Given a (finite dimensional) Riemannian manifold $(M,g)$, a given point $x_0\in M$ and a positive number $\beta$, is the function $x\in M\mapsto e^{-\beta d(x,x_0)}$ necessarily integrable with respect to the Riemannian volume on $M$? Here, $d$ is the distance inherited from the Riemannian metric.
In dimension 1, one can do all the computations and the answer is positive. Of course, when the manifold is compact, the answer is also positive. Now, assume that the exponential map at $x_0$ is a diffeomorphism on the whole tangent space at $x_0$ and let $f$ be the density of the pullback measure of the Riemannian volume by the exponential map at $_0$. In that case, one simply needs to consider integrability of the function $e^{−\beta \|u\|}f(u), u\in T_{x_0}M$, with respect to the Lebesgue measure (here, $\|\cdot\|$ is the Euclidean norm in $T_{x_0}M$ inherited from the metric $g$. I don't see why $f$ couldn't have exponential growth, but don't have counterexamples in mind.
There is no reason for $e^{-\beta d(\cdot,x_0)}$ to be an integrable function: the Riemannian volume form can grow faster than that (in some sense that should be clear after the study below).
Consider the $(n+1)$-dimensional hyperbolic space $\Bbb H^{n+1}$. Let $x_0\in \Bbb H^{n+1}$ be arbitrary and let $r(x) = d(x,x_0)$ be the Riemannian distance to $x_0$. $\DeclareMathOperator{\d}{d} \newcommand{\dx}{\d \!} \newcommand{\vol}{\mathrm{vol}}$ In Riemannian polar coordinates $\Bbb H^{n+1}\setminus \{x_0\} \simeq (0,+\infty) \times S^n$, the hyperbolic metric reads $$ g_{\Bbb H^{n+1}} = \dx r^2 + \sinh^2(r) g_{S^n}, $$ where $g_{S^n}$ is the round metric of the unit sphere. The Riemannian volume form then reads $$ \dx\mu_{\Bbb H^{n+1}} = \sinh^n(r) \dx r\wedge \dx\mu_{S^n}. $$ Hence, $$ \int_{H^{n+1}} e^{-\beta r} \dx\mu_{\Bbb H^{n+1}} = \vol(S^n)\int_0^{\infty} e^{-\beta r}\sinh^n(r)\dx r. $$ Note that $$ e^{-\beta r} \sinh^n(r) \underset{r\to \infty}{\sim} \frac{1}{2^n}e^{(n-\beta)r}, $$ from which it follows that $e^{-\beta r}$ is $\mu_{\Bbb H^{n+1}}$-integrable if and only if $\beta > n$.
Here is an example for which $\exp(-\beta r)$ is not integrable for any $\beta$. Consider $\Bbb R^{n+1}$ with the metric in polar coordinates $$ g = \dx r^2 + \sinh^2(r^2) g_{S^n}. $$ Its Riemannian volume form is then $$ \dx \mu_g = \sinh^n(r^2) \dx r \wedge \dx \mu_{S^n}. $$ Notice that $e^{-\beta r} \sinh^n(r^2) \sim_{r \to \infty} \frac{1}{2^n}e^{nr^2-\beta r}$, which is never integrable, so that $$ \int_{\Bbb R^{n+1}} e^{-\beta r} \dx \mu_g = \vol(S^n) \int_0^{\infty} e^{-\beta r} \sinh^n(r^2) \dx r = +\infty. $$ In this example, the curvature is very negative, in the sense that the curvature is rapidly decaying to $-\infty$ with the geodesic distance $r$. The volume of geodesic balls has growth strictly bigger than exponential.