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Math Help - finding Riemann sums with 64 equal subintervals

  1. #1
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    finding Riemann sums with 64 equal subintervals

    use upper and lower Riemann sums for the integral sqrt(x),from 0 to 64, on the interval [0,64] with 64 equal subintervals to find upper and lower bounds for SIGMA n=1~64 sqrt(n).

    what are the answers for upper and lower??

    if the question were 5 equal intervals then i could solve it by hand, but 64!!! are there any shortcut or something that i can do?

    thankyou
    Last edited by mr fantastic; November 4th 2009 at 05:08 PM. Reason: Removed excessve ?'s in post title
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  2. #2
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    Quote Originally Posted by Hikari Clover View Post
    use upper and lower Riemann sums for the integral sqrt(x),from 0 to 64, on the interval [0,64] with 64 equal subintervals to find upper and lower bounds for SIGMA n=1~64 sqrt(n).

    what are the answers for upper and lower??

    if the question were 5 equal intervals then i could solve it by hand, but 64!!! are there any shortcut or something that i can do?

    thankyou
    I don't understand: what's the difference between 5 , 64 or 475482934 intervals? And even better: all of the same length! It's the same: the function f=\sqrt{x} is monotone ascending and thus the lower sum is evaluating the function at the left extreme of each interval, and the upper one at the right extreme:

    L(f,[0,64],n=64)=\sum\limits_{k=i}^{64}f(k-1)(k-(k-1))=\sum\limits_{k=i}^{64}\sqrt{k-1} =\sqrt{0}+\sqrt{1}+...+\sqrt{63}

    The upper Riemann sum with n = 64 is very simmilar to the above one.

    Tonio
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  3. #3
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    Quote Originally Posted by Hikari Clover View Post
    use upper and lower Riemann sums for the integral sqrt(x),from 0 to 64, on the interval [0,64] with 64 equal subintervals to find upper and lower bounds for SIGMA n=1~64 sqrt(n).

    what are the answers for upper and lower??

    if the question were 5 equal intervals then i could solve it by hand, but 64!!! are there any shortcut or something that i can do?

    thankyou
    If (and only if) you are interested in an approximate value of the two sums and if you are allowed to use a calculator then use with a TI83/84 the following command:

    sum(seq(√(N),N,0,63)) for the lower sum and

    sum(seq(√(N),N,1,64)) for the upper sum.

    The values you get have to differ by 8 (ähemm ... why? )
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    the question was asking the lower estimate and upper estimate , the answers should be 1024/3 and 1048/3 , wonder how the answers came out
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    Quote Originally Posted by Hikari Clover View Post
    the question was asking the lower estimate and upper estimate , the answers should be 1024/3 and 1048/3 , wonder how the answers came out

    Define "estimate" in relation with Riemann sums, and then define lower and upper "estimates".

    Tonio
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    the original question is here
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  7. #7
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    Quote Originally Posted by Hikari Clover View Post


    the original question is here

    This doesn't help since there's no definition of "estimate", but anyway: the lower "estimate", if it is something close to what I could define "estimate" , is correct, since it is lower than the lower bound. The upper one's definitely wrong, though.
    Now, perhaps they mean by "estimate" the exact value that you get when carrying on the lower and upper sums, and then both "estimates" are wrong.

    Tonio
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