Let $f(x) \in C[0,1]$, and $f(x)>0$ over $[0,1]$. Prove $$\ln \int_0^1 f(x)dx \geq \int_0^1 \ln f(x) dx.$$
If we denote $$F(x):=\ln \int_0^x f(t){\rm d}t-\int_0^x \ln f(t){\rm d}t, ~~~x \in[0,1]$$ Differentiate the both sides with respect to $x$， we obtain $$F'(x)=\frac{f(x)}{\int_0^x f(t){\rm d}t}-\ln f(x)=\frac{f(x)-\ln f(x)\int_0^x f(t){\rm d}t}{\int_0^x f(t){\rm d}t},$$ which is helpful?
This is just Jensen's inequality. We have $\exp(\int g(x)dx) \leq \int e^{g(x)}dx$ because $e^{x}$ is a convex function. Take $g(x)=\ln (f(x))$ in this.
Proof
Since $y=\ln x$ is convex upward over $(0,+\infty)$，by Jensen inequality，we have $$\ln\left(\frac{1}{n}\sum_{k=1}^n a_k \right)\geq \frac{1}{n}\sum_{k=1}^n \ln a_k,$$where $a_k>0(k=1,2,\cdots,n)$. As per the definition of definite integral (by Riemann), we have $$\begin{aligned}\ln\int_0^1 f(x)dx&=\ln\lim_{n \to \infty} \frac{1}{n}\sum_{i=1}^n f\left(\frac{i}{n}\right)\\ &=\lim_{n \to \infty} \ln \left[\frac{1}{n}\sum_{i=1}^n f\left(\frac{i}{n}\right)\right]\\ &\geq\lim_{n \to \infty} \frac{1}{n}\sum_{i=1}^n\ln f\left(\frac{i}{n}\right)\\ &=\int_0^1\ln f(x)dx,\end{aligned}$$which is desired.