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Definite Integral of the Natural Log of a Quadratic using Riemann Sums

  1. Jan 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Use the form of the definition of the integral to evaluate the following:

    lim (n [itex]\rightarrow[/itex] ∞) [itex]\sum^{n}_{i=1}[/itex] x[itex]_{i}[/itex][itex]\cdot[/itex]ln(x[itex]_{i}[/itex][itex]^{2}[/itex] + 1)Δx on the interval [2, 6]



    2. Relevant equations

    x[itex]_{i}[/itex] = 2 + [itex]\frac{4}{n}[/itex]i
    Δx = [itex]\frac{4}{n}[/itex]
    Ʃ[itex]^{n}_{i=1}[/itex]i[itex]^{2}[/itex] = [itex]\frac{n(n+1)(2n+1)}{6}[/itex]
    Ʃ[itex]^{n}_{i=1}[/itex]i = [itex]\frac{n(n+1)}{2}[/itex]
    Ʃ[itex]^{n}_{i=1}[/itex]c = cn


    3. The attempt at a solution

    So far I've gotten it down to: lim (n [itex]\rightarrow[/itex] ∞) [itex]\frac{4}{n}[/itex][itex]\cdot[/itex] [itex]\sum^{n}_{i=1}[/itex] [(2+[itex]\frac{4}{n}[/itex]i)[itex]\cdot[/itex]ln([itex]\frac{16}{n^{2}}[/itex]i[itex]^{2}[/itex] + [itex]\frac{16}{n}[/itex]i + 5)]

    but I am unsure how to distribute the sigma into the natural log, or how to get the i's out of the log... in short I can't think of a way to move forward!

    I hope I wrote this clearly enough, can anyone assist me?

    EDIT: Just to be clear, I am trying to put it in a form such that I can get rid of the sigma(Ʃ)
     
    Last edited: Jan 12, 2012
  2. jcsd
  3. Jan 13, 2012 #2
    For more clarification on what i'm trying to do, watch this video:
     
    Last edited by a moderator: Sep 25, 2014
  4. Jan 13, 2012 #3

    I like Serena

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    Well, your video is quite a bit easier than what you have here.
    But let's see what we can do...

    The logarithm is indeed in the way, so it needs to move out.
    You would need to apply the log laws to do that:
    b ln a = ln(a^b)
    ln a + ln b = ln(ab)

    This turns the summation into a product with the logarithm on the outside.
     
  5. Jan 13, 2012 #4
    I understand these laws, but am unsure how to apply them in this example.

    Lets set a = (2 + [itex]\frac{4}{n}[/itex]i)
    and lets set b = ln([itex]\frac{16}{n^{2}}[/itex]i[itex]^{2}[/itex]+[itex]\frac{16}{n}[/itex]i+5)

    So by your laws I would turn it into ln[([itex]\frac{16}{n^{2}}[/itex]i[itex]^{2}[/itex]+[itex]\frac{16}{n}[/itex]i+5)^(2 + [itex]\frac{4}{n}[/itex]i)]

    And I'm not exactly sure how that will help me get rid of the ln()

    because there are no laws for ln(a+b), just for ln(a) + ln(b)
     
  6. Jan 13, 2012 #5

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    Good.

    Remember that your ##\sum\limits_{i} \ln(a_i)## is actually ##\ln a_1 + \ln a_2 + ...##.
     
  7. Jan 13, 2012 #6
    Ok. So I take it there is some way I can rewrite it to make my life easier... but it's not clicking for me.

    Can I get another nudge?
     
  8. Jan 13, 2012 #7

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    Okay.
    So you have ##\ln[({16i^2 \over n^2}+{16i \over n} +5)^{(2 + {4i \over n})}]##.
    Lets call this ##\ln a_i##.

    So ##a_i = ({16i^2 \over n^2}+{16i \over n} +5)^{(2 + {4i \over n})}##.

    Your summation is then:
    $$\sum_{i=1}^n \ln a_i = \ln a_1 + \ln a_2 + ... + \ln a_n = \ln(a_1 \cdot a_2 \cdot ... \cdot a_n) = \ln \prod_{i=1}^n a_i$$
     
  9. Jan 13, 2012 #8
    I've never worked with that symbol before, but I take it it's the same as Sigma except for multiplication?

    Ok, and if so, the whole problem has been narrowed down to that final term you have at the end... but I am still lost as to how to get an actual number from that.

    I may be trying to delve into these things too soon... this problem was not assigned yet, but it's in the book and gosh darnit with all the time i've spent on it so far I need to understand it!

    So from here... I need some way to get rid of that new symbol don't I? So I can get an actual number?
     
  10. Jan 13, 2012 #9

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    Yep. You've got it right.

    The challenge is to get the term at the end in an appropriate form and determine its limit if n goes to infinity.
    This is not easy and is quite a bit of work still.
    TBH I haven't worked through it myself yet.
     
  11. Jan 13, 2012 #10

    Dick

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    I don't think they want you to evaluate the integral by finding the limit of the Riemann sum. I think they want you to find the limit of the Riemann sum by evaluating the integral. Just do the integral.
     
    Last edited: Jan 13, 2012
  12. Jan 14, 2012 #11
    We are not allowed to use the Fundamental Theorem of Calculus for this problem, we have to solve it by definition, which in this case is finding the limit of the Riemann Sum. Sorry.
     
  13. Jan 14, 2012 #12

    Dick

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    I am sorry too. I really can't see any point to doing a problem like that by explicit sums, and have no idea how. But I wish you luck!
     
  14. Jan 14, 2012 #13

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    The method that I see, is to expand the product into Taylor series.
    This will lead to a point where the higher order terms will approach zero when n approaches infinity.
    Quite a bit of work though and I do not know if the OP is familiar with Taylor expansions yet.

    Actually, I was wondering if there isn't an easier method, but I fail to see that as yet.


    As to the point of the problem, isn't it just a mathematical exercise?
     
  15. Jan 14, 2012 #14

    Dick

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    Now you've got me wondering again. The first line of the problem statement is "Use the form of the definition of the integral to evaluate the following". Expanding the series terms in a power series and trying to sum them doesn't 'use the definition of the integral'. And trying to expand is going to lead to higher and higher powers i^n, and you'll have to sum them all to get an exact answer. I really think somebody is misinterpreting the intent of this question. That's my opinion.
     
  16. Jan 14, 2012 #15

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    The powers are typically of the form ##c^{i \over n^2}## and the terms also contain typically factors like ##{i \over n^2}##.

    (Note that there is a factor ##4 \over n##, which is as yet on the outside of the complete expression, that is, outside the logarithm.)


    As to the intent of the problem, I'm just following the questions of the OP.
    Seeing the complexity of the solution, I believe you are right.
    The intent of the problem will be to go the other way and use the definition of the integral to find result of the summation by integration.

    Either way, it is good practice.
     
  17. Jan 14, 2012 #16
    Guys... I just figured out, Dick is right... I did misinterpret the question. Thanks to my dyslexia I was reading the instructions from the section below mine to answer the questions above it...

    I offer my sincerest of apologies, but at the same time I am still kind of interested to see where this goes. If you guys want to keep talking about it, I will stay involved and try to help out, but if you guys are done with it, then I mark this question resolved.
     
  18. Jan 14, 2012 #17

    Dick

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    No need to apologize! I don't think it's worth chasing this one any farther. If you HAD to you could expand it using a series like log(1+x)=x-x^2/2+x^3/3+... and sum a few low powered terms, but that would only give you an approximation. It's hardly worth it since the integral gives you the exact answer.
     
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