Integrating equations relating infinitely small changes I've been studying classical mechanics recently and I have two (related) questions on calculus:


*

*In the first chapter of R. Shankar's book Fundamentals of Physics, he derives $vdv = adx$ (in a time interval $[t, t + dt]$ as $dt \rightarrow 0$) from the definitions of $a$ and $v$. Then he writes:
$$\int_{v_1}^{v_2} vdv = a\int_{x_1}^{x_2} dx$$
for the situation when $v$ and $x$ change in a time interval $[t_1, t_2]$, and I don't quite understand how that follows (as the quantities are equal in the same time intervals, while the integration is with respect to $v$ and $x$; and $v$ as a function of $v$ is completely different than $v$ as a function of $t$).

*How to justify the variable change done here? It probably flows from the answer to the first question, but maybe there's something more to be said.
 A: First of all, note that I use much of the material found here (with some extra added for the fact that you have a Definite Integral, not simply an antiderivative). Also, I highly recommend reading up on this PDF justifying the Kinematic Equations rigorously. Anyway, now to the proof  
Let's say we have a first order, separable differential equation (which yours is)
$$m(x) + n(y)\frac{dy}{dx} = 0$$
This is usually solved by seperating both sides, i.e. by niavely treating $\frac{dy}{dx}$ as a fraction and then integrating both sides by a different variable (which is exactly the problem you are struggling with)
$$\int n(y) dt = -\int m(x) dx$$
Instead we can justify the process as such: First, we rewrite the original equation as such:
$$n(y)\frac{dy}{dx} = -m(x)$$
We now integrate both sides with respect to $x$
$$\int_{x_1}^{x_2} n(y)\frac{dy}{dx} dx = -\int_{x_1}^{x_2}m(x) \,dx$$
Now we let $N(y)$ be the antiderivative of $n(y)$, and note that, by the chain rule, 
$$\frac{d\,N(y)}{dx} = n(y)\frac{dy}{dx}$$
Substituting stuff above, we get that
$$\int_{x_1}^{x_2} n(y)\frac{dy}{dx} \,dx = \int_{x_1}^{x_2} \frac{d\,N(y)}{dx} \,dx = N(x_1)-N(x_2)$$
$$N(x_1)-N(x_2) = \int_{x_1}^{x_2} n(y)\,dy$$
Note that this is a little different than what your professor wrote, as using the process above would give
$$\int_{x_1}^{x_2} v\,dv = a\int_{x_1}^{x_2} dx$$
$$\frac{v^2}{2}\bigg|_{x_1}^{x_2} = ax\bigg|_{x_1}^{x_2}$$
We now have to note that $v$ is a function of $x$ the way we wrote it, so we get
$$\frac{v(x_2)^2}{2} - \frac{v(x_1)^2}{2} = a(x_2 - x_1)$$
$$\implies v(x_2) = v(x_1) + 2a\Delta x$$
Now we see your professor’s reasoning - since $y$ is a function of $x$, then we can simply replace $v(x_1)$ with $v_1$ and $v(x_2)$ with $v_2$, which seemingly removes the dependency on $x$ (which is why your professor's integral on the left seemed to not involve $x$) and also puts the above Kinematic Equation in standard form
A: What confused me was that $a$ must be a function of $x$ (i.e. there can't be more than one value of acceleration for a given position) for the second equation to work (or, for that matter, to even allow one to put an integral before $adx$), which I didn't take into account when I tried to visualize it in my mind. This is of course the case in this context, as $a$ is constant. Here's how to justify it more formally.
