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!
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? ...
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 $
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.