Bayesian statistical inference problem continuous case This question is about updating distribution of unknown random variable with observed data x. I took a photo of what I have done and my reasoning. I am newbie to probability and statistics and I do not know every term. My problem is with denominator of posterior pdf. I do not understand how to find it and incorporate realized value of X into it.
the problem
 A: Let $\Theta \sim \mathcal{U}[2,5]$ and assume the conditional distribution of $X_1,X_2$ given $\Theta=\theta$ is $\mathcal{U}[0,\theta]$. Let's derive an expression for your posterior pdf $f_{\Theta|X_1,X_2}(\theta|x_1,x_2)$ for arbitrary observations $x_1,x_2$. Set $m=\max\{x_1,x_2\}$. Then your likelihood $f_{X_1,X_2|\Theta}(x_1,x_2|\theta)$ is  \begin{equation*}
f_{X_1,X_2|\Theta}(x_1,x_2|\theta)=f_{X_1|\Theta}(x_1|\theta)f_{X_2|\Theta}(x_2|\theta)= \left\{
        \begin{array}{ll}
            1/\theta^2 & \quad \theta\in [m,\infty) \\
            0 & \quad \theta \in (-\infty,m)
        \end{array}
    \right.
\end{equation*} Notice that product $f_{X_1,X_2|\Theta}(x_1,x_2|\theta)f_{\Theta}(\theta)\neq 0$ only when $\theta \in [2,5] \cap [m,\infty)$ whose intersection is non$-$empty whenever $m \leq 5$. For our practical purposes let's assume $2<m<5$ so that $[2,5]\cap [m,\infty)=[m,5]$. Then your posterior pdf $f_{\Theta|X_1,X_2}(\theta|x_1,x_2)$ equals \begin{equation*}
f_{\Theta|X_1,X_2}(\theta|x_1,x_2)=\frac{f_{X_1,X_2|\Theta}(x_1,x_2|\theta)f_{\Theta}(\theta)}{\int_{m}^{5}f_{X_1,X_2|\Theta}(x_1,x_2|\theta)f_{\Theta}(\theta)d\theta}= \left\{
        \begin{array}{ll}
            \frac{5m}{(5-m)\theta^2} & \quad \theta\in [m,5] \\
            0 & \quad \theta \notin [m,5]
        \end{array}
    \right.
\end{equation*} If we take $m \rightarrow 5^{-}$ then this posterior pdf  becomes $\delta(\theta -5)$ where $\delta$ is the dirac$-$delta function. This is what you get when your observations are $x_1=3,x_2=5$.
