Diagonalize $T$ with eigenvalues $\lambda_a$ and use the periodic <Fourier series>
$$
\eta^a(t)=\sum_{k\in\mathbb Z}\eta_k^ae^{2\pi ikt},
\qquad
\bar\eta^a(t)=\sum_{k\in\mathbb Z}\bar\eta_k^ae^{-2\pi ikt}.
$$
Each <Berezin integral> contributes its quadratic coefficient, so
$$
Z=\prod_{a=1}^n\prod_{k\in\mathbb Z}(2\pi ik+\lambda_a).
$$
Pairing $k$ with $-k$ and using the infinite product for the hyperbolic sine gives, up to the local regularization factor,
$$
\prod_{k\in\mathbb Z}(2\pi ik+\lambda)
=2\sinh\frac\lambda2
=e^{\lambda/2}(1-e^{-\lambda}).
$$
The product of the prefactors is $e^{\operatorname{tr}T/2}=1$ because $T$ is traceless. Thus
$$
\boxed{Z=\prod_a(1-e^{-\lambda_a})=\det(1-e^{-T})}.
$$
This agrees with the direct identity that the <supertrace> of an induced linear map on an exterior algebra is its characteristic determinant.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 307 / 2 / d / Solution

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.

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

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The Berezin integral is differentiation with respect to a Grassmann variable: $\int d\theta\,1=0$ and $\int d\theta\,\theta=1$. A finite-dimensional fermionic Gaussian integral gives a determinant or Pfaffian.


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