I was hoping to gain some intuition as to the process of integration by substitution and how exactly it works when evaluating a definite integral in one dimension.
So when evaluating a definite integral in one dimension, $\int_a^b f(x) dx$, you divide the $x$-axis between $a$ and $b$ into $n$ equal sections of length $\delta x$ and then you have that $\int_a^b f(x) dx = lim_{n \rightarrow \infty} \sum_{i=1}^{n}f(x_i)\delta x$
So then when you use integration by substitution to evaluate the equivalent integral, you use a change of variables, $u(x)$ which means you are now integrating along a new $u$-axis between limits $u(b)$ and $u(a)$. You compute the integral in the exact same way to obtain $\int_{u(a)}^{u(b)} f(x(u)) du$. However you then have to multiply by the 'scale factor' $\frac{dx}{du}$, and I'm having difficulty understanding this. I guess that the problem is that since the difference between $u(b)$ and $u(a)$ on the $u$-axis is in general different from the difference between $b$ and $a$ on the $x$-axis that this means that $du$ and $dx$ are not equal in size and thus multiplying by the 'scale factor' basically has the effect of making every section of length $du$ back into the 'proper length' $dx$. Is this the case?
Yes, intuitively you're right.
So if $u = g(x)$.
Then $\frac{du}{dx} = g'(x)$
Then we can intuitively think in terms of differentials
$du = g'(x)dx$
So $dx = \frac{du}{g'(x)}$
So you can substitute this $dx$ back into the integrand to get
$\int_a^b f(x) dx$
$\int_{u(a)}^{u(b)} f(x(u)) dx$
$=\int_{u(a)}^{u(b)} f(x(u)) \frac{du}{g'(x)}$
If you want to be completely rigorous though, we're using the chain rule in reverse... the "integration by substitution" theorem is proven here:
https://en.wikipedia.org/wiki/Integration_by_substitution
In reality the cross multiplying by $dx$ is just a mnemonic... a book-keeping tool that works well to accomplish what the theorem is saying. I guess there is a theory of differentials that could make this multiplying by dx rigorous.