Principle of Least Action Question Let's say we have a particle with no forces on it. The path that this classical particle takes is the one that minimizes the integral 
$$\frac{1}{2}m\int_{t_i}^{t_f}v^2dt.$$ 
So if we graph this for the actual path a particle takes it is a straight, horizontal line on the $(t,v^2)$ plane. But couldn't we lessen the integral if we first slow down and then speed up near the end to create a sort of parabolic line that has less area under the $(t,v^2)$ plane? So why doesn't the particle take this path? What am I missing in my thinking?
 A: You have to minimize the integral subject the the constraint that the initial and final positions $x(t_i)$ and $x(t_f)$ are held fixed.  In particular, $\Delta x = \int_{t_i}^{t_f} v(t)\, dt$ is held fixed.  If the particle slowed down than sped up as you suggested, the action would be less, but it wouldn't have a high enough average speed to cover the full $\Delta x$ in time.  You can play around with a few specific trajectories and check for yourself.
A: You're missing that Dirichlet boundary conditions 
$$ x(t_i)~=~x_i  \quad\text{and} \quad x(t_f)~=~x_i $$
are implicitly implied. The stationary action principle is not well-posed without boundary conditions. 
A: Your proposed path has a VERY LARGE action. As @tparker pointed out, you have to minimize the path subject to the constraint that the average velocity doesn't change. Now, the action is quadratic in velocity. A little fiddling around with the math should convince you that to minimize the integral of $v^2$ subject to the constraint that the average velocity doesn't change, you want the velocity to be constant. Intuitively, going fast for a bit raises the action a LOT (because velocity is squared) while going slow for a bit doesn't do much to lower the action (because velocity is squared).
