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Definite integral as Riemann sums

  1. Nov 8, 2016 #1
    1. The problem statement, all variables and given/known data

    Determine ##\int_{0}^{2}\sqrt{x}dx## using left riemann sums

    2. Relevant equations

    ##\int_{a}^{b}f(x)dx = \lim_{n\rightarrow \infty}\sum_{i=0}^{n-1}(\frac{b-a}{n})f(x_i)##

    3. The attempt at a solution

    ##\frac{b-a}{n}=\frac{2-0}{n}=\frac{2}{n}##
    ##\int_{0}^{2}\sqrt{x}dx = \lim_{n\rightarrow \infty}\frac{2}{n}(0+ \frac{\sqrt{2}}{n}+\frac{\sqrt{4}}{n}+...) =\lim_{n\rightarrow \infty}\frac{2}{n} \frac{\sqrt{2}}{n}(\sqrt{0} + \sqrt{1}+\sqrt{2}+\sqrt{3}+...) ##

    I'm stuck now. Please help. I don't know how to get the sums of the infinite radical series
     
  2. jcsd
  3. Nov 8, 2016 #2

    LCKurtz

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    I think you may want a square root on that ##n## I colored red. But to answer your question, that is not a trivial sum to calculate, and I wouldn't expect to see this problem in a typical calculus exercise set. See, for example,
    http://math.stackexchange.com/questions/1241864/sum-of-square-roots-formula
     
  4. Nov 8, 2016 #3

    Mark44

    Staff: Mentor

    The sums in the last two expressions above do not have an infinite number of terms. What's the last term in each sum?
     
  5. Nov 9, 2016 #4
    ##\lim_{n\rightarrow \infty }\frac{2}{n}\sqrt{\frac{2}{n}}(\sqrt{0} + \sqrt{1} + \sqrt{2} + ... + \sqrt{\frac{2}{n}}(n-1)))##

    Alright, I've fixed my mistakes. And, I'm still confused how to get the sum of that series
     
  6. Nov 9, 2016 #5

    Ray Vickson

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    You should go back and read post #2, which explains why you probably cannot sum the series, no matter how hard you try. Of course, you may be able to find good approximations to the sum, but that is not the same as finding an exact value.
     
  7. Nov 10, 2016 #6

    lurflurf

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    Doing the sum is overkill for finding the limit.
    I think you have an error I think you should have
    $$\int_0^2 \! \sqrt{n} \, \mathrm{d}x=\lim_{n\rightarrow\infty} {\left( \frac{n}{2} \right)}^{-3/2}\sum_{i=1}^n \sqrt{n}$$
    Instead of finding the sum we will make a few observations that suffice
    we can replace the sum in the limit by a simple expression
    $$\sum_{i=1}^n \sqrt{n}\sim {C} {\sqrt{n}}^3$$
    The exponent is the only one that allows the limit to converge to a nonzero value
    we find C by computing the limit
    $$C=\lim_{n\rightarrow\infty}\frac{{\sqrt{n}}}{{\sqrt{n+1}}^3-{\sqrt{n}}^3}$$
    because we must have
    $$\sum_{i=1}^{n+1} \sqrt{n}-\sum_{i=1}^n \sqrt{n}\sim {\sqrt{n}}$$
     
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