Sorry, its been a while and my calculus was never good. This is really a very elementary question which I am unable to un -complicate from its shroud of notation. My difficulty is how does this parametrization thing take place,i.e how does the integral in terms of $t$ come to be, and how to convince myself that the coordinates of any curve in D dimensions can be written as single valued functions of a single parameter.
If $dl^2 = \delta_{ij} dx^i dx^j$ then why is $\int_{P_1}^{P_2} dl = \int_{t_1}^{t_2} \sqrt{\delta_{ij} \dot{x}^i \dot{x}^j} dt\qquad \dot{x}^i\equiv \frac{dx^i}{dt}$
Additions. $x^i$ are the rectangular coordinates in Euclidean space. So a position vector $\vec{r} = ( x^1,x^2,\cdots x^D)$. $\delta_{ij}$ is the Euclidean metric tensor defined in the same way as the Kronecker delta. An infinitesimal line element is defined as $dl^2 = \delta_{ij} dx^i dx^j$. $P_1$ and $P_2$ are two points in this space then the length of some curve $C$ between $P_1$ and $P_2$ $\Delta L =\int_{P_1}^{P_2} dl$
After this, the problematic bit is that "A curve in D dimensional euclidean space can be described as a subspace of the D dimensional space where the D coordinates $x^i$ are given by single valued functions of the parameter $t$. So this can be written as $P_1 = x(t_1)$ and $P_2 = x(t_2)$
$\Delta L =\int_{t_1}^{t_2} \sqrt{\delta_{ij} \dot{x}^i \dot{x}^j} dt\qquad \dot{x}^i\equiv \frac{dx^i}{dt}$
I can relate to what I did in two dimensions for parametrizing a circle in a plane by $x^0 = R \cos t$ and $x^1 =R\sin t$, but I just reasoned by taking the derivative as a fraction and cancelling numerators and denominators.
Note on convention: $\displaystyle g^i_j a_ib^j = \sum_{1\leq i\leq D}\quad \sum_{1\leq j\leq D}g^i_j a_i b^j$