Infinitesimals as interval limits in integration

[cubic]

Ok so what I want to know is, is this valid? If so what does it mean?

 

[mathman]

The notation is unusual. It might simply mean F(0)dx.

 

[cubic]

The integral comes from finding the work done by a force over a distance dx. The force may or may not be variable so I needed to prove the variableness of F did not matter over an infinitesimal for the purposes of determining the amount of work done.

So dw = F⋅dx => F0⋅dx

So integrating over the distance dx would prove this.

I'm doing a proof that I'd like to make rigorous. This makes sense to me visually - we are simply looking at the area of the first strip of dx - which would be the initial value of F. However I cannot find a way to prove it mathematically.

 

[mathman]

You first need to define an integral with an upper limit dx.

 

[cubic]

An upper limit of dx would be the area of the first strip of dx. I need to prove the above.

 

[cubic]

Doh. The area of that first strip would be the height (F₀) by the width (dx.) Ok problem solved.

 

[mathman]

Notation quibble: dx is used for the differential. If you are trying to describe a small non-zero width, use Δx.

 

[cubic]

It has to be along a distance dx.

 

[Mark44]

It has to be along a distance dx.

Like mathman said, use Δx. dx has its own meaning. Δx can represent some small distance along the x-axis.

 

[cubic]

dx is the width of each strip. dx would then be the width of the first strip. Δx is the width of a finite number of strips, which is not what I am looking for.

 

[pwsnafu]

dx is the width of each strip. dx would then be the width of the first strip. Δx is the width of a finite number of strips, which is not what I am looking for.

Thirding what the others said: the width of the first strip is $\Delta x_0$, the width of the second strip is $\Delta x_1$ etc. The width of a finite number of strips is $\Sigma \Delta x_n$.

dx is not used for the width of a strip.