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Math Help - Integration using Riemann sums

  1. #1
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    Integration using Riemann sums

    The question:
    Suppose that f(x) = \frac{1}{\sqrt{1 - x^2}}

    i) Show that f is increasing on the interval [0, 1/2].
    ii) Find the upper Riemann sum for f with respect to the partition
    { \frac{0}{2n}, \frac{1}{2n}, \frac{2}{2n}, \frac{3}{2n}, ..., \frac{n}{2n} } of [0, 1/2]
    iii) Hence evaluate limit n -> infinity ( \frac{1}{\sqrt{4n^2 - 1^2}} + \frac{1}{\sqrt{4n^2 - 2^2}} + \frac{1}{\sqrt{4n^2 - 3^2}} + ... + \frac{1}{\sqrt{4n^2 - n^2}})

    My attempt:
    i) I took the derivative of f, and found the stationary points by equating it to 0. I noticed that the only stationary point was at 0 itself, thus the graph may only change gradient at this point. I substituted for 1/2 and found that the gradient was positive, so the interval [0, 1/2] is increasing.

    ii) I worked out the width of each partition is \frac{1}{2n}, and the height of a given partition 'k' is f(\frac{k}{2n}). My sum then looked something like this:

    \sum^n_{k=0}{\frac{1}{\sqrt{1 - \frac{k^2}{4n^2}}}

    This looks different to what's in part iii). Is it incorrect? Also, how do I solve part iii)? Thanks.
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  2. #2
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    Any ideas?
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  3. #3
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    Quote Originally Posted by Glitch View Post
    The question:
    Suppose that f(x) = \frac{1}{\sqrt{1 - x^2}}

    i) Show that f is increasing on the interval [0, 1/2].
    ii) Find the upper Riemann sum for f with respect to the partition
    { \frac{0}{2n}, \frac{1}{2n}, \frac{2}{2n}, \frac{3}{2n}, ..., \frac{n}{2n} } of [0, 1/2]
    iii) Hence evaluate limit n -> infinity ( \frac{1}{\sqrt{4n^2 - 1^2}} + \frac{1}{\sqrt{4n^2 - 2^2}} + \frac{1}{\sqrt{4n^2 - 3^2}} + ... + \frac{1}{\sqrt{4n^2 - n^2}})

    My attempt:
    i) I took the derivative of f, and found the stationary points by equating it to 0. I noticed that the only stationary point was at 0 itself, thus the graph may only change gradient at this point. I substituted for 1/2 and found that the gradient was positive, so the interval [0, 1/2] is increasing.

    ii) I worked out the width of each partition is \frac{1}{2n}, and the height of a given partition 'k' is f(\frac{k}{2n}). My sum then looked something like this:

    \sum^n_{k=0}{\frac{1}{\sqrt{1 - \frac{k^2}{4n^2}}}
    But the integral is \frac{1}{2n}\sum_{k=0}^n\frac{1}{\sqrt{1- \frac{k^2}{4n^2}}}= \sum_{k=0}^n\frac{1}{2n\sqrt{1-\frac{k^2}{4n^2}}} = \sum_{k=0}^n\frac{1}{\sqrt{4n^2(1- \frac{k^2}{4n^2}})}

    This looks different to what's in part iii). Is it incorrect? Also, how do I solve part iii)? Thanks.
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  4. #4
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    Oh! I see what you did! I must have forgotten to multiply by width! >_<

    Now that I understand that bit, how do I calculate the limit?
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  5. #5
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    Well, you have now that the Riemann sum is \sum_{k=0}^n\frac{1}{\sqrt{4n^2(1- \frac{k^2}{4n^2}})}= \sum_{k=0}^n\frac{1}{\sqrt{4n^2- k^2}} and the limit of that, as n goes to infinity is the integral from 0 to 1/2.

    But the limit as n goes to infinity is exactly the sum you want to find.

    So, go ahead and integrate from 0 to 1/2: \int_{0}^{1/2} \frac{dx}{\sqrt{1- x^2}}.
    I recommend a trignometric substitution.
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  6. #6
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    Oh, so I don't have to take a limit, I just solve it as an integral?
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