I am interested in the value of the integral:
$$\int_{-1}^2 x^\prime(t)(1 + t^2x^\prime(t)) dt$$
subject to the constraint that $x$ is a piecewise smooth function $[-1,2]$ and $x(-1) = 1$ and $x(2) = 1$.
For clarity, $x$ being piecewise smooth on $[-1,2]$ means that $x$ is continuous on $[-1,2]$, has a finite number of corner points on $[-1,2]$, $x$ is $C^1$ away from those corner points, and at $-1$, $2$, and any corner points, the left- and/or right-hand derivatives of $x$ exist (but of course may not be equal).
I see that $x(t) = 1$ for all $t \in [-1,2]$ satisfies the constraints and gives the integral a value of $0$. However, I suspect that $x(t) = 1$ is a global minimum of this integral. That is:
$$\int_{-1}^2 x^\prime(t)(1 + t^2x^\prime(t)) dt \geq 0$$
for all $x$ satisfying the constraints. However, I haven't made much progress proving this.
For context, this is a problem I encountered in calculus of variations.
$\int_{-1}^2 x^\prime(t)(1 + t^2x^\prime(t)) dt = \int_{-1}^2 x^\prime(t)dt + \int_{-1}^2t^2{\left(x^\prime(t)\right)}^2 dt \geq \int_{-1}^2 x^\prime(t)dt = x(2)-x(-1) = 0.$ The inequality is due to the fact that $t^2{\left(x^\prime(t)\right)}^2 \ge 0$ for $-1\le t\le 2$. Precise reasoning of $\int_{-1}^2 x^\prime(t)dt = x(2)-x(-1)$ would require enumeration of your corner points, but the result will hold.