Let $L(\theta,\delta)=(\theta-\delta )^2e^{\frac{(\theta-100)^2}{900}}$ with $X\sim N(\theta,100)$ and $\theta\sim N(100,225)$ find the bayes rule.
I already founded that posterior is $f(\theta|x)\sim N(\frac{9x+400}{13},\frac{900}{13})$
Then I need to calculate
$$\delta^\pi (x)=\frac{E^{f(\theta|x)}[w(\theta)g(\theta]}{E^{f(\theta|x)}[w(\theta)]}$$
where $w(\theta)=e^{\frac{(\theta-100)^2}{900}}$. Thus
$$E^{f(\theta|x)}[w(\theta)g(\theta)]=\int_{-\infty}^\infty \theta e^{\frac{(\theta-100)^2}{900}}\frac{1}{\sqrt{2*900/13}}(\theta-\frac{9x+400} {13})^2d\theta$$
and
$$E^{f(\theta|x)}[w(\theta)]=\int_{-\infty}^\infty e^{\frac{(\theta-100)^2}{900}}\frac{1}{\sqrt{2*900/13}}(\theta-\frac{9x+400} {13})^2d\theta$$
probably should be some simplification, but I could not do.