Suppose we have an arclength parametrization of a curve in the $xy$-plane given by $x(s)$, $y(s)$ where $0 \leq s \leq L$. We want to integrate a scalar function $f(x,y)$ along this line. Since we are working with an arclength paraemtrization it seems that the integral to evaluate should intuitively be
$ \int_0^L f(x(s),y(s)) ds$.
Going through the formalism though we have that the integral is
$ \int_0^L f(x(s),y(s)) \sqrt{x'(s)^2 + y'(s)^2} ds$.
Is the first representation of the integral correct as well? If so, is there a way of seeing why $\sqrt{x'(s)^2 + y'(s)^2} = 1 $ for an arclength parametrization?
It seems that your problem is the following: Given a parametrization $$\gamma:\quad t\mapsto\bigl(x(t),y(t)\bigr)\qquad(0\leq t\leq T)\ ,\tag{1}$$ how can I recognize that it is in fact arc-length-parametrization? Whether or not $t$ is arc length we can compute the arc length $s(t)$ of the piece of $\gamma$ described in the time interval $[0,t]$ by means of the following formula: $$s(t)=\int_0^t\sqrt{x'^2(\tau)+y'^2(\tau)}\ d\tau\qquad(0\leq t\leq T)\ .$$ When $t$ is in fact arc length then $s(t)\equiv t$, which implies $$s'(t)=\sqrt{x'^2(t)+y'^2(t)}\equiv1\ .$$