Rules for integrating a Grassmann quantity So I've read that for Grassmann numbers, integration is the same as differentiation. Indeed, on Wikipedia, integration of Grassmann quantities are defined so this is true:
$$\int 1 \, d \theta = 0$$
$$\int \theta \, d \theta = 1$$
The article claims that these rules follow from requiring linearity and the so-called partial integration formula:
$$\int \left[ \frac{\partial}{\partial \theta} f(\theta) \right] d \theta = 0$$
but I don't see how this works. 
The first rule seems to make sense, since if $f(\theta) = \theta$, then according to the partial integration formula, we should have
$$\int \left[ \frac{\partial}{\partial \theta} \theta \right] d \theta = \int 1 \, d \theta = 0$$
On the other hand, if $f(\theta) = \theta^2 = 0$, then the partial integration formula should give:
$$\int \left[ \frac{\partial}{\partial \theta} \theta^2 \right] d \theta = \int 2 \theta \, d \theta = 0$$
and so $\int \theta \, d \theta = 0$.

To answer your example, your taking of the derivative is incorrect: to differentiate a product of Grassmann variables, the rule is
$$ \partial_{\theta} (\alpha \beta) = (\partial_
{\theta} \alpha )\beta - \alpha (\partial_{\theta} \beta), $$
because (e.g. when thinking of the derivative as a difference quotient) we have to commute the $\theta$ and the $\alpha$ to move the derivative quotient to $\beta$. Then if $\alpha=\beta=\theta$, you find
$$ 0 = \partial_{\theta} (\theta \theta) = (\partial_
{\theta} \theta )\theta - \theta (\partial_{\theta} \theta) = \theta-\theta=0, $$
consistently.
