What is the Limit of Summation Notation for a Given Interval?

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Homework Help Overview

The discussion revolves around the limit of summation notation within a specified interval, specifically involving a Riemann sum approximation to an integral. Participants are examining the validity of approaches to evaluate the limit as \( n \) approaches infinity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to evaluate a limit by substituting \( n \) with infinity, leading to a discussion about the correctness of this method. Some participants question this approach, suggesting that it may not be appropriate to directly substitute infinity. Others propose expressing the problem in terms of a Riemann sum to facilitate a transition to integral calculus.

Discussion Status

There is a mix of agreement and disagreement among participants regarding the initial approach. Some suggest that the original poster's method is incorrect, while others provide guidance on how to reformulate the problem using Riemann sums. The conversation is ongoing, with multiple interpretations of the problem being explored.

Contextual Notes

Participants note that the multiple choice answers provided may not align with the correct evaluation of the limit, indicating potential confusion or misinterpretation of the problem setup.

Biosyn
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Homework Statement



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Homework Equations


The Attempt at a Solution



I just plugged in ∞ for n

[2+\frac{3}{∞}]2 (\frac{3}{∞}) =

[2+0]2 (0) = 0Did I do the problem correctly? I might need a refresher on summation notations.

Here are the multiple choice answers:

a. 0
b. 1
c. 4
d. 39
e. 125
 
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Seems right to me.
 
It's not right at all. You can't plug n=infinity into that. It looks like a Riemann sum approximation to an integral to me. None of the multiple choice answers that you've shown are correct either.
 
Dick said:
It's not right at all. You can't plug n=infinity into that. It looks like a Riemann sum approximation to an integral to me. None of the multiple choice answers that you've shown are correct either.


I couldn't fit all of the choices into the frame.

The choices:

a. 0
b. 1
c. 4
d. 39
e. 125
 
Biosyn said:
I couldn't fit all of the choices into the frame.

The choices:

a. 0
b. 1
c. 4
d. 39
e. 125

I'm not going to pick an answer for you. Show me how to get it. Put x=k/n and express that as a limiting sum for a Riemann integral over x.
 
Okay , I think I did it.


(3/n) = ΔX
ΔX = (b-a)/n

so, b=3 ; a=0

xi = a + [i(b-a)]/n

xi = [0 + (3k)/n + 2]2

f(xi) = (x+2)2


The integral would be ^{3}_{0}∫(x+2)2
 
Biosyn said:
Okay , I think I did it.


(3/n) = ΔX
ΔX = (b-a)/n

so, b=3 ; a=0

xi = a + [i(b-a)]/n

xi = [0 + (3k)/n + 2]2

f(xi) = (x+2)2


The integral would be ^{3}_{0}∫(x+2)2

One of the listed answers is correct.

RGV
 
Biosyn said:
Okay , I think I did it.


(3/n) = ΔX
ΔX = (b-a)/n

so, b=3 ; a=0

xi = a + [i(b-a)]/n

xi = [0 + (3k)/n + 2]2

f(xi) = (x+2)2


The integral would be ^{3}_{0}∫(x+2)2

Right.
 

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