Let us consider uniformly distributed generated nummbers in $[0, \vartheta]$, where $\vartheta > 0$ is unknown.
Thus, we have that $\mathbb{X}_n = [0,\infty)^n, \mathbb{P}_{\vartheta} = Uniform([0, \vartheta]), \mathbb{P}_{\vartheta}(A) = \frac{1}{\vartheta} \int_A 1 dx$.
Now let us consider the point estimators $$T_n(x) = \frac{2}{n}\sum_{k=1}^n x_k$$ $$\overset{*}{T_n}(x) = max(x_1,...,x_n)$$
If I now want to determine $$\mathbb{E}_{\vartheta}^n(T_n) = \int_{\mathbb{X}_n} T_n \: d \mathbb{P}_{\vartheta}^n = \frac{2}{n} \sum_{k=1}^n \frac{1}{\vartheta} \int_0^{\vartheta} x_k \: dx_k = \frac{2}{n} \sum_{k=1}^n \frac{1}{\vartheta} \frac{\vartheta^2}{2} = \vartheta$$ $$\mathbb{E}_{\vartheta}^n(\overset{*}{T_n}) = \int_{\mathbb{X}_n} \overset{*}{T_n}\: d \mathbb{P}_{\vartheta}^n = \int t \cdot f_{\overset{*}{T_n}}(t) dt = \int_0^{\vartheta} t \cdot \frac{n}{\vartheta^n} \cdot t^{n-1} dt = \frac{n}{n+1}\vartheta$$
It seems to me that for the first point estimator I have to use the formular $\mathbb{E}(g(X)) = \int g(t) f_X(t) dt$ and for the second point estimator the formular $\mathbb{E}(X) = \int t \cdot f_X(t) dt$. Why do I have to use different formulars if both are functions of $X$?