now, if they did the harder version and left it more-or-less as is, I'm fine with taht
problem is, some people tried to have it both ways
to wit: You've got $$\Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{i (kx-\hbar k^2 t/2m)}\phi(k)\,dk$$
that's the general case, at least.
the special case is when, instead of taking a sum over infinitely many wavenumbers $k$, you just suppose that there's a single $k$ and so write $\Psi(x,t)=Ae^{i(kx-\hbar k^2 t/2m)}$
in that case, it's simple to show that the probability current, defined as $$J(x,t)=\frac{i\hbar}{2m}\left(\Psi^\star \frac{\partial \Psi}{\partial x}-\Psi \frac{\partial \Psi^\star}{\partial x}\right)$$
(might be off by a minus sign, blarg)