Consider a particle constrained to a manifold $\mathbb Q$ embedded within standard Euclidean 3-dimensional space that experiences no forces other than constraint forces keeping it within $\mathbb Q$. The Lagrangian in such a case is simply the kinetic energy $T$, so therefore the action $\mathcal S$ minimized by the subsequent motion is as follows, where we take the integration to be along the path taken by the particle.
$$\mathcal S= \int dt\ T$$
The resulting motion of a particle will both be (1) a path of stationary length and (2) constant in $T$. I'm am having trouble visualizing why both of these two properties follow from the action integral stated above.
Even in the simplest case, where $\mathbb Q$ is the one-dimensional $x$-axis of the Cartesian coordinate system, the motion will of course be described by $x(t) = x_0 + v_0 t$, but the action integral reads
$$\mathcal S = \frac{m}{2} \int dt\ \dot x^2.$$
It is not clear to me that minimizing $\int dt\ \dot x^2$ implies that $x(t)$ is linear. Heck, I can't even seem to evaluate that antiderivative in terms of $x$, $\dot x$, $\ddot x$, etc. (nor can WolframAlpha, so I'm not sure it can be evaluated with this level of generality.)
Any suggestions on how to make sense of this both in this special case as well as in general? I feel like there must be something simple here I'm missing given how simplified of a case this is. Yes, I could whip out the Euler-Lagrange equations and easily find $m\ddot x = 0$ in this specific case, but I'm looking for as direct and conceptual of a sense of why this is so as possible.