Limit of Summation and Integral Solution Verification

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Homework Help Overview

The discussion revolves around the limit of a summation and its relation to an integral. The original poster presents a limit involving a summation and attempts to evaluate it using a substitution method to convert it into an integral.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the correctness of the original poster's solution, particularly focusing on the variable substitution and the limits of integration. Questions arise regarding the proper bounds after the substitution and the implications of these changes.

Discussion Status

Some participants provide feedback on the original poster's approach, suggesting that there may be an error in the bounds of the integral after the substitution. The conversation explores different interpretations of the limits and the necessity of adjusting them accordingly.

Contextual Notes

There is mention of a potential oversight in changing the limits of integration during the variable substitution process, which is a key point of discussion among participants.

azatkgz
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Can someone check this solution.

Homework Statement


[tex]\lim_{n\rightarrow\infty}\sum^{n}_{i=1}\sqrt{\frac{1}{n^2}+\frac{2i}{n^3}}[/tex]

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]
 
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That is correct.
 
My prof.says that I forgot to change something,while changing variables.
 
O perhaps the bounds on the integral? Sorry I am having a bad day..
 
yes, I think problem is on Integral.
 
I am sure of it now, you forgot the change the bounds. New limits of integration should be 3 and 1.
 
and the answer is
[tex]\int^{3}_{1}\sqrt{1+2x}dx=\frac{3^{\frac{3}{2}}-1}{3}[/tex]
yes?
 
Correct, though it may look more pretty as [tex]\frac{3\sqrt{3}-1}{3}[/tex] lol.
 
Can you explain, please,why we must choose 3 and 1.
 
  • #10
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'.
 
  • #11
ok,thanks
 

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