Field of a uniformly charged disk: integration question In my book (University Physics by Young and Freedman), during solving the common example of finding the electric field along the x-axis from a uniformly charged disk, they arrive at this differential expression:

$$dE_x = {1 \over 4\pi\epsilon_0}{2\pi\sigma rx dr \over (x^2+r^2)^{3/2}}$$

Then they say:

To find the total field due to all the rings, we integrate $dE_x$ over $r$ from $r=0$ to $r=R$ (not from $-R$ to $R$):
$$E_x = \int_0^R {1 \over 4\pi\epsilon_0}{(2\pi\sigma r dr)x \over (x^2+r^2)^{3/2}}$$

I am confused as to how they arrived at this line. The right hand side suggests that they integrated both sides of the differential equation with a definite integral, but if that were the case, the left hand side would have been
$$\int_0^R dE_x = [E_x]_0^R = R - 0 = R$$
Instead, it looks like they integrated the left hand side of the differential equation with an indefinite integral, so they would get
$$\int dE_x = E_x$$
Certainly, integrating one side with an indefinite integral and the other side with a definite integral cannot be a valid step because you are not doing the same thing to both sides.
How can their result be rigorously justified and what is wrong with my reasoning?
 A: The left hand integral limits should be with respect to the field you get as you add up the contributions to the field due to each charged ring. Therefore, it should start at $0$ and end with the final field. i.e.
$$\int_0^{E_x}\text dE_x' = \int_0^R{1 \over 4\pi\epsilon_0}{2\pi\sigma rx\,\text dr \over (x^2+r^2)^{3/2}}$$
Trivially, the integral on the left side is just the total field $E_x$.

Seeing some of your comments, you seemed to be confused as to what $r$ and $R$ actually represent. It looks like $x$ is the constant distance between the center of the disk and the point at which you are calculating the field at. $r$ is the radius of one of the rings of thickness $\text dr$, and $R$ is the radius of the entire disk. Note that you are not determining the field at $r$, since $r$ is not a position in this case.
Your differential relation
$$\text dE_x = {1 \over 4\pi\epsilon_0}{2\pi\sigma rx\,\text dr \over (x^2+r^2)^{3/2}}=\alpha(r)\,\text dr$$
is essentially telling you each ring of radius $r$ and thickness $\text dr$ contributes to the field an amount $\alpha(r)\,\text dr$. It makes sense that we just need to add up (integrate) all of these values (right integral) to determine the total field (left integral). 
A: 
But then what does $E_x(r)$ represent? The problem only said that $dE_x$
  represents the field component at the point due to a ring

$E_x(r)$ is a poor notation here since it certainly looks like the $x$ component of the electric field as a function of the radial coordinate $r$. Further, it is clear from the integral that $E_x = E_x(x)$.
Here, $dE_x$ or $dE_x|_r$ is the contribution to the total electric field (at coordinate $x$ on the $x$-axis) due an infinitesimal ring of charge between $r$ and $r + dr$.
If it helps, consider the approximation by finite rings of charge located between $r$ and $r + \Delta r$. Let there be $N$ such rings such that
$$\Delta r = \frac{R}{N}$$
and
$$r_n = (n-1)\Delta r$$
To find the total electric field due to the contributions $E_{x,n}$ from the $N$ rings making up the disk, we simply sum the contributions (linearity of electric field).
$$E_x = \sum_{n=1}^NE_{x,n}$$
In the limit $N\rightarrow\infty$, the contributions become infinitesimal $E_{x,n}\rightarrow dE_x|_r$, and the sum becomes an integral.
