake in the evaluation at lim n-> Ʃ i^3/n * 1/n is not lim n-> Ʃ i^3/n * 1/n is lim n-> Ʃ i^3/n^3 * 1/n
     
  10. Mar 15, 2012 #9
    I was having a hard time finding how did he got that 1/n^4.... LOL it was an ibvious mistake from part (a) even part (a) is wrong.
     
  11. Mar 15, 2012 #10
    deltax= 1-0/n=1/n than according to definition 2 the R endpoints formula is (a+ideltax)deltax so is i/n*1/n than substituting from x^3 we have (i/n)^3*1/n=i^3/n^3*1/n=i^3/n^4 which makes more sense. Now i just get rid of the 1/n^4 and put it in the other side of the summation and ta chan! problem solved LOL
     
  12. Mar 15, 2012 #11

    Dick

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    It's pretty clear he meant (i/n)^3 not i^3/n since the n^4 appears in the next line. Why are you resurrecting posts over to a year old to complain about notational mistakes?
     
  13. Mar 15, 2012 #12
    is not clear in part (a).
     
  14. Mar 15, 2012 #13

    Dick

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    I'll admit the way it was written isn't clear. Using an extra parentheses would have helped.
     
  15. Mar 15, 2012 #14
    Thank you.
     
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