# Thread: Notation problems

1. ## Notation 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!

2. ## Re: Notation problems

If you want to calculate:
$\int \frac{dx}{x}$, the primitive is indeed $\ln(x)+C$

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

Do you know what $dx$ 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? ...

3. ## Re: Notation problems

dx represents the change in x,
so it's just a number right?

4. ## Re: Notation problems

Originally Posted by elieh
dx represents the change in x,
so it's just a number right?
That is certainly historically correct. It maybe that $dx$ is just an artifact of history. I knew an American analyst who refused to use it.
He would simply write $\int_a^b f$ whereas most would write $\int_a^b {f(x)dx}$.

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

So in your question $\int_a^b {dx}=\int_a^b {1dx}=\left. x \right|_{x = a}^b$

5. ## Re: Notation problems

Makes sense, so I'm assuming that dx is negligible.

6. ## Re: Notation problems

What I want to say was that $dx$ 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 $f$ and you want to calculate the area in an interval [a,b] then you've to calculate:
$\int_{a}^{b}f(x)dx$
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 $\Delta (x_i)=\frac{b-a}{n}$ and height $f(x'_i)$, 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:
$\sum f(x'_i)\cdot \Delta(x_i)$
But you want a very good approximation of the area under the curve so limits will be a solution for that:
$\lim_{n\to \infty} \left(\sum_{i=1}^{n} f(x'_i)\cdot \Delta(x_i)\right)$
Because the approximation will be better if the length of the part intervals is very small that means you want $\Delta(x_i)$ 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 $n \to \infty$ then $\Delta(x_i)=\frac{b-a}{n} \to 0$. So $dx$ is the length of a very small part interval.