Proving a Simple Integral with Exponents Let $f$ be differentiable in $[a,b]$.
How can I show that

$$\exp\left(\frac{1}{b-a} \int_a^b f(x)dx \right) \le \left(\frac{1}{b-a}\right) \int_a^b \exp(f(x)) dx $$

 A: Hint: Since the function $φ(\cdot)=\exp(\cdot)$ is convex, use Jensen's inequality.

Solution:  In particular, Jensen's integral inequality (see also here) states that if $φ$ is a convex function on $C\subseteq\mathbb R$ and $f$ is a function such that $f\left([a,b]\right)\subseteq C$, then $$\color{blue}{φ}\left(\int_{a}^{b}λ(x)f(x)dx\right)\le \int_{a}^{b}λ(x)\color{blue}{φ}\left(f(x)\right)dx\tag{1}$$ if $λ(x)\ge 0$ for $x \in [a,b]$ such that $$\int_{a}^{b}λ(x)dx=1$$ 
In your case let $$\begin{cases}λ(x)&:=&\dfrac{1}{b-a}\\\\φ(x)&:=&\exp(x)\end{cases}$$ and $f(x):=f(x)$. Observe that $λ(x)$ satisfies the condition above (positive and integral equal to 1) and that $\exp(x)$ is convex in $\mathbb R$  since $$\frac{\partial^2}{\partial x^2}\left(\exp(x)\right)=\exp(x)> 0$$ for all $x \in \mathbb R$. Thus, substituting in $(1)$ you obtain that $$\color{blue}{
\exp}\left(\int_{a}^{b}\frac{1}{b-a}f(x)dx\right)\le \int_{a}^{b}\frac{1}{b-a}\color{blue}{\exp}\left(f(x)\right)dx$$ or equivalently (since $\frac{1}{b-a}$ does not depend on $x$) that $$\color{blue}{
\exp}\left(\frac{1}{b-a}\int_{a}^{b}f(x)dx\right)\le \left(\frac{1}{b-a}\right)\int_{a}^{b}\color{blue}{\exp}\left(f(x)\right)dx$$
