How do you interpret definite-integrating of both sides of an equation? Consider a particle is moving at an acceleration of $a=f(s)$ [$s$ stands for particle's location) if we have the initial velocity of a particle then what is the final velocity?
So... we know that $ads=vdv$ after replacing $a$ with $f(s)$ then: $f(s)ds=vdv$ and now we can integrate the both sides to reach the final velocity as far as I know from calculus, $f(s)ds$ and $vdv$ are the small changes in the functions which have the derivatives $f(s)$ and $v$ respectively and by summing these changes (I mean integrating) how can we maintain the equation?
Can someone explain this this to me rigorously and physically? What is the meaning of functions which have derivative $f(s)$ and $v$?
 A: Let me try and answer your question by applying it to the integration of:
$$ads=vdv.$$
$$f(s)ds=vdv.$$
Multiplying both sides with $m$ (mass of particle), we get:
$$mf(s)ds=mvdv.$$
Here $mf(s)=ma$, so it's the force acting on the particle.
So far this means that if we take the force $mf(s)$ applied to effectuate an infinitesimal displacement $ds$ and multiply with it, then that product equals the instantaneous momentum $p=mv$ multiplied by the infinitesimal change in $v$, i.e. $dv$.
The meaning of the equation will become clearer on integration:
$$\int_0^s mf(s)ds=m\int_0^s f(s)ds=\int_0^vmvdv=m\int_0^vvdv=\frac{mv^2}{2}.$$
(I've assumed the initial and end state to be resp. $0,0$ and $s,v$ and constant $m$)
So now it should be clear that this is a work - kinetic energy relation:
$$W = K,$$
With:
$$W=m\int_0^s f(s)ds.$$
And:
$$K=\frac{mv^2}{2}.$$
So the work $W$ performed by $f(s)$ on the particle equals the change in kinetic energy $K$ of the particle. Since as we've assumed no friction or drag, energy is fully conserved.
