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## Main Question or Discussion Point

Within specified limits, the area under the curve of an equation and the submission of all values on that curve will be different right?...just confirming.

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Within specified limits, the area under the curve of an equation and the submission of all values on that curve will be different right?...just confirming.

- #2

- 655

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What do you mean when you say "submission of all values"?

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Submission of all values of f(x) at infinitely small intervals.

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- 655

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http://hyper-ad.com/tutoring/math/calculus/Construction of the Riemann Integral.html

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What I mean to say here is that there are 2 things that can be computed as a submission of 'something' in the function.

This 'something' might be infinitely small area components determined by dx, and it might be the submission of all the values of f(x) made at infinity small intervals.

I've read many relating this submission of area to the submission of f(x), but what I think, this is not possible.

So I wanna confirm this.

- #6

Gib Z

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- #7

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Sorry about the spelling.

The summation will different from the area you mean to say right?

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HallsofIvy

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What do **you** mean by the sum of an infinite collection of numbers?

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Its not an infinite collection of numbers...its a...sum of numbers in some sorta progression.

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HallsofIvy

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I mean...sum of all f(x) within a defined limit, for instance between a to b.

- #12

- 655

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Give a concrete example of what you mean for a specific function. Say, integrating f(x)=x from 0 to 1.

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Take a function y = x (simplest possible)

Suppose I want summation of f(x) (or y) from limits 5 - 11 on x.

So this will be 5+5.1+5.2+5.3.......................10.5+10.6+10.7+10.8+10.9+11

This wont give an accurate result...the intervals at which the addition should occur should be infinity small to give 100% accuracy.

So what I mean to say is when we integrate a function, this summation is not the result right? The summation will be the result in cause of line integral right?

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So its not a Riemann sum.

- #15

- 655

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So this will be 5+5.1+5.2+5.3.......................10.5+10.6+10.7+10.8+10.9+11

This wont give an accurate result...the intervals at which the addition should occur should be infinity small to give 100% accuracy.

That summation is not the result for anything. Not regular integrals or line integrals or anything. If the spacing gets smaller and smaller, the sum will just get bigger and bigger unbounded, so in the limit as the spacing becomes small, the sum becomes infinity!

However, the sum is very very close to the reimann integral. In fact, all you have to do to make your sum into a riemann sum is to divide by the spacing. eg:

(5+5.1+5.2+5.3.......................10.5+10.6+10.7 +10.8+10.9+11)/(0.1)

In this way, even as the sum becomes infinitely big, the denominator becomes infinitely small in just the right proportions such that the riemann sums converge to the integral.

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HallsofIvy

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No, it wouldn't. That is the sum of x-values, not f(x) and it is done with a "step" of 0.1. You had said nothing about a particular step interval before.aaa...thanks for notifying me about that.

Take a function y = x (simplest possible)

Suppose I want summation of f(x) (or y) from limits 5 - 11 on x.

So this will be 5+5.1+5.2+5.3.......................10.5+10.6+10.7+10.8+10.9+11

The Riemann sum, of f(x) with a step of 0.1, between 5 and 11 would be (f(5)+ f(5.1)+ ...+ f(10.9))(0.1). Is that what you are asking about?This wont give an accurate result...the intervals at which the addition should occur should be infinity small to give 100% accuracy.

So what I mean to say is when we integrate a function, this summation is not the result right? The summation will be the result in cause of line integral right?

- #17

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- #18

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Ok I got the answer...thanks a lot everyone!

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