B Where Did My Neglect of High-Order Terms Go Wrong in Integral Sums?

[Mike_bb]

Hello.

As is known, we can neglect high-order term in expression $f(x+dx)-f(x)$. For $y=x^2$: $dy=2xdx+dx^2$, $dy=2xdx$.

I read that infinitesimals have property: $dx+dx^2=dx$

I tried to neglect high-order terms in integral sum ($dx^2$ and $4dx^2$ and so on) and I obtained wrong result in the end.

Where is my mistake?

Thanks.

 

[PeroK]

Hello.

As is known, we can neglect high-order term in expression $f(x+dx)-f(x)$. For $y=x^2$: $dy=2xdx+dx^2$, $dy=2xdx$.

I read that infinitesimals have property: $dx+dx^2=dx$

I tried to neglect high-order terms in integral sum ($dx^2$ and $4dx^2$ and so on) and I obtained wrong result in the end.

View attachment 304873

Where is my mistake?

Thanks.

I don't really understand what you've done. The $dx$ in an integral isn't an infinitesimal.

In fact, to be precise, there is no such thing as an "infinitesimal" in standard calculus. Although, there is the concept of the "differential" $dx$.

The $dx$ in the integral is not a differential either, but a notation specific to integration.

 

[Mike_bb]

I don't really understand what you've done. The $dx$ in an integral isn't an infinitesimal.

In fact, to be precise, there is no such thing as an "infinitesimal" in standard calculus. Although, there is the concept of the "differential" $dx$.

The $dx$ in the integral is not a differential either, but a notation specific to integration.

I read "A treatise on infinitesimal calculus" Bartholomew Price. Integral sum, for $2xdx$:

 

[PeroK]

Okay. Infinitesimal calculus is not something I know anything about, I'm sorry to say!

 

[Mark44]

@Mike_bb, what you're doing is evaluating the definite integral using a Riemann sum.
In post 1 you said this:

For $y = x^2, dy = 2xdx + dx^2, dy = 2xdx$

This doesn't make any sense to me. If the function is defined by the formula $y = f(x) = x^2$, then by definition, the differential of f (df) or differential of y (dy) is given by $dy = \frac{df}{dx} dx = 2xdx$. I don't see where you're getting the $dx^2$ term from.

In evaluating the def. integral $\int_{x_0}^{x_n} x^2 dx$, the author of the book you're working from is using a Riemann sum to evaluate the integral. BTW, from the language used, my guess is that this is a very old textbook (maybe over 100 years old!). I would advise getting something a bit more modern and that uses illustrations.

The author's use of dx in the Riemann sum is not at all the way things are usually done these days. See https://en.wikipedia.org/wiki/Riemann_sum for a more modern explanation.

In the Riemann sum, the interval $[x_0, x_n}$ is divided into subintervals of equal length. This is usually written as $\Delta x = \frac{x_n - x_0} n$ and not as dx.

In each subinterval, an estimate is calculated of the area under the curve, within that subinterval, and then all n subintervals are added to get the total. This is easier to follow from a drawing, several of which are shown with examples in the wiki article I linked to.