Integrate both sides with different limits, is it legit? My question arise from physics, however I'm interested in the mathematical justification of the following integration.
In the derivation of a equation of motion with constant acceleration, $a$, (link to page) the author integrate both sides of an equation with different limits.
Is this integration really mathematically correct (my equation $3$)?

\begin{align}
a&=\frac{dv(t)}{dt}\tag 1\\
\implies dv &= a\, dt \tag 2\\
\int_{v_i}^{v} dv &= \int_0^t  a \, dt \tag 3\\
v-v_i&=a(t-0)=at \tag 4\\
v&=v_i+at \tag 5
\end{align}

Aren't we supposed to integrate both sides with the same limits?
 A: First, you shouldn't use the same variable both for the upper limits and the differentials. Thus, I'll make the equation $$\int_{v_i}^{v_f}{\mathrm d v}=\int_0^T{a \mathrm dt}.$$
Note that as $t$ goes from $0$ to $T,$ $v$ also varies correspondingly between $v_i$ and $v_f.$
What's just happened follows from the differential equation $$\mathrm d v =a \mathrm dt.$$ This equation says the differentials are equal for each $v$ corresponding to $t.$ It then follows that their integrals between corresponding intervals would also be equal.
A: Let's start with the claim that the limits should be the same. If we take this quite literally the units do not work out properly. ($t = \frac{l}{t}$) We are then forced to ask what we mean when we say the limits should be the same.
I believe equation (2) is a bit of short hand that requires you to know what's going on behind the scenes. When solving (1) you could integrate both sides with respect to $t$:
$$\int_0^Ta\cdot dt = \int_0^T\frac{dv}{dt}dt$$
at this point you can do a $u$-substitution on the right hand side where $u=v(t)$. This would give us:
$$\int_0^Ta\cdot dt = \int_{v(0)}^{v(T)} dv$$
Another take you have on solving your original problem is that you could start with the indefinite integrals to get:
$$v(t)+A= a t + B$$
$$v(t)= a t + (B-A)$$
you could then solve for $(B-A)$ given that they are constants of integration and you should know $v(t=0)$.
A: This is basically $u$-substitution spelled out in such a way that it reveals the notational quirks in the way we teach calculus. Start with the right-hand side and set $u = v(t)$. Crank the $u$-sub machine in the mechanical way you learned from school and you get the left-hand side. The bounds are secretly exactly the same!
A: When the differential is $dt$, you integrate on $t$ and the bounds are $0$ and $t$.
When the differential is $dv$, you integrate on $v$ and the bounds are the corresponding values of $v$, i.e. $v(0)$ and $v(t)$. Just like with a change of variable.
A: you should write v1=v(0) and vf=v(T)
also on the left side you add velocities, on the right side, you add time-intervalls times accelerations .
trula
