Let us consider for example, the riemannian metric $g=e^xdx^2+dy^2$ (it is symmetric and definite positive), with associated matrix $\begin{pmatrix} e^x & 0\\ 0 & 1 \end{pmatrix}$.
Consider the vector $(1,1)$.
Then $||(1,1)||=d((1,1),(0,0))=\int_0^1 \sqrt{g(\gamma',\gamma')}dt$, where $\gamma(t)=(t,t)$.
NB: Here I'm using the definition of length of a curve, and the curve is one that passes through $(0,0)$ and $(1,1)$. I do not know if the result depends on the chosen curve...
As $\gamma'(t)=(1,1)$ we get $||(1,1)||=\int_0^1\sqrt{e^t\cdot1^2+1^2}dt=\ldots$
Is there a fast way to do this? Thanks!
This is actually a good question and weirdly not mentioned explicitly very often.
A Riemannian manifold $(M,g)$ has a tangent space $T_pM$ defined for every point $p\in M$, which is a vector space attached to each $p$. For instance, a 2D manifold in 3D has a tangent plane associated to every point of the surface (so $T_pM=\mathbb{R}^2$). The Riemannian metric $g$, only defines distances and angles in the tangent space. So $g$ doesn't let you compute distances directly, it only lets you compute infinitesimal distances.
In particular, if you have two tangent vectors $v$ and $u$, then the "distance" between them is defined by $g$ via: $$ d(u,v) = g_{ij}v^i u^j $$ or, in matrix terms, $ d(u,v)=u^Tgv $. (For you, $g=\text{diag}([e^x,1])$).
In Euclidean space, $g=I$ or $g_{ij}=\delta_{ij}$, so this reduces to the standard inner product, the dot product. This is why people say that $g$, the metric tensor, defines an inner product on the manifold (it actually defines one on each $T_pM$).
Now to norms. For any metric $d(p_1,p_2)$ you can define a norm $||s||=d(s,s)$. Riemannian manifolds are no different. So we define: $$ ||v||_g^2 = d(v,v) = g_{ij}v^iv^j = v^T g v $$ as our Riemannian norm.
Also, it turns out computing "infinitesimal" distances in $T_pM$ lets us define "actual" distances between points in $M$. Let $\gamma(t)$ be a curve on $M$. To get the length of $\gamma$, you just sum up the infinitesimal distances: $$ \ell_g(\gamma;a,b) = \int_a^b ||\gamma'(t)||_g dt =\int_a^b \sqrt{g_{ij} \dot{\gamma}(t)^i\dot{\gamma}(t)^j} dt $$ Now you can define the distance between $p$ and $q$, where $p,q\in M$, as the length of the shortest curve between them: $$ D(p,q) = \inf_{\gamma} \ell_g(\gamma;p,q) $$