Why $E[f(X_U)] = \lambda \int_0^{\infty} e^{- \lambda t} E[ f(X_t)]dt$ ? (Where $U$ is exponentially distributed)

Question

Let $X_t$ Be a stochastic process that is independent of exponentially distributed random variable $U$ (With parameter $\lambda$). Is it true that (if so, why?) $$ E[f(X_U)] = \lambda \int_0^{\infty} e^{- \lambda t} E[f(X_t)]dt $$ I think the reason why I don’t get this is that I don’t really understand how to independence assumption works with stochastic processes. The $\lambda$ In the expression for sure comes from the exponential distribution but to get somewhere I think the independence has to be somehow used. I appriciate if someone can be really precise in explaining this.