Nick89 
Quote by Nick89
We might have a different print of Griffiths, but I cannot find the second conclusion in bold in my book, nor does it make much sense to me.

The first quote (as you know) is from Griffiths)
Sorry  the second was found on a professor's web notes  University of Texas I think  where I was searching for further help on this topic.
Nick89, Cyosis, Bob_for_short, muppet and others 
followup question 
I agree that if you "know" that [tex] \psi^2[/tex] is a probability density (or behaves like a probability density  I know wording can be sticky) then the area under the curve has to be 1.0.
 If you didn't know how [tex][\psi^2] [/tex]behaves, could you make the statements:
“But [tex] \psi[/tex] must go to zero as x goes to infinity."  Griffiths
“The above equation is satisfied provided goes to zero as goes to zero."  some professor's notes
by seeing this alone: [tex] \frac {d}{dt} \int_ {\infty}^{\infty}\psi^2 dx= \frac{ih}{2m}(\psi*\frac{d\psi}{dx}  \frac{\psi*}{dx}\psi ) [/tex]
Meaning by analysis alone  does the math tell me that “But [tex] \psi[/tex] must go to zero as x goes to infinity." (without knowing the probability density behavior)?
Does my question make sense?  I don't see how the statement can be made "by inspection of the resulting "derivation".
Thanks