I'm a little confused regarding the way Total work = Change in kinetic energy is derived using calculus. My issue can be seen at 3:26 of this video: https://youtu.be/2dqO4sy4Njg?t=3m20s
Why can the limits of the integral just be changed like that? how did it go from final/initial displacement to final/initial velocity?
Writing the work-KE theorem starting from (1D case for simplicity)
$$\int_{x_i}^{x_f} ma dx$$
is in principle wrong, for this already assume $a$ is a function of position.
The correct way should be
$$\int_{t_i}^{t_f}ma v dt=\int_{t_i}^{t_f}m\frac{dv}{dt}vdt=\int_{t_i}^{t_f}\frac{d}{dt}(\frac{1}{2}mv^2)dt=\frac{1}{2}mv(t_f)^2-\frac{1}{2}mv(t_i)^2$$
There is no need to do change of variable. Always stick to the independent variable $t$, which the integrand must be a function of. Otherwise, if you use $x$ or $v$, you have to worry whether the integrand is a function of them, which is a prerequisite that the change of variable is valid.
The change in limits is called substitution. Initially they assumed a to be a function of x. Therefore the work done was the change In position from x(i) to x(f) . But once they canceled dx the integral was a function of dv. In other words when x ranges from x(i) to x(f) v ranges from v(i) to v (f). This is just a shorter method u can in fact keep the limits as it is and integrate but once we integrate and this is just harder
d E = F dx
F = ma
a = dv/dt use the chain rule , given that dx/dt =v
a = [ dv/dt ] = [ dv/dx][dx/dt]
a = v[ dv/dx] , put all this information into dE
dE = F dx = ma dx = m [(v)(dv/dx)]dx , cancel dx then
dE = mv dv , now integrate and
E(1) - E(0) = [1/2] m v(1)^2 - [1/2]m v(0)^2
