Why distance between point and image of exponential map is travel time? Let $(M, g)$ be a compact manifold with strictly convex boundary. The distance function $d_{g}: M \times M \rightarrow \mathbb{R}$ is given by
$$
d_{g}(x, y)=\inf _{\gamma \in \Lambda_{x, y}} \ell_{g}(\gamma)
$$
where $\Lambda_{x, y}$ denotes the set of smooth curves $\gamma:[0,1] \rightarrow M$ such that $\gamma(0)=x$ and $\gamma(1)=y$ and
$$
\ell_{g}(\gamma):=\int_{0}^{1}|\dot{\gamma}(t)|_{g} d t.
$$
Now we know that on a simple manifold, the exponential map
$$
\exp _{x}: D_{x} \rightarrow M
$$
is a diffeomorphism. So for any pair of points in $M$ there is a unique geodesic between them, and this geodesic minimizes length. Now if we consider $t v \in D_{x}$ and $|v|_{g}=1$, then I need to show following
$$
f\left(\exp _{x}(t v)\right)=d_{g}\left(x, \exp _{x}(t v)\right)=t.
$$
What I was thinking is following:
By definition:
\begin{align*}
d_{g}\left(x, \exp _{x}(t v)\right)&=\int_{0}^{1}|\dot{\exp_{x}}(tv)|_{g} d t\\
&=\int_{0}^{1}|v|_g|\dot\gamma_{x,v}(t)|_g dt\\
&=\int_{0}^{1}|\dot\gamma_{x,v}(t)|_g dt
\end{align*}
Now $\gamma$ is geodesic so it has unit length so $||\dot\gamma_{x,v}(t)|_g|=|\dot\gamma_{x,v}(0)|_g=|v|_g=1$.
But I need to show that  equals to $t$. where is I am making mistakes
Please help me.
Any help/hint/reference will be highly appreciated.
 A: As pointed out in the comments, your notations for the last integrals are messed up. Also, you used the facts that radial geodesics are locally minimizing and almost finished the computation, but you missed a couple things: first, the integral which appears in the distance equality can be taken from $0$ to $1$ only when the geodesic is not unit speed (so either you drop the assumption that $\|v \| = 1$ or you change the bounds of the integral - you can't do both), and also that geodesics have constant speed. So let us establish that $\gamma_v$ is given by $\gamma_v(u) = \exp_p(u v)$ where $0 \leq u \leq s$. Then
$$\mathrm{dist}(x, \exp_p(s v) ) =  \int_{0}^{s} \| \gamma'_{v}(t) \| \ \mathrm{d} t  =  \int_{0}^{s} \| \gamma'_{v}(0) \| \ \mathrm{d} t = \int_{0}^{s} \| v\| \ \mathrm{d} t = s \cdot \|v\| $$
Notice also that this is only true in the same local sense that geodesics are locally minimizing (i.e, it is only true for small enough $s$) - we crucially used the locally minimizing property of geodesics in the first equality. If $\|v\|$ is of unit length you get the result your desire. Of course, if you really wanted to parameterize the geodesic from $0$ to $1$, you could just define $\alpha: [0, 1] \to M$ by $\alpha_v(t) = \gamma(s \cdot t)$ for all $0 \leq t \leq 1$. Then we would again get:
$$\mathrm{dist}(x, \exp_p(s v) ) =  \int_{0}^{1} \| \alpha'_{v}(t) \| \ \mathrm{d} t  =  \int_{0}^{1} \| \alpha'_{v}(0) \| \ \mathrm{d} t = \int_{0}^{1} s \cdot \| v\| \ \mathrm{d} t = s \cdot \|v\| $$
