Bound of an Integrable function (Analysis)

Question

For non-negative Riemann integrable function f in [a,b], and dissection $\mathcal D= {x_0,x_1,...,x_n } $, if $p(f,\mathcal D) $ is defined as $$p(f,\mathcal D)=\prod_{k=1}^n [1+(x_k-x_{k-1}) \inf_{x\in[x_{k-1},x_{k}]}f(x)]$$ How do you proof that $p(f,\mathcal D) < e^{\int_{a}^b f(x)dx} $?

Answer 1

Hint (using $\exp(x) = e^x$ for greater clarity): $$ \prod_{k=1}^n[1 + (x_k-x_{k-1})\inf_{x\in[x_{k-1},x_{k}]}f(x)]\le \prod_{k=1}^n\exp\left((x_k-x_{k-1})\inf_{x\in[x_{k-1},x_{k}]}f(x)\right) = $$ $$ = \exp\left(\sum_{k=1}^n{(x_k-x_{k-1})\inf_{x\in[x_{k-1},x_{k}]}f(x)}\right) $$ (Why the $\le$ and the $=$?)