Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2016 / S7 / 2 / 1

A constant of motion has zero total derivative on every solution, and may explicitly depend on time. For an infinitesimal transformation $\delta q_i=ϵ{\xi }_i$ with $\delta L=ϵdF/dt$, the on-shell first variation is $\delta L=ϵd({\sum }_i{p}_i{\xi }_i)/dt$. Therefore $\boxed{I=\sum _i{p}_i{\xi }_i-F}$ is conserved. Including a time shift $\delta t=ϵ\tau$ gives the corresponding $I={\sum }_i{p}_i{\xi }_i-H\tau -F$. This proves Noether's theorem in its usual point-transformation form. A symmetry of the action up to a boundary term is sufficient; exact invariance of the displayed Lagrangian is stronger than necessary. Discrete transformations by themselves do not give this continuous-charge construction.