Question

\[f(u)=A_0e^{-u^2/2}, g(u)=2uA_1e^{-u^2/2}\]
Where both A's have a value so that\[\int\limits_{-\infty}^{\infty}f(u)du=\int\limits_{-\infty}^{\infty}g(u)du=1\]
Show that\[A_1=\frac{A_0}{\sqrt{2}}\]
From where I'm reading, it says: "Do not evaluate at A_0, but give a value for the particle'smean square position when in the second (quantum) state: \[=\int\limits_{-\infty}^{\infty}u^2g(u)^2du\]"