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Math Help - Can someone explain this Riemann sum?

  1. #1
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    Can someone explain this Riemann sum?

    Can someone explain this Riemann sum?

    "Evaluate the integral by computing the limit of riemann sums"

    Interval: 0-1 (bottom to top, I dont know how to do the integral symbol)
    2x dx

    first thing I do is figure out the change in x, which is 1/n (1-0)/n

    next thing, I figure out x sub i, which is i/n (dxi)

    then my book has me write out the formula, bear with me

    (1/n)(summation symbol i=1 with the upperbound as n)(2(i/n))

    then, seemingly through magic, it becomes (2/n^2)(n(n+1))/2

    I say magic, because it just pretends like its immediately obvious whats going on. I know that im supposed to use that summation formula, no problem understanding that, but I dont understand how it got to 2/n^2

    Also, how do I know which summation formula to use?
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  2. #2
    MHF Contributor Amer's Avatar
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    Re: Can someone explain this Riemann sum?

    \sum_{i = 1}^{n} \frac{i}{n} = \frac{1}{n}+ \frac{2}{n} + \frac{3}{n} + ... + \frac{n-2}{n} + \frac{n-1}{n} + \frac{n}{n}

    if n is even add the last one with the first one then add the second first with the second last and so on
    that will give you \frac{n+1}{n} times \frac{n}{2}

    \frac{1}{n}+ \frac{2}{n} + \frac{3}{n} + ... + \frac{n-2}{n} + \frac{n-1}{n} + \frac{n}{n} = \frac{n+1}{n} \times \frac{n}{2}

    if n is odd can you think about how we can find this sum ?? it is the same why ?

    you can try that with limited terms like
    1+2+3+4+...+10 for example and see
    or 1+2+...+11 try
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  3. #3
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    Re: Can someone explain this Riemann sum?

    hi the riemnann sum is \text{Limit}\left[\frac{1}{n}\sum _{i=1}^{n-1} \frac{i}{n},n\to \infty \right]
    and the result it can be calculate as \int_0^1 x \, dx equal to 0.5
    chao
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