I have been given the following problem:
Investigate whether the functional $I=\int_{t_1}^{t_2}t\dot{x}^2dt$ is invariant under the transformation $\bar{t}=t+\epsilon$ and $\bar{x}=x$, with $\epsilon$ being an arbitrary constant.
All I have so far is the formula: $$ L(\bar{t},\bar{x},\bar{\dot{x}})d\bar{t}=L(t,x,\dot{x})dt+d\Phi(t,x,\alpha_s)\\ $$with$$ \Phi(t,x,\alpha_s)=\Phi_s(t,x)\alpha_s $$Where $\alpha_s$ is the transformation parameter.
I am also given: The Function $\Phi_s$ can also be zero, in which case we have invariance in its usual meaning.
So, am I correct in saying that in order for me to investigate whether or not the functional is invariant, I need to solve for $I$ and $\bar{I}$, with $\bar{I}=\int_{t_1}^{t_2} L(\bar{t},\bar{x},\bar{\dot{x}})d\bar{t}$?
In which case, this functional is not invariant because $\bar{t}$ has an arbitrary constant, whilst $t$ does not.
Well, given a variational problem $$ J[x(t)]=\int_{\Omega}L(t,x(t),\frac{dx}{dt}(t))dt $$ a Noether symmetry or divergence symmetry is a is a one-parameter local group of transformations $$ t'=t'(t,x,s)\quad \quad x'=x'(t,x,s) $$ such that $$ L\left(t',x',\dot{x}'\right)dt'=L\left(t,x,\dot{x}\right)dt+d\Phi(t,x,s). $$
It is the same that you have written. But I don't get why you use $\epsilon$ for the parameter and then change to the strange $\alpha_s$.
To see that your transformation is not a Noether symmetry, observe that $\bar{\dot{x}}=\dot{x}$ and then $$ L(\bar{t},\bar{x},\bar{\dot{x}})d\bar{t}=\bar{t}\bar{\dot{x}}^2d\bar{t}=(t+\epsilon) \dot{x}^2 dt= $$ $$ =t\dot{x}^2 dt+\epsilon\dot{x}^2dt=L(t,x,\dot{x})dt+\epsilon\dot{x}^2dt, $$ and it can be shown that the $\epsilon\dot{x}^2dt\neq d \Phi$ for any $\Phi$, since $ \epsilon\dot{x}^2dt $ is not closed.