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

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The discussion focuses on the mathematical concepts of upper and lower sums in calculus, specifically in the context of the function f(x) = x² over the interval [0, 2]. The upper sum is calculated using the right endpoints (Mi = 0 + i(2/n)), while the lower sum utilizes the left endpoints (mi = 0 + (i-1)(2/n)). This distinction arises because the left endpoint for any interval is the right endpoint of the previous interval, necessitating the use of (i-1) for the lower sum. Understanding these concepts is crucial for accurately calculating the area under curves using Riemann sums.

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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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