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Area under the curve and its submission.

  1. Feb 28, 2009 #1
    Within specified limits, the area under the curve of an equation and the submission of all values on that curve will be different right?...just confirming.
  2. jcsd
  3. Mar 1, 2009 #2
    What do you mean when you say "submission of all values"?
  4. Mar 1, 2009 #3
    Submission of all values of f(x) at infinitely small intervals.
  5. Mar 1, 2009 #4
  6. Mar 1, 2009 #5
    Nope that doesn't form any analogy, or at least explicitly state the ambiguity.

    What I mean to say here is that there are 2 things that can be computed as a submission of 'something' in the function.

    This 'something' might be infinitely small area components determined by dx, and it might be the submission of all the values of f(x) made at infinity small intervals.

    I've read many relating this submission of area to the submission of f(x), but what I think, this is not possible.
    So I wanna confirm this.
  7. Mar 1, 2009 #6

    Gib Z

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    You know, I think you mean summation ! For the original question - as there are an infinite number of points of the curve, the summation of their values will diverge.
  8. Mar 2, 2009 #7
    Re: Area under the curve and its summation.

    Sorry about the spelling.

    The summation will different from the area you mean to say right?
  9. Mar 2, 2009 #8


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    What do you mean by the sum of an infinite collection of numbers?
  10. Mar 3, 2009 #9
    Line integral?:tongue2:

    Its not an infinite collection of numbers...its a...sum of numbers in some sorta progression.
  11. Mar 3, 2009 #10


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    Then what are you talking about? What "summation of all values on a curve" are you associating with an integral?
  12. Mar 3, 2009 #11
    Yes of course but normal integral...not line.

    I mean...sum of all f(x) within a defined limit, for instance between a to b.
  13. Mar 3, 2009 #12
    No one here knows what you mean by the "sum of all f(x)". I'm guessing that you are thinking along the lines of the riemann sum, but it is not clear. You are going to have to explain exactly what you mean by "summing all values" for us to be able to help you.

    Give a concrete example of what you mean for a specific function. Say, integrating f(x)=x from 0 to 1.
  14. Mar 4, 2009 #13
    aaa...thanks for notifying me about that.

    Take a function y = x (simplest possible)

    Suppose I want summation of f(x) (or y) from limits 5 - 11 on x.

    So this will be 5+5.1+5.2+5.3.......................10.5+10.6+10.7+10.8+10.9+11

    This wont give an accurate result...the intervals at which the addition should occur should be infinity small to give 100% accuracy.

    So what I mean to say is when we integrate a function, this summation is not the result right? The summation will be the result in cause of line integral right?
  15. Mar 4, 2009 #14
    So its not a Riemann sum.
  16. Mar 4, 2009 #15

    That summation is not the result for anything. Not regular integrals or line integrals or anything. If the spacing gets smaller and smaller, the sum will just get bigger and bigger unbounded, so in the limit as the spacing becomes small, the sum becomes infinity!

    However, the sum is very very close to the reimann integral. In fact, all you have to do to make your sum into a riemann sum is to divide by the spacing. eg:
    (5+5.1+5.2+5.3.......................10.5+10.6+10.7 +10.8+10.9+11)/(0.1)

    In this way, even as the sum becomes infinitely big, the denominator becomes infinitely small in just the right proportions such that the riemann sums converge to the integral.
  17. Mar 4, 2009 #16


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    No, it wouldn't. That is the sum of x-values, not f(x) and it is done with a "step" of 0.1. You had said nothing about a particular step interval before.

    The Riemann sum, of f(x) with a step of 0.1, between 5 and 11 would be (f(5)+ f(5.1)+ ...+ f(10.9))(0.1). Is that what you are asking about?
  18. Mar 4, 2009 #17
    in my above post there is an error. It should be multiplying by the spacing as Halls says ,not dividing by it, of course.
  19. Mar 5, 2009 #18
    Ok I got the answer...thanks a lot everyone! :approve:
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