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Riemann sum

  1. Apr 24, 2007 #1
    1. The problem statement, all variables and given/known data
    The following sum
    [tex]\sqrt{9 - \left(\frac{3}{n}\right)^2} \cdot \frac{3}{n} + \sqrt{9 - \left(\frac{6}{n}\right)^2} \cdot \frac{3}{n} + \ldots + \sqrt{9 - \left(\frac{3 n}{n}\right)^2} \cdot \frac{3}{n}[/tex]
    is a right Riemann sum for the definite integral. Solve as n->infinity
    [tex]\int_0^b f(x)\, dx[/tex]

    2. Relevant equations
    [tex]\int_0^b f(x)\, dx[/tex]


    3. The attempt at a solution
    I can't seem to get this one. My work is a bit long to show but I get
    (9/n^3) *Sigma(i=1,n) [sqrt(n^2+i^2)]
    not sure what to do here, do i substitute Sigma(i=1,n)(i^2=(n(n+1))/2?c
     
    Last edited: Apr 24, 2007
  2. jcsd
  3. Apr 24, 2007 #2

    HallsofIvy

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    A don't see a question here! What is it you are trying to "get"?

    You certainly cannot "substitute Sigma(i=1,n)(i^2=(n(n+1))/2" because you are not summing i2!
     
  4. Apr 24, 2007 #3
    my mistake, I am suppose to solve the sum as n-> infinity
     
  5. Apr 24, 2007 #4
    help
    anyone?
     
  6. Apr 24, 2007 #5
    :cry: :cry: :cry:
     
  7. Apr 24, 2007 #6

    quasar987

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    Hi, there is a result I'm sure you're aware of that says that if a function f is continuous on [a,b], then

    [tex]\lim_{n\rightarrow\infty}\frac{b-a}{n}\sum_{i=1}^{n}f \left( a+i\frac{b-a}{n}\right) =\int_a^bf(x)dx[/tex]

    Well this is what you gotta use. You just have to find the proper f(x) and what a and b are.
     
    Last edited: Apr 24, 2007
  8. Apr 24, 2007 #7

    quasar987

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    Maybe start by rewriting the sum as

    [tex]\frac{3}{n}\sum_{i=1}^{n}\sqrt{9-\left(\frac{3i}{n}\right)^2}[/tex]

    just to fit the mold of the formula above.
     
  9. Apr 24, 2007 #8
    ok so I've had that originally but I some how got to get i by itself(to the right of the sigma) and sub in
    [tex]\frac{n(n+1)}{2}[/tex]
    that will get rid or the sigma, then I can rearrange and solve for the limit. But how to do that?
     
  10. Apr 24, 2007 #9

    quasar987

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    I didn't understand a word you said.

    Did you read my post? There are no limits to solve. You only have to find f, a and b such that the formula of post #7 becomes that of post #8 (modulo the limit symbol).
     
  11. Apr 24, 2007 #10
    ok so [tex]f(x)=\sqrt(9-x^2)[/tex] a=0 and b=3
    giving:
    [tex]\int_0^3 \sqrt(9-x^2) dx=\lim_{n\rightarrow\infty} \frac{3}{n}\sum_{i=1}^{n}\sqrt{9-\left(\frac{3i}{n}\right)^2}[/tex]
    Is that what your sayin is the answer?
    My homework problem says" The limit of these Riemann sums as [tex]n \to \infty[/tex] is...(blank).
    And I am suppose to type something in.
     
    Last edited: Apr 24, 2007
  12. Apr 24, 2007 #11

    quasar987

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    Yes, bravo, you've got it.

    Now that you know that this weird looking sum is actually equal to an integral, you can use the fundamental theorem of calculus to evaluate the integral and get yourself a nice numerical value for the sum.
     
  13. Apr 25, 2007 #12
    Thanks quasar.
    If I could ask you a hint on one more problem I would be ever more greatful.
    Suppose f(x) is continuous and decreasing on the closed interval [tex]3\le x\le 11, that f(3)=7, f(11)=4 and that \displaystyle \int_{3}^{11}f(x)\,dx=46.918141.
    Then \displaystyle \int_{4}^{7}f^{-1}(x)\,dx[/tex]=?
     
  14. Apr 25, 2007 #13

    quasar987

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    This is easily done graffically knowing that the integral is the area under the curve. All you have to do is convince yourself that the integral of f^-1 is the area btw the curve and the y axis.
     
  15. Apr 25, 2007 #14
    ok, but we are not really given the function, we have the area on the interval 3,11. not sure where to go there
     
  16. Apr 25, 2007 #15

    quasar987

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    Like I said it's easy once you graphically represent the situation. You'll be able to extract the area btw the curve and the y axis by substracting a small 3x4 square and 46.918141 to a big 7x11 square.
     
  17. Apr 25, 2007 #16
    Haha..thats not hard at all, I guess I was expecting something different with the
    f(x)-1 but I see now. thanks Quasar.
     
  18. Jul 10, 2007 #17
    ew, gross :P Riemann sum, we learned a bit about it in precalculus this year. well i have a lot to look forward to next year *eyeroll* i'm sure it won't be that bad when i learn it
     
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