Is a substitution always required to change the limits of an integral? For example:
Suppose we have an impulsive force $f(t)$ lasting from $t=t_0$ until $t=t_1$ which is applied to a mass $m$. Then by Newtons Second law we have $$\int_{t=t_0}^{t=t_1}f(t)\,\mathrm{d}t=\int_\color{red}{t=t_0}^\color{red}{t=t_1}m\color{blue}{\frac{\mathrm{d}v}{\mathrm{d}t}}\mathrm{d}t=\int_\color{red}{v=v_0}^\color{red}{v=v_1}m\,\mathrm{d}v=m(v_1-v_0)\tag{1}$$
What I can't understand is; What substitution was made to allow the limits marked $\color{red}{\mathrm{red}}$ to change from $t$ to $v$. 
I thought it might be due to $$\color{blue}{\frac{\mathrm{d}v}{\mathrm{d}t}}=\frac{\mathrm{d}v}{\mathrm{d}x}\cdot \underbrace{\frac{\mathrm{d}x}{\mathrm{d}t}}_{\Large{\color{#180}{=v}}}=v\frac{\mathrm{d}v}{\mathrm{d}x}$$ by the chain rule.
But it is something much simpler than this, and I believe I am over-thinking it too much. Could someone please tell me what substitution was made to change the limits marked $\color{red}{\mathrm{red}}$ in equation $(1)$?

Edit:
Comments below seem to indicate that one can simply change the limits of integration to make the integral dimensionally correct. But I consider this to be a less rigorous approach, and I was taught that integral limits must be changed via a substitution. So I still need to know what substitution was made?
Thanks again.
 A: The fundamental reason to change the limits of integration is that
the variable of integration has changed.
Substitution is an obvious case in which this is likely to occur.
For example, substitute $u = x - 2$ in $\int (x - 2) dx$:
$$
\int_0^2 (x - 2) dx = \int_{-2}^0 u\; du.
$$
The intuition I follow on this is that the start of the integral occurs
"when $x=0$" and ends "when $x=2$".
But "when" $x=0$, it must also be true that $u=-2$,
and "when" $x=2$, it must also be true that $u=0$.
So in terms of $u$, the integral needs to start 
"when $u=-2$" and end "when $u=0$".
A more rigorous treatment would take $x - 2$ as a function over
the domain $[0,2]$ and transform it; but transforming the function
also transforms its domain, so $x - 2$ over the domain $[0,2]$
transforms to $u$ over the domain $[-2,0]$.
Anything that changes the integration variable of a definite integral
also has to be reflected in the limits of integration,
because just as with any substitution, you're integrating a (possibly)
different function over a (possibly) different domain.
That is, if the integrand changes from $f(t)dt$ to $h(v)dv$
(even if $h(v)$ is a constant function, as it is in the question),
the integral over $v$ needs to start and end at $v$-values that are
correctly matched to the $t$-values at which the integral over $t$
started and ended.

Note that in any change of variables, regardless of whether we achieve it
by first writing down an explicit substitution formula (such as $u = x - 2$),
has to account for the derivative of the new variable of integration
with respect to the old one. For a substitution from $t$ to $u$ via the 
equation $u = h(t)$, the derivative of the new w.r.t. the old is 
$\dfrac{du}{dt} = h'(t)$
and it is accounted for in the rule
$$
\int g(h(t))\, h'(t)\, dt = g(u)\, du.
$$
As far as I know, a change of variables must not break this rule,
so there must somehow be a substitution $u = h(t)$ that can explain it.
In the integral in the question, 
$f(t) = m \dfrac{d^2 x}{dt^2}$. 
If $\dfrac{dx}{dt} = v = h(t)$ 
and if $g$ is the constant function with value $m$, 
then $\dfrac{d^2 x}{dt^2} = \dfrac{dv}{dt} = h'(t)$ and $g(h(t)) = m$, so
$$
\int m \frac{dv}{dt} \, dt 
 = \int g(h(t))\, h'(t)\, dt
 = \int g(v)\, dv = \int m \, dv.
$$
The definite integral follows the same rule but also has to make the
corresponding change to the interval of integration:
$$
\int_{t_0}^{t_1} g(h(t))\, h'(t)\, dt
 = \int_{h(t_0)}^{h(t_1)} g(v)\, dv;
$$
setting $v_0=h(t_0)$ and $v_1=h(t_1)$,
$$
\int_{t_0}^{t_1} m \frac{dv}{dt} \, dt 
 = \int_{h(t_0)}^{h(t_1)} m \, dv
 = \int_{v_0}^{v_1} m \, dv.
$$
A: Apart from your prefactors (like mass), when you integrate $dv/dt$ you integrate the derivative of a function, hence we all know the answer is the function itself. Hence: $$\int_{t_0}^{t_1}  v^{\prime}(t)\, dt = v(t_1) - v(t_0)$$
