When u-substituted, what does the transformed integral represent? I'm really trying to get a grasp on what the definite integral represents after u-substitution. Suppose we're integrating:
$$\int_a^b f(x) \mathrm{d}x$$
and perform the substitution: $$ u = x^2 $$
$$ \mathrm{d}u = 2x\mathrm{d}x$$
and the integral becomes:
$$\int_{a^2}^{b^2} f(\sqrt{u}) \frac{\mathrm{d}u}{2\sqrt{u}}$$
(1) What is the relationship between $\mathrm{d}u$ and $\mathrm{d}x$? 
(2) Is the ratio between $\mathrm{d}u$ and $\mathrm{d}x$ always $2x$? Can I choose the width of $\mathrm{d}u$ "slicing" arbitrarily (where it would then violate the above relationship of $\mathrm{d}u = 2x\mathrm{d}x$?)? If not, is it just because $u$ is a dependent variable and $x$ is an independent variable in:
$$ \mathrm{d}u = g'(x) \mathrm{d}x $$
if $$ u = g(x) $$
or is there something deeper going on?
For instance, in:
$$\int_{a^2}^{b^2} f(\sqrt{u}) \mathrm{d}u$$
Can I partition $u$ such that:
$$ \mathrm{d}u = \mathrm{d}x = \frac {b-a} {N}  \:(\text{chosen equal partitions for dx}) $$
such that the Riemann sums of original integral and u-substituted integral be equal without the $g'(x)$ factor?:
$$  \lim_{N \to \infty} \sum_{n = 1}^{N} f(x_n)(x_{n+1}-x_{n}) = \lim_{M \to \infty} \sum_{m = 1}^{M} f(\sqrt{u_m})(u_{m+1}-u_{m}) $$
 A: *

*What you are doing mathematically is applying [Integration_by_substitution][1].

You are using a different notation common for physics. Its not quite so clean but i try to formalize it.
We define a function $u(x) = x^2$ and we derive it like $\frac{du}{dx} = 2x$. This function $\frac{du}{dx}$ is defined like so:
$$
u'(\bar x) := \frac{du}{dx}(\bar x) := \lim_{x\to \bar x} \frac{u(x) - u(\bar x)}{x - \bar x}
$$
which then means for every sequence $x_n \to \bar x$ the limit $\lim_{n\to \infty} \frac{u(x_n) - u(\bar x)}{x_n - \bar x}$ exists and is equal to $u'(\bar x)$. Lets look at one particular sequence $x_n$ now. Since we know that
$$
\lim_{n\to \mathbb N} \frac{u(x_n) - u(\bar x)}{x_n - \bar x} = u'(\bar x)
$$
is converging we find for every $\epsilon > 0$ some $N\in \mathbb N$ so that for every $n \ge N$ we have
$$
\frac{u(x_n) - u(\bar x)}{x_n - \bar x} \le u'(x) + \epsilon
$$
Now the $x_n - \bar x$ term is you $dx$ and the $u(x_n) - u(\bar x)$ is your $du$. We can now treat it the way the physicists often do and say its equivalent to
$$
\iff u(x_n) - u(\bar x) \le (\underbrace{2x}_{u'(x)} + \epsilon)(x_n - \bar x)
$$
The right side is asymptotically the same as $2x(x_n - \bar x)$ which means if $n$ gets large you can ignore the $\epsilon$. The $\ge$ follows analogously.
Now this whole conclusion works for every single sequence $x_n$ and to get rid of all this formality one can just pretend that for every such sequence $x_n$. $dx = x_n - \bar x$ and $du = u(x_n) - u(\bar x)$. Then you multiply and add as usual and simply forget that you would actually need all that formality about sequences to move $dx$ and $du$ around.
So that is the relationship behind it.


*If you write it like that then $dx$ is the change in $x$ and $du$ is the corresponding change in $u(x)$. Actually the whole information in this $dx$ and $du$ is their ratio since in the above sence they would ultimately both be $0$. Some people consider them as infinitesimals. But in differential geometry one uses different meanings for them.


*if $x_m(M) = x_m$ is some partition of $[a,b]$ which gets endlessly fine with growing $M$ and $u_m = u(x_m) = g(x_m)$ then one as
$$
\lim_{M\to\infty} \sum_{m=1}^M f(\sqrt{u_m})(u_{m+1} - u_m) = \sum_{m=1}^M f(\sqrt{u_m})u'(x_m)(x_{m+1} - x_m)
$$
so you actually still have the $u'$ term in there
A: We start with
$$\int_a^b f(x)\, \mathrm{d}x.$$
Assuming $f$ is a continuous function on $[a,b],$ this formula says we integrate the function $f(x)$ over the $x$ values $a \leq x \leq b.$
Now you say $u = g(x)$ and you want an integral of some function of $u$ over the $u$ values $g(a) \leq u \leq g(b).$
But furthermore, the purpose of this substitution is to find the value of the original integral, so whatever function we end up integrating,
let's call it $h(u),$ we want to ensure that
$$\int_{g(a)}^{g(b)} h(u) \, \mathrm{d}u = \int_a^b f(x) \, \mathrm{d}x.$$
A way to do this is to ensure an even stronger result: for any $x$ in $[a,b],$
$$\int_{g(a)}^{g(x)} h(v) \, \mathrm{d}v = \int_a^x f(t) \, \mathrm{d}t. \tag1$$
Now let
\begin{align}
F(x) &= \int_a^x f(t) \, \mathrm{d}t,\\
H(u) &= \int_{g(a)}^u h(v) \, \mathrm{d}v.
\end{align}
Then the Fundamental Theorem of Calculus says that
$F'(x) = f(x)$ and $H'(u) = h(u).$
But the relationship $u=g(x)$ together with Equation $(1)$ says that
$$ H(g(x)) = F(x). $$
These two functions of $x$ are equal, so their derivatives with respect to $x$ are equal. Using the Chain rule to differentiate $H(g(x))$ with respect to $x,$
$$ \frac{\mathrm d}{\mathrm dx} H(g(x)) = H'(g(x)) g'(x) = h(g(x)) g'(x) = f(x). $$
So in order to make this work we, need $h$ such that
$$ f(x) = h(g(x)) g'(x). $$
In the specific case where $u = g(x) = x^2,$ that means
$$ f(x) = h(u) \times 2x. $$
That is to say, it means
$$ h(u) = \frac{f(x)}{2x} = \frac{f\left(\sqrt u\right)}{2 \sqrt u}. $$
We can conclude that
$$\int_{g(a)}^{g(b)} \frac{f\left(\sqrt u\right)}{2 \sqrt u} \, \mathrm{d}u
 = \int_a^b f(x) \, \mathrm{d}x.$$
The only role for partitions in all of this is in order to make the integrals well defined throughout it all.
For each integral, if you take Riemann sums over any sequence of partitions whose mesh size converges to zero, you'll always get the same answer for every sequence of partitions.
A function for which this is not true would not have a Riemann integral.
If you happen to be using a regular mesh of $N$ intervals, then yes, you can say the size of each interval is $\frac{b - a}{N}$; but
$$  \lim_{N \to \infty} \frac {b-a} {N}  = 0, $$
simply and exactly.
The symbol $\mathrm dx$ has nothing to do with a mesh, except to tell you which variable to take  meshes over in the definition of an integral, and it is certainly not equal to $\lim_{N \to \infty} \frac{b-a}{N}$, because if it were it would be zero, and $f(x)\,\mathrm dx$ would be zero.
Likewise the symbol $\mathrm du$ has nothing to do with a mesh, except for naming a variable to take meshes over.
Unless you've developed enough of a theory of differentials for the notations
$\mathrm du$ and $\mathrm dx$ to mean something when there is no integral sign in front of each one, the "equation" $\mathrm du = 2x\,\mathrm dx$ is just a formal manipulation, meaning it has the form or appearance of a real equation but does not actually represent mathematical objects.
In short, the formal equation $\mathrm du = 2x\,\mathrm dx$ is not dictating how you have to make meshes over $u$ relative to how you make meshes over $x$;
it is merely a reminder to you to insert the factor $g'(x)$ in the correct place in the $u$-substitution.
And the factor $g'(x)$ also has little to tell you about any mesh; rather, it comes from the chain rule.
