I've recently been told that dy and dx can be treated as very small numbers since they're a ratio.

If this is true then how is the integral of dx/x = lnx ?

it would mean that x' = dx

Which means that dx=1

Could anyone explain?

Thanks!

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- Aug 1st 2011, 01:19 PMeliehNotation problems
I've recently been told that dy and dx can be treated as very small numbers since they're a ratio.

If this is true then how is the integral of dx/x = lnx ?

it would mean that x' = dx

Which means that dx=1

Could anyone explain?

Thanks! - Aug 1st 2011, 01:47 PMSironRe: Notation problems
If you want to calculate:

$\displaystyle \int \frac{dx}{x}$, the primitive is indeed $\displaystyle \ln(x)+C$

What do you mean with x'=dx and so dx=1?

Do you know what $\displaystyle dx$ is? if you don't I think it can be useful to look at the meaning of the limit definition of the definite integral. Are you familiar with that? ... - Aug 1st 2011, 02:19 PMeliehRe: Notation problems
dx represents the change in x,

so it's just a number right? - Aug 1st 2011, 02:31 PMPlatoRe: Notation problems
That is certainly historically correct. It maybe that $\displaystyle dx$ is just an artifact of history. I knew an American analyst who refused to use it.

He would simply write $\displaystyle \int_a^b f $ whereas most would write $\displaystyle \int_a^b {f(x)dx} $.

I have been guilty of telling students that artifact simply reminds what the variable of integration is.

So in your question $\displaystyle \int_a^b {dx}=\int_a^b {1dx}=\left. x \right|_{x = a}^b $ - Aug 1st 2011, 02:37 PMeliehRe: Notation problems
Makes sense, so I'm assuming that dx is negligible.

- Aug 2nd 2011, 12:28 AMSironRe: Notation problems
What I want to say was that $\displaystyle dx$ has definitely a meaning if you take a closer look to the geometric aspect of definite integrals. But in the case of indefinite integrals I agree with what Plato said.

Meaning in case of definite integrals:

Why do you calculate integrals? Probably you know that (calculating the area under a curve). If you have given a curve of the function $\displaystyle f$ and you want to calculate the area in an interval [a,b] then you've to calculate:

$\displaystyle \int_{a}^{b}f(x)dx$

If you're going to take a closer look tot the geometric aspect you're going to divide the interval [a,b] in n equal part intervals with length $\displaystyle \Delta (x_i)=\frac{b-a}{n}$ and height $\displaystyle f(x'_i)$, if you want to calculate the area then you've to make the sum of the area's of all the part intervals that means:

$\displaystyle \sum f(x'_i)\cdot \Delta(x_i)$

But you want a very good approximation of the area under the curve so limits will be a solution for that:

$\displaystyle \lim_{n\to \infty} \left(\sum_{i=1}^{n} f(x'_i)\cdot \Delta(x_i)\right)$

Because the approximation will be better if the length of the part intervals is very small that means you want $\displaystyle \Delta(x_i)$ as small as possible (so you'll get more part intervals and so the approximation of the area under the curve will be better)

Because if $\displaystyle n \to \infty$ then $\displaystyle \Delta(x_i)=\frac{b-a}{n} \to 0 $. So $\displaystyle dx$ is the length of a very small part interval.