In the paper "The Berry-Keating operator on a lattice" by J. Bolte, S. Egger, and S. Keppeler (this can be found on arXiv), they perform an action calculation of an orbit with $2E=\xi^2-x^2$ in the fundamental domain $(-\frac{\ell_x}{2},\frac{\ell_x}{2})\times (-\frac{\ell_\xi}{2},\frac{\ell_\xi}{2})$. They then go on to find the action, $S_p$ of a periodic orbit $p$. The periodic orbit consists of two sections of the hyperbola, $2E=\xi^2-x^2$, between the points $(x_{-},\ell_\xi/2)$ and $(x_{+},\ell_\xi/2)$, and $(x_{+},-\ell_\xi/2)$ and $(x_{-},-\ell_\xi/2)$, where $x_{\pm}=\pm\sqrt{\ell_\xi^2/4+2E}$ (which I believe to be $x_{\pm}=\pm\sqrt{\ell_\xi^2/4-2E}$). The result they get is: $$ S_p = 2\int_{x_{-}}^{x_{+}} \sqrt{x^2+2E}\ dx +\ell_\xi(x_{+}-x_{-}).$$
I am a bit confused by this: I understand where the $\ell_\xi(x_{+}-x_{-})$ comes from since the area of $[x_{-},x_{+}]\times[-\ell_\xi/2,\ell_\xi/2]$ is the lengths of the intervals multiplied together. Then to find the area enclosed we calculate the area underneath the orbit and find the difference. I was told that to find the action you integrate from left to right and if the orbit opposes your direction you take the negative of this. So the top integral is $$\int_{x_{-}}^{x_{+}}\sqrt{x^2+2E}dx$$ (since the orbit is in the direction of the way you take the integral). This is the area underneath the top hyperbola, and above the $y=0$ axis. For the other hyperbola we again go from left to right but since the orbit direction opposes this, we get a minus, as well as the minus from the branch of the solution to $2E=\xi^2-x^2$. Thus, $$-\int_{x_{-}}^{x_{+}}(-\sqrt{x^2+2E}dx)=\int_{x_{-}}^{x_{+}}\sqrt{x^2+2E}dx.$$ This is then the area between the bottom hyperbola and the line $y=0$. So the area enclosed by the orbits is: $$ \begin{split} S_p &= \ell_\xi(x_{+}-x_{-}) - \left(\int_{x_-}^{x_+}\sqrt{x^2+2E}\ dx + \int_{x_-}^{x_+}\sqrt{x^2+2E}\ dx \right) \\ &= \ell_\xi(x_+ - x_-) - 2\int_{x_-}^{x_+} \sqrt{x^2+2E}\ dx. \end{split} $$ I've tried to figure this out but I can't seem to get the areas to contribute a positive sign. Am I missing something or is the paper wrong?