# Basic Grassmann/Berezin Integral Question

Is there a reason why $\int\! d\theta~\theta = 1$ for a Grassmann integral? Books give arguments for $\int\! d\theta = 0$ which I can follow, but not for the former one.

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If the integral $I:=\int d\theta$ on the algebra ${\cal A}$ of superfunctions $f(\theta)=\theta a + b$ should be

2) translation invariant, i.e., $\int d\theta ~f(\theta+\theta') =\int d\theta~f(\theta)$,
3) and if the output $\int d\theta~ f(\theta)$ should not depend on the integration variable $\theta$,
Interestingly, this means that Berezin integration $\int d\theta$ is the same as differentiation $\partial / \partial\theta$.