calculus - The integral of a acceleration

If $v_1$ is the x-velocity of the body at time $t_1$ and $v_2$ is the x-velocity at time $t_2$,

Then apparently,

$v_2 - v_1 = \int^{v_2}_{v_1} dv_x = \int^{t_2}_{t1_1}a_x dt$

I am confused about the second equality, $\int^{v_2}_{v_1} dv_x$ What does this mean? How is it equal to the integral of the acceleration with respect to t? The integral of the acceleration with respect to t will be a velocity, but if we take the integral of the velocity with respect to itself, would we not get a value of $\frac{m^2}{s^2}$? Of course, this unit is wrong for velocity!

Answer

You might be over-thinking things a bit. If you can understand the integral,

$$\int_a^bdx=b-a,$$

and the integral,

$$\int_{x_1}^{x_2}dx=x_2-x_1,$$

then you can understand

$$\int_{v_1}^{v_2}dv_x=v_2-v_1.$$

I can see where there is potential for confusion though. The subscript $x$ in $v_x$ is meant to indicate that $v_x$ denotes the $x$-component of the velocity, as opposed to the $y$ or $z$ components. On the other hand, the subscripts $1$ and $2$ in $v_1$ and $v_2$ are merely labels for distinguishing between different values of the $x$-component of velocity at different times. So we see that the $x$ subscript and the $1,2$ subscripts are used to represent two completely different concepts. Yet when both usages pop up in the same term, as in $\int_{v_1}^{v_2}dv_x$, you might be tricked into trying to view the $x$ subscript as a kind of variable ranging over the values $x=1$ and $x=2$. This is an illusion, since the symbol $x$ here is just a label, not a variable.

In my opinion, the expression $\int_{v_1}^{v_2}dv_x$ is an example of bad notation. If the problem context is one-dimensional motion, I'd opt for dispensing with the $x$ subscript altogether since the information providing is redundant, and it just clutters up the formula. If you do need it though, then notation such as

$$\int_{v_{x,1}}^{v_{x,2}}dv_x,$$

would be less of an abuse of notation.