I Finding the sum of heights under a curve

[yosimba2000]

In integral calc, you add up very small areas to find the total area under the curve. So it would be f(x1)Δx + f(x2)Δx+ ..., summed up. But what if you wanted to find out the sum of all heights under the curve? So it would be something like f(x1) + f(x2) + ...

I'm thinking the formulation would be this:

Take bounds [a,b], and divide that into many infinite n parts to get a step size of (b-a)/n, lim n -> inf.
Using right end points, evaluate f(x) where x = a + i*(b-a)/n, and i goes from 1 to inf
So it then becomes Σ_i=1ⁱ⁼ⁿ f(x_i)

For f(x) = 2x, from 0 to 5
5-0/n, lim n ->inf
x_i = 0 + 5i/n = 5i/n

Σ_i=1ⁱ⁼ⁿ 2x_i and i = n at the end since you need to add up as many segments as you made (n segments)
Σ_i=1ⁱ⁼ⁿ 2(5i/n)
(10/n)Σ_i=1ⁱ⁼ⁿ i
(10/n)(1+2+3+4+5...+n)
and since (1+2+3+4+5...+n) is greater than n by an infinite amount
then (10/n)(1+2+3+4+5...+n) = infinity?

I think this result makes sense in that there are infinitely many x points to choose under the curve, and they all have a positive height (except for at x=0), so infinitely many positive heights added up = infinity.

What do you think?

 

[Mark44]

In integral calc, you add up very small areas to find the total area under the curve. So it would be f(x1)Δx + f(x2)Δx+ ..., summed up. But what if you wanted to find out the sum of all heights under the curve? So it would be something like f(x1) + f(x2) + ...

This doesn't make any sense. In any interval [a, b] of nonzero length, there are an uncountable infinity of points.

First off, how are you going to list them? You can use x1, x2, etc., with indices from the integers, because the integers, while infinite, are only countably infinite.

I'm thinking the formulation would be this:

Take bounds [a,b], and divide that into many infinite n parts to get a step size of (b-a)/n, lim n -> inf.
Using right end points, evaluate f(x) where x = a + i*(b-a)/n, and i goes from 1 to inf
So it then becomes Σ_i=1ⁱ⁼ⁿ f(x_i)

For f(x) = 2x, from 0 to 5
5-0/n, lim n ->inf
x_i = 0 + 5i/n = 5i/n

Σ_i=1ⁱ⁼ⁿ 2x_i and i = n at the end since you need to add up as many segments as you made (n segments)
Σ_i=1ⁱ⁼ⁿ 2(5i/n)
(10/n)Σ_i=1ⁱ⁼ⁿ i
(10/n)(1+2+3+4+5...+n)
and since (1+2+3+4+5...+n) is greater than n by an infinite amount
then (10/n)(1+2+3+4+5...+n) = infinity?

I think this result makes sense in that there are infinitely many x points to choose under the curve, and they all have a positive height (except for at x=0), so infinitely many positive heights added up = infinity.

What do you think?

I think it doesn't make any sense for precisely the reason you say. As you evaluate the function at more and more points, you get larger and larger numbers. How is that useful?

 

[yosimba2000]

This doesn't make any sense. In any interval [a, b] of nonzero length, there are an uncountable infinity of points.

First off, how are you going to list them? You can use x1, x2, etc., with indices from the integers, because the integers, while infinite, are only countably infinite.

The limit of n to inf of the interval (b-a)/n is the step size where the function is evaluated. It's the same way a normal integral is calculated, but all I've done is remove multiplying the width by (b-a)/n AKA width of very small size.
https://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/defintdirectory/DefInt.html

 

[Mark44]

The limit of n to inf of the interval (b-a)/n is the step size where the function is evaluated. It's the same way a normal integral is calculated, but all I've done is remove multiplying the width by (b-a)/n AKA width of very small size.

This is a crucial difference. In a Riemann integral you're taking the limit of a sum of products that could be interpreted as areas, with each of them being some small but finite width. What you're doing is just adding up more and more numbers. There's really no comparison.

https://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/defintdirectory/DefInt.html