Solving for kinematics equations with calculus I am trying to solve for kinematics equations using calculus. I am trying to get to this equation:
$$v= v_0+at$$
The explanation to get there is the following:

Now, I am still a bit new to calculus so I struggle with the manipulation of the notation (dv/dt)
So I totally get why acceleration is dv/dt. It is the rate of change of velocity with respect to time, which must be dv/dt, instead of v/t, in order to get the rate of chance at any point in the line, or the acceleration at any point.
I get that, but what I don't get is when you start manipulating that notation. So when you go from the original a =(dv/dt) to:
$$dv=(a)dt$$
1) What does this mean? Does it mean that the change in velocity = the acceleration times the change of time?
Now, when you continue, I also understand the definite integral. I understand why you must have v and v initial and t and 0 (t initial). However, my second question is:
$$\int{dv}$$
2) How can you get the antiderivative of something that is not even a derivative? is dv a derivative? I thought the derivative was (dv/dt), not just dv by itself?
I just don't understand how dv can be anything by itself?
 A: Taking the second part of the question first, let's consider how antiderivatives are defined in terms of derivatives.
If $f(x) = F'(x),$ that is $f$ is the derivative of $F,$ then $F$ is an antiderivative of $f,$ and we can write an indefinite integral
$$  \int f(x) \, dx = F(x) + C, $$
where $C$ is a constant of integration.
We can also write a definite integral
$$ \int_a^b f(x) \, dx = F(b) - F(a).$$
Now consider what happens if we define $F$ as $F(x) = x.$
then $f(x) = F'(x) = 1,$ and
$$ \int_a^b 1 \, dx = \int_a^b f(x) \, dx = F(b) - F(a) = b - a.$$
But we usually just write $\int dx$ when we mean $\int 1\,dx.$ So
$$ \int_a^b  dx = b - a.$$
The equation
$$ \int_{v_0}^v dv = v - v_0 $$
is just the same thing with different symbols for the variables.
There is one potentially confusing thing about the equation in this form,
which is that the same letter ($v$) is used for two different things:
we see it used as the variable of integration (in $dv$)
and at the upper bound of the integral.
Technically, these are two different variables that just happen to have
the same name; the $v$ in $dv$ is meaningful only within the integrand
(in this case, just $dv$), not at the bounds of the definite integral
nor outside the integral, and we could change that name from $v$ to
anything else without changing the bounds or the value of the integral.
For the first part of the question, the notation involved could be differentials, which are somewhat like change in velocity and change in time, and indeed the person explaining this may intend you to think of them that way. (Physicists tend to care more about whether their calculations get the results they should than on the finer details of the mathematical explanation.)
Another interpretation is as a change of variables.
Writing the velocity $v(t)$ as a function of $t,$
we have
$$
 \int_0^{\Delta t} a \, dt = \int_0^{\Delta t} \frac{dv}{dt} \, dt
= \int_0^{\Delta t} v'(t) \, dt.
$$
We change the integration variable from $t$ to $v,$
which means the integration runs from $v(0)$ to $v(\Delta t)$
instead of $0$ to $\Delta t$
and we substitute $dv$ for $v'(t) \,dt$ in the integrand:
$$
\int_0^{\Delta t} v'(t) \, dt = \int_{v(0)}^{v(\Delta t)} dv.
$$
From this, we use what we know about antiderivatives to write
$$
\int_{v(0)}^{v(\Delta t)} dv = v(\Delta t) - v(0).
$$
But going back to the original integral of $a,$ if $a$ is constant we can factor it out:
$$
 \int_0^{\Delta t} a \, dt =  a \int_0^{\Delta t} dt 
  = a\left(\Delta t - 0 \right) = a \Delta t.
$$
Following the chain of equations around, we have found that
$$
v(\Delta t) - v(0) = a \Delta t,
$$
therefore
$$
v(\Delta t) = v(0) + a \Delta t.
$$
If we then set $v_0 = v(0)$ and write $v$ to mean $v(\Delta t),$
we get the form shown in the explanation you read.

As a personal opinion, I actually find the way notation was used in the text that you read to be quite annoying.
For one thing, why take all this trouble to prove a formula that only works
for constant accelerations starting at time zero, when you could just as easily (in fact, I think easier and with less chance of confusion)
prove a formula for constant acceleration starting at any time?
That is, writing the velocity as a function of time, $v(t),$
we accelerate from velocity $v(t_0)$ at time $t_0$
to velocity $v(t_1)$ at time $t_1$ with constant acceleration $a,$
with the result
$$
v(t_1) - v(t_0) = \int_{v(t_0)}^{v(t_1)} dv 
 = \int_{t_0}^{t_1} a\,dt = a(t_1 - t_0),
$$
that is,
$$ v(t_1) = v(t_0) + a(t_1 - t_0). $$
That is a perfectly useful general formula, which we can rewrite 
by substituting the symbol $t$ for $t_1,$
$$ v(t) = v(t_0) + a(t - t_0), $$
and then (if we are so inclined) defining $v_0 = v(t_0)$
and $\Delta t = t - t_0$ 
and abbreviating $v(t)$ to $v$
when it is convenient to do so for some purpose.
A: Good questions, and there are lots of nuances at play, meaning that you might get differing answers (all correct) based on mathematical maturity. 
Regarding your first question, it's perhaps best to think about the "d"'s as deltas - change. So, the change in velocity equals that acceleration times the change in time. So velocity increases or decreases based upon the acceleration (positive or negative), and the amount of time that has occured. (Liked, if you increased your velocity at a rate of $2 \frac{m}{s^2}$, then your velocity changes more over the duration of $5$ seconds than, say, 1 second.
Regarding your second question, "dv" is a differential with respect to v. So, think of it as a tiny little sliver of v (velocity). Integration is simply a fancy way to sum up infinitely many tiny things. Thus, $$ \int dv = v $$ In English, the sum of all little bits of the velocity is equal to the velocity. 
I hope this helps without being too rigorous or abstract. :)
A: For the first question, 
$dv=(a)dt$.
This means that, for a small interval of time, the change in velocity can be calculated by multiplying the time interval with the acceleration in that small interval of time (provided that acceleration doesn't change suddenly in that interval, which would make the function non differentiable at that point).
For the second part, 
Think of $\displaystyle \int dv$ as $\displaystyle \int 1.dv$.   So, 1 is the derivative of $v$ with respect to $v$ hence the antiderivative of 1 with respect to $v$ would be $v$
