Proof: if $f(x)\ge c$ on $[a,b]$, then $\int_{a}^{b}{f(x)dx} \ge c(b-a)$,where $a,b,c \in{R}$ and $a<b$, for some real-valued continuous function $f$.

Question

This proof is very simple but I have a little problem: I start with $f(x) \geq c$, which is obvious. But in order to turn the left side ($f(x)$) into the form of the limit of a Riemann sum, I need to change $x$ to $x_{i}$. How do I achieve this?

Answer 1

$[\inf_{x \in [a,b]} f(x) ] \cdot (b-a)$ is a lower Darboux-sum for the partition $x_0=a,x_1=b$.

Then

$c(b-a) \leq [\inf_{x \in [a,b]} f(x) ] \cdot (b-a) \leq \int_a^b f(x) dx$

Answer 2

Let $f(x)=c+g(x)$ where $g(x)\ge0$ on $[a,b]$. Therefore $\int_a^bf(x)dx=\int_a^b(c+g(x))dx=c(b-a)+\int_a^bg(x)dx$, Well, if $g(x)$ is non-negative then $\int_a^bg(x)dx$ is non-negative also.