Let's assume $A>0$ and $\lambda\neq0$. Evaluate $\int\limits_{0}^{\infty}Ax^2e^{-\lambda^2x^2}dx$. (Hint: Use the properties of the density function of the normal distribution)
Let $Y$ obey $\mathcal{N}(0,\frac{1}{2})$. We know that
$ \begin{align*} &\mathbb{V}(Y)=\frac{1}{\sqrt{2\pi\left(\frac{1}{\sqrt2}\right)^2}}\int\limits_{-\infty}^{\infty}x^2e^{-\frac{x^2}{2\left(\frac{1}{\sqrt{2}}\right)^2}}dx=\frac{1}{2}\\ &\implies \int\limits_{-\infty}^{\infty}x^2e^{-x^2}dx=\frac{\sqrt{\pi} }{2}. \end{align*}$
Because of the symmetry around $0$ we know that $ \int\limits_{0}^{\infty}x^2e^{-x^2}dx=\frac{1}{4}\sqrt{\pi}$.
Hence,
\begin{align*} \int\limits_{0}^{\infty}Ax^2e^{-\lambda^2x^2}dx=\frac{A}{\lambda^3}\int\limits_{0}^{\infty}z^2e^{-z^2}dz=\frac{A\sqrt{\pi}}{\lambda^3 4}. \end{align*}
Is this correct? I am not sure because the sample solution says $\frac{A\sqrt{\pi}}{\lambda^3}$.