I'm reading a paper and I have a doubt: Suppose we have a compact Riemannian manifold and an eigenfunction of the Laplace operator $\Delta u=-\lambda u$. Suppose in addition that we have the following gradient estimate $$ \frac{|\nabla u|}{\sqrt{2-u^2}}\leq \sqrt{\lambda}. $$ The paper says: Take now two points $x_1$ and $x_2$ such that $u(x_1)=0$ and $u(x_2)=1$ and connect them by the shortest geodesic $\gamma(t)$ with $|\dot\gamma(t)|=1$. Then, integrating over $\gamma$, we have \begin{align} \int_0^1\frac{du}{\sqrt{2-u^2}} & \leq \int_{\gamma}\frac{1}{\sqrt{2-u^2}}|\frac{du}{dt}|dt\\ & \leq \int_{\gamma}\sqrt{\lambda}dt\\ & \leq \sqrt{\lambda}d. \end{align} Could someone please, explain me the reason why are those the limits of integration? I think my doubt is because I'm confuse about the definition of "integration over a curve".
My attempt was the following: I think that we should take a parametrization of $\gamma:[0,l]\to M$ and then take the composition $\varphi(t):=\arcsin(\frac{u(\gamma(t))}{\sqrt 2})$. We would get
\begin{align} |\dot \varphi(t)| & =\Big|\frac{1}{\sqrt{2-u(\gamma(t))^2}}\Big\langle\nabla u_{\gamma(t)},\dot\gamma\rangle\Big|\\ & \leq \frac{|\nabla u_{\gamma(t)}|}{\sqrt{2-u(\gamma(t))^2}} \end{align} Then we use the Fundamental Theorem of Calculus and previous inequality to estimate the result \begin{align} |\varphi(l)-\varphi(0)| & =|\int_0^l\dot \varphi(s)ds|\\ & \leq \int_0^l |\dot \varphi(s)|ds\\ & \leq \int_0^l|\dot \varphi(s)|ds\\ & \leq \int_0^l\frac{|\nabla u|}{\sqrt{2-u^2}}dt\\ & \leq \int_0^l\sqrt \lambda dt. \end{align} Which would give a wrong answer. Could someone please, point out to me what I am doing wrong?