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 $f$.

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?

• Welcome to MSE! Please use the basic tutorial and quick reference guide to assist with formatting going forward. Commented Feb 1, 2021 at 3:20
• Hint: if you subtract $c$ from $f$, you'll have to integrate $f(x)-c$ which is a non-negative function. What do you know about the integral of a non-negative function..? Commented Feb 15, 2021 at 15:53

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