Suppose I have a model with a continuum of agents denoted by $i \in [0;1]$. In a symmetric equilibrium, i.e. in a equilibrium where $x_i = x_j$ for all $x,i$, would I be able to conclude that: and some expression which I know holds: $$ \int^1_0 f(x_i) di = f \left( \int^1_0 x_i di \right) $$ The reasoning being that $\int^1_0 f(x_i) di$ can heuristically be thought of as a normalized sum over agents, i.e. as: $$ \sum_i^n \frac{1}{n} f(x_i) $$ Which in a symmetric equilibrium equals $$ \frac{n}{n} f(x_i) = f(x_i) = f\left(\sum_i^n \frac{1}{n} x_i\right) $$ Which can again heuristically be thought of as $f \left( \int^1_0 x_i di \right) $.
Are my heuristics wrong? Also if someone has a more rigorous approach to understanding these types of integrals please tell me. I have been unable to find a better answer than "think of the integral as a sum" while searching this website and others.
EDIT: I believe this equality does indeed hold. We can think of $x$ as a function of $i$. In a symmetric steady-state the function $x(i)$ is simply the constant function which returns the constant $x_{ss}$, thus $\int_0^1 f(x_i) di = \int^1_0 f(x_{ss}) di = f(x_{ss}) \int^1_0 di = f(x_{ss})$
