= Solution
For the <wavefunction> of a particle of <mass> $m$, the <probability density> and <probability current> are
$$
\boxed{\rho=\Psi^*\Psi,\qquad j=\frac{\hbar}{2mi}(\Psi^*\Psi_x-\Psi\Psi_x^*)=\frac\hbar m\operatorname{Im}(\Psi^*\Psi_x).}
$$
For a real potential $V$, the <Time-dependent Schrödinger equation> and its <complex conjugate> read
$$
i\hbar\Psi_t=-\frac{\hbar^2}{2m}\Psi_{xx}+V\Psi,\qquad -i\hbar\Psi_t^*=-\frac{\hbar^2}{2m}\Psi_{xx}^*+V\Psi^*.
$$
Multiplying by $\Psi^*$ and $\Psi$ respectively yields
$$
\rho_t=\frac{i\hbar}{2m}(\Psi^*\Psi_{xx}-\Psi\Psi_{xx}^*)=-j_x.
$$
The potential terms cancel because $V$ is real. This is the <continuity equation>, expressing local conservation of probability.
For the given <superposition> the common time phase cancels in $\Psi^*\Psi_x$, and
$$
\Psi^*\Psi_x=ik(1-|R|^2)+ik(R^*e^{2ikx}-Re^{-2ikx}).
$$
The expression in the second parentheses is purely imaginary, so the second term is real and contributes nothing to the <probability current>. Thus
$$
\boxed{j=\frac{\hbar k}{m}(1-|R|^2).}
$$
It is the incoming <probability current> minus the reflected <probability current>.
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