Upper & Lower Sums: Why (i-1) vs i?

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Homework Help Overview

The discussion revolves around understanding the differences between upper and lower sums in the context of Riemann sums, specifically focusing on the use of the indices in the equations for these sums. The original poster is questioning why the lower sum uses (i-1) while the upper sum uses i, particularly in relation to the function f(x) = x^2 over the interval [0,2].

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the reasoning behind the choice of indices for left and right endpoints in calculating lower and upper sums. There is a focus on understanding the implications of using (i-1) versus i in the context of partitioning the interval.

Discussion Status

Some participants have provided insights into the nature of left and right endpoints, with one suggesting that the left endpoint corresponds to the right endpoint of the previous interval. There is ongoing exploration of these concepts, but no consensus has been reached regarding the original poster's confusion.

Contextual Notes

The problem is set within the context of calculating Riemann sums for the function f(x) = x^2, and participants are working within the constraints of a partitioned interval [0,2] with n subintervals. There is a noted lack of clarity about the definitions and roles of the endpoints in this context.

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In an equation, the upper sum is Mi = 0+i(2/n)
and the lower sum is mi = 0+(i-1)(2/n)
So the question is why is it (i-1) for the lower sum and only i for the upper sum?

Any help is highly appreciated! ^_^
 
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State the entire problem.
 
it's not a problem, but a question.

Set in the interval [0,2], it asks me to explain why I need to have i minus 1 in finding the lower sum (the left endpoints) where as in finding the upper sum, it is just i.
 
Lower/Upper sum of what?
 
ok. here is everything. i don't know if it helps.

Find the upper and lower sum for the region bounded by the graphy of f(x) = x^2 and x-axis between x=0 and x=2. To begin, partition the interval [0,2] into n sublevels, each of length (triangle X) = (b-a)/n = (2-0)/n = 2/n

Left endpoints:
https://www.physicsforums.com/latex_images/39/394830-1.png
Right endpoints:
https://www.physicsforums.com/latex_images/39/394830-3.png

QUESTION: why does the equation need (i-1) for finding the left endpoints when it only needs i in finding the right endpoints?
 
Last edited by a moderator:
Hi

First, those are the positions of the left/right endpoints, not the upper/lower sums.

Now, think about this, is 0 a left or a right endpoint? and 2?
What happens when i=1 and i=n? (remeber that i=1,2,...n)
 
Last edited:
I tried to understand what you are asking me but I'm getting confused... again.
So i asked a friend and she said:

"Mi and mi are the right and left endpoints so to find its exact value, we take the i (which names the specific interval) and multiply it by the value of the subintervals (delta x)and since the area of the subinterval is height times width, then height is found by f(mi) or f(Mi) and width is delta x. mi is (i-1) because let's say Mi is at i
then the left endpoint (mi) is the right endpoint of the previous interval so that is why it is i-1."

She isn't really sure about her answer but i can't think of anything else so I'm just going to accept it. Also if this is the answer, it's like common sense, so I'm going to be feeling really stupid.

Thanks for all your help though. I really appreciate your time and hard effort in attemping to free me from my dilemma. ^_^
 
You finally told us that the function in question was x2!

That is an increasing function. If you draw a horizontal line at xi-1 and xi which is lower? Which gives a rectangle that is smaller?
 

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