calculus - Integration operation, and its relation to differentials

I need to get a few things straight about the integration operation (as an intro calc student). I understand that integration is a process that takes a function and returns its antiderivative. We can think of it as an operator, where $\displaystyle \int...dx$ is kind of like an opening and a closing bracket for an input function. This is how I interpret integration: the summa and the differential (the closing "bracket") are inseparable, since they are part of the same notation. However, my professor confused me with his derivation of velocity under constant acceleration:

$$\displaystyle \frac{dv}{dt} = a$$
$$\displaystyle v\frac{dv}{dt} = va$$
$$\displaystyle v\frac{dv}{dt} = a\frac{dx}{dt}$$
Then he "cancels" the $dt$
$$vdv = adx$$

The next part is what confuses me:
$$\int_{v_1}^{v_2}vdv = a\int_{x_1}^{x_2}dx$$
which comes out to be
$$\frac{1}{2}v_2^2 - \frac{1}{2}v_1^2 = a(x_2 - x_1)$$

My primary question is, how did the two summas appear, if there were no corresponding differentials at the time of application? This is disconcerting because the same operation is supposed to be applied to both sides of the equation, and those look like two different operations in terms of two different variables. Shouldn't he have done something like $\displaystyle \int_{t_1}^{t_2}vdv~dt = \int_{t_1}^{t_2}adx~dt$?

If it's no trouble, I have two additional questions. Why does he use $\displaystyle \int_{v_1}^{v_2}...dv$ rather than $\displaystyle \int...dv$? Also, what justifies that he "cancels" the dt differentials?

Answer

So you can get to here without trouble:

$$\displaystyle v\frac{dv}{dt} = a\frac{dx}{dt}$$

The next step is simply to integrate both sides with respect to $t$ from $t_1$ to $t_2$:

$$\int_{t_1}^{t_2} \displaystyle v\frac{dv}{dt}dt = \int_{t_1}^{t_2} a\frac{dx}{dt}dt$$

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Now we just need to show that the $dt$'s "cancel". To do so we just need to use $u$-substition. Forgetting about the RHS for the moment, we start with

$$\int_{t_1}^{t_2} v(t)\frac{dv(t)}{dt}dt$$

Let $u=v(t)$, then $du = \frac{dv(t)}{dt}dt$. So

$$\int_{t_1}^{t_2} v(t)\frac{dv(t)}{dt}dt = \int_{u(t_1)}^{u(t_2)} udu$$ or changing the variable name back to $v$, we have $$\int_{t_1}^{t_2} v(t)\frac{dv(t)}{dt}dt = \int_{v(t_1)}^{v(t_2)} vdv$$

Then do the same thing for the RHS.