Curve segment satisfying $x_{\ell} = x_0 + \int_0^{\ell} \cos (\theta) ds$ and $y_{\ell}= y_0 + \int_0^{\ell} \sin(\theta) ds$ - why?

Question

I want to know why the statement

Then a curve segment of length $\ell$ satisfies $(x_0, y_0)$ and ending at $(x_{\ell}, y_{\ell})$ satisfies $x_{\ell} = x_0 +\int_0^{\ell} \cos(\theta) ds$ and $y_l = y_0 + \int_0^{\ell} \sin(\theta) ds$

is true in context of the following:

Answer 1

\begin{align} \gamma(s) &= \begin{bmatrix} \xi(s) \\ \eta(s) \end{bmatrix} \\ \gamma'(s) &= \begin{bmatrix} \xi'(s) \\ \eta'(s) \end{bmatrix} \tag{tangent vector} \\ |\gamma'(s)| &= 1 \tag{arc length parametrization} \\ \xi'(s)^2+\eta'(s)^2 &= 1\\ \tan \theta &= \frac{\eta'(s)}{\xi'(s)} \tag{slope of tangent} \\ \begin{bmatrix} \xi'(s) \\ \eta'(s) \end{bmatrix} &= \begin{bmatrix} \cos \theta \\ \sin \theta \end{bmatrix} \\ \xi(s) &= x_0+\int_0^{s} \cos \theta \, ds \\ \eta(s) &= y_0+\int_0^{s} \sin \theta \, ds \\ x_{\ell} &= \xi(\ell) \\ y_{\ell} &= \eta(\ell) \\ \end{align}