Expand $X^a=x^a+\theta\psi^a$, $DX^a=\psi^a-i\theta\dot x^a$, and $g_{ab}(X)=g_{ab}(x)+\theta\psi^c\partial_cg_{ab}$. Performing the <Berezin integral> gives
$$
S=\int dt\left[\frac12g_{ab}\dot x^a\dot x^b
+\frac i2g_{ab}\dot\psi^a\psi^b
+\frac i2(\partial_cg_{ab})\psi^c\dot x^a\psi^b\right].
$$
After a fermionic integration by parts, this is the manifestly covariant expression
$$
\boxed{S=\frac12\int dt\left[g_{ab}\dot x^a\dot x^b-i g_{ab}\psi^a\frac{D\psi^b}{dt}\right]},
\qquad
\frac{D\psi^a}{dt}=\dot\psi^a+\Gamma^a{}_{bc}\dot x^b\psi^c.
$$
Under a <diffeomorphism>, $x$ is a point of $M$, $\psi\in T_xM$ is a tangent vector, $D_t\psi$ is its <covariant derivative along a curve>, and every index is contracted with the <Riemannian metric>; the action is therefore invariant.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 307 / 3 / a / Solution

A connection differentiates a vector field $V(t)$ along a curve $\gamma(t)$ by $D_tV=\nabla_{\dot\gamma}V$. In coordinates, $(D_tV)^a=\dot V^a+\Gamma^a{}_{bc}\dot\gamma^bV^c$.


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 Incoming links: Covariant derivative along a curve