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Summation and Integral

  1. Oct 7, 2007 #1
    Can someone check this solution.
    1. The problem statement, all variables and given/known data
    [tex]\lim_{n\rightarrow\infty}\sum^{n}_{i=1}\sqrt{\frac{1}{n^2}+\frac{2i}{n^3}}[/tex]

    3. The attempt at a solution
    [tex]=\lim_{n\rightarrow\infty}\frac{1}{n}\sum^{n}_{i=1}\sqrt{1+\frac{2i}{n}}=\int^{1}_{0}\sqrt{1+2x}dx[/tex]
    for u=1+2x->du=2dx
    [tex]\int^{1}_{0}\frac{\sqrt{u}du}{2}=\frac{1}{3}[/tex]
     
  2. jcsd
  3. Oct 7, 2007 #2

    Gib Z

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    That is correct.
     
  4. Oct 7, 2007 #3
    My prof.says that I forgot to change something,while changing variables.
     
  5. Oct 7, 2007 #4

    Gib Z

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    O perhaps the bounds on the integral? Sorry im having a bad day..
     
  6. Oct 7, 2007 #5
    yes, I think problem is on Integral.
     
  7. Oct 7, 2007 #6

    Gib Z

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    I am sure of it now, you forgot the change the bounds. New limits of integration should be 3 and 1.
     
  8. Oct 7, 2007 #7
    and the answer is
    [tex]\int^{3}_{1}\sqrt{1+2x}dx=\frac{3^{\frac{3}{2}}-1}{3}[/tex]
    yes?
     
  9. Oct 7, 2007 #8

    Gib Z

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    Correct, though it may look more pretty as [tex]\frac{3\sqrt{3}-1}{3}[/tex] lol.
     
  10. Oct 7, 2007 #9
    Can you explain, please,why we must choose 3 and 1.
     
  11. Oct 7, 2007 #10

    Gib Z

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    O simple. What the limits on the original integral mean are 'sum for values of x between 1 and 0'. You made the substitution u= 2x+1. So when the original integral says sum for x between 1 and 0, the new integral, where u is the new variable, must say 'sum for x between 1 and 0, and since u=2x+1, sum for u between 3 and 1'.
     
  12. Oct 7, 2007 #11
    ok,thanks
     
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