Supersymmetry transformation: why does the Lagrangian transform as total derivative? There is something I don't understand at page 36 of these lecture notes (Author: Fiorenzo Bastianelli from the university of Bologna, title: Path integrals for fermions and supersymmetric quantum mechanics.)
I'll summarize it here but I linked them anyway in case someone want to check them. 
So we're trying to build a supersymmetric action, we work in the super space $D=1$ and $N=2$ with one spacetime coordinate $t$ and two Grassman coordinates $\theta$ and $\bar{\theta}$.
The generator of time translation is 
$$H= i \frac{\partial}{\partial t}$$ 
The generators of supersymmetry transformation, that are translations in the anticommuting directions are 
$$Q= \frac{\partial}{\partial \theta} + i \bar{\theta}  \frac{\partial}{\partial t} $$ and 
$$\bar{Q}= \frac{\partial}{\partial \bar{\theta}} + i \theta  \frac{\partial}{\partial t} $$
We define a scalar, Grassman even superfield $X(t,\theta, \bar{\theta})$ which, under supersymmetry transformation transforms in this way 
$$\delta_S X(t,\theta, \bar{\theta}) = (\epsilon \bar{Q} + \bar{\epsilon} Q)\, X(t,\theta, \bar{\theta}) $$
With $\epsilon$ and $\bar{\epsilon}$ Grassmann parameters.
Now we define covariant derivatives
$$
D= \frac{\partial}{\partial \theta} - i \bar{\theta}  \frac{\partial}{\partial t}$$
$$
\bar{D}= \frac{\partial}{\partial \bar{\theta}} - i \theta  \frac{\partial}{\partial t}
$$
so that the covariant derivative of a superfield is still a superfield, which means
$$
\delta_S DX = (\epsilon \bar{Q} + \bar{\epsilon} Q)\, DX
$$
All commutators and anticommutators are null beside these ones 
$$ \{ Q,\bar{Q} \} = 2H$$ $$\{D,\bar{D}\} = -2i \partial_t$$
Now we Say that a Lagrangian $L=L(X,DX,\bar{D}X)$ that depends only implicitly on the coordinates of the superspace through the superfield and its covariant derivatives can give you a supersymmetric action. And this is  because it transforms under supersymmetry transformation as a total derivative. The exact form of the Lagrangian variation under supersymmetry transformation is this:
$$
\delta_S L(X,DX, \bar{D}X) = (\epsilon \bar{Q} + \bar{\epsilon} Q) \, L(X,DX, \bar{D}X)
$$
Now the things I don't understand are these two:


*

*Why does the Lagrangian transforms like that under supersymmetry transformation? I am not able to prove it, I can provide a sketch of my attempt of working out its transformation if requested, but it doesn't really anything I think.

*Assuming that's the right transformation law of the Lagrangian, why is that a total derivative? It looks to me that it just transforms like a super field, but I don't see why that's a total derivative.
 A: *

*Use that the covariant derivatives $D$ and $\bar{D}$ anticommute with the generators $Q$ and $\bar{Q}$ of SUSY$^1$
$$ \delta_S L
~=~\delta_SX~ \frac{\partial_L L}{\partial X} 
+D\delta_SX ~\frac{\partial_L L}{\partial DX} 
+\bar{D}\delta_SX~ \frac{\partial_L L}{\partial \bar{D}X} $$
$$~=~(\epsilon \bar{Q} + \bar{\epsilon} Q)X~ 
\frac{\partial_L L}{\partial X} 
+D(\epsilon \bar{Q} + \bar{\epsilon} Q)X ~
\frac{\partial_L L}{\partial DX} 
+\bar{D}(\epsilon \bar{Q} + \bar{\epsilon} Q)X~ 
\frac{\partial_L L}{\partial \bar{D}X} $$
$$~=~(\epsilon \bar{Q} + \bar{\epsilon} Q)X~ 
\frac{\partial_L L}{\partial X} 
+(\epsilon \bar{Q} + \bar{\epsilon} Q)DX ~
\frac{\partial_L L}{\partial DX} 
+(\epsilon \bar{Q} + \bar{\epsilon} Q)\bar{D}X~ 
\frac{\partial_L L}{\partial \bar{D}X} 
~=~ (\epsilon \bar{Q} + \bar{\epsilon} Q)L.$$

*In the SUSY varied action $$\delta_SS~=~\int \!\mathrm{d}t~\mathrm{d}\theta~\mathrm{d}\bar{\theta}~(\epsilon \bar{Q} + \bar{\epsilon} Q)L $$ perform the Grassmann-odd Berezin integrations (which are the same as Grassmann-odd differentiations) to see that only a total time-derivative survives. (Recall that if we differentiate wrt. the same Grassmann-odd variable twice we get zero.)
--
$^1$ The subscript "$L$" on a partial derivative indicates left  derivatives, i.e. a differentiation acting from left. 
