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Math Help - estimate integral

  1. #1
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    estimate integral

    Estimate

    ln 1.5 = (integral from 1 to 1.5) dt/t


    using the approximation (1/2)[Lf(P) + Uf(P)] with
    P = {1 = (4/4) , (5/4), (6/4) , (7/4) , (8/4), (9/4), (10/4) = (5/2)} .

    sorry everyone! the book is online so i had copied and pasted this question but for some reason the numbers changed and i didnt notice. im sorry for the confusion. thank you all for letting me know.
    [note- the f's in Lf and Uf are like L(sub)f and U(sub)f]
    Last edited by runner07; August 25th 2007 at 06:16 PM.
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  2. #2
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    Quote Originally Posted by runner07 View Post
    P = {1 = 8 , 9 , 10 , 11 , 12 = 1. 5}.
    1) I cannot imagine what that might mean.
    2) If you invented this notation, please don't invent new notation. There is already way too much.
    3) Can you provide the EXACT wording of the question and some indication of your efforts? If your textbook's author invented the notation, the explanation has to be in there, somewhere.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by runner07 View Post
    Estimate

    ln 1.5 = (integral from 1 to 1.5) dt/t

    using the approximation (1/2)[Lf(P) + Uf(P)] with
    P = {1 = 8 , 9 , 10 , 11 , 12 = 1. 5}.
    I believe he is describing a Darboux integral

    P describes the partitions and L(f,P) is the lower Darboux sum and U(f,P) is the upper Darboux sum
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  4. #4
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    That may be, but I'm still wondering how 1 = 8 and 12 = 1.5!

    Perhaps this means the number of subintervals?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by TKHunny View Post
    That may be, but I'm still wondering how 1 = 8 and 12 = 1.5!

    Perhaps this means the number of subintervals?
    yeah, he lost me there too
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  6. #6
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    Quote Originally Posted by runner07 View Post
    Estimate

    ln 1.5 = (integral from 1 to 1.5) dt/t

    using the approximation (1/2)[Lf(P) + Uf(P)] with
    P = {1 = 8 , 9 , 10 , 11 , 12 = 1. 5}.
    Your partition makes no sense.

    Say the partition is P=\{1,1.1,1.2,1.3,1.4,1.5\}.

    Then, U(f,P) = \sup_{[1,1.1]}\{ f\}\cdot  (1.1-1) + \sup_{[1.1,1.2]}\cdot (1.2-1.1) + ... + \sup_{[1.4,1.5]}\cdot (1.5-1.4)
    Since f is non-increasing the supremum is the left endpoint.
    Thus,
    U(f,P) = f(1)(.1)+f(1.1)(.1)+...+f(1.4)(.1) = .1 \left( 1+\frac{1}{1.1}+...+\frac{1}{1.4} \right)
    Similarly,
    L(f,P) = f(1.1)(.1)+f(1.2)(.1)+...+f(1.5)(.1) = .1\left( \frac{1}{1.1}+\frac{1}{1.2}+...+\frac{1}{1.5} \right)
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  7. #7
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    sorry everyone! the book is online so i had copied and pasted this question but for some reason the numbers changed and i didnt notice. im sorry for the confusion. thank you all for letting me know.
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  8. #8
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    No worries. Use TPH's very clear demonstration and let's see what you get.
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