= Solution
{c}
For the local variation $\delta\psi=i\alpha(x)\psi$, $\delta\bar\psi=-i\alpha(x)\bar\psi$, the terms proportional to $\alpha$ cancel and the differentiated phase gives
$$
\begin{aligned}
\delta\mathcal L
&=i(-i\alpha\bar\psi)\gamma^\mu\partial_\mu\psi
+i\bar\psi\gamma^\mu\partial_\mu(i\alpha\psi)
-m\,\delta(\bar\psi\psi)\\
&=-\bar\psi\gamma^\mu\psi\,\partial_\mu\alpha.
\end{aligned}
$$
Thus, for a compactly supported $\alpha$, <integration by parts> gives
$$
\boxed{j^\mu=\bar\psi\gamma^\mu\psi,\qquad
\delta S=-\int d^4x\,j^\mu\partial_\mu\alpha
=\int d^4x\,\alpha\partial_\mu j^\mu.}
$$
This <Dirac current> is the <Noether current> in the convention $\delta S=-\int j^\mu\partial_\mu\alpha$. Using the convention with a plus sign reverses the <Noether current> and the <conserved charge> together. There is one independent <global symmetry in field theory>, with four spacetime components of its <Noether current>. The associated <conserved charge> is $Q=\int d^3x\,j^0$, with vacuum subtraction when required.
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