Let have the Rayleigh density given by $$f(r)= re^{-0.5r^2}$$ where $r\geq0$. Let $Z$ have the standard normal density. Use the mean and variance of $Z$ to find $E[R]$ without any further integration.
I'm having trouble finding the connection between $R$ and $Z$.
On
Since $var(Z) = 1$,$E(Z) = 0$, and $var(Z) = E(Z^2) - E(Z)^2$, we know $E(Z^2) = 1$. That is,
$$\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} z^2 e^{-z^2/2} \; dz = 1$$
Hence $$ \int_0^{\infty} z^2 e^{-z^2/2} \; dz = \sqrt{\frac{\pi}{2}}$$ Replacing $z$ with $r$ in the integral above, we have $$E(R) = \sqrt{\frac{\pi}{2}}$$
You can use the folded normal distribution. Let $Y$ be the folded distributed variable and $Z$ the standard normal distribution. Then $Y=|Z|$. $Y$ has a mean of $\mu_y=\sqrt{\frac{2}{\pi}}$ and a variance of $\sigma_y^2=1-\mu_y^2=1-\frac{2}{\pi}$
Now we use the definition of $E(Y^2)$ in terms of the expected value and variance to calculate it. $$\mathbb E(Y^2)=\sqrt{\frac2{\pi}}\cdot \int_0^{\infty} y^2\cdot e^{-\frac{y^2}{2}} \, dy=\sqrt{\frac2{\pi}}\cdot \int_0^{\infty} y\cdot \underbrace{y\cdot e^{-\frac{y^2}{2}}}_{f_R(y)} \, dy=\sqrt{\frac2{\pi}}\cdot \mathbb E(R)$$
$$\mathbb E(Y^2)=Var(Y)+\mathbb [E(Y)]^2=1-\frac{2}{\pi}+\frac{2}{\pi}=1$$
Finally use the results of the two last lines and deduce $\mathbb E(R)$.