1. ## Integral Order

Theorem. Let $a,b,M$ and $m$ be real numbers with $a and $m \leq M$. Let $f$ and $g$ be real-valued functions that are Riemann integrable on $[a,b]$. Then the following hold:

1. If $f(x) \geq 0$ for all $x \in [a,b]$ then $\int_{a}^{b} f \geq 0$.

2. If $f(x) \leq g(x)$ for all $x \in [a,b]$ then $\int_{a}^{b} f \leq \int_{a}^{b} g$.

3. If $m \leq f(x) \leq M$ for all $x \in [a,b]$, then $m(b-a) \leq \int_{a}^{b} f \leq M(b-a)$.

Outline of Proof. Basically we consider the values $f$ and $g$ can take, and look at $\mathcal{R}(f,P)$ and $\mathcal{R}(g,P)$? For instance, in #2 we look at $g-f$ (which is Riemann integrable) and show that $\int_{a}^{b} g-f \geq 0$?

2. Originally Posted by manjohn12
Theorem. Let $a,b,M$ and $m$ be real numbers with $a and $m \leq M$. Let $f$ and $g$ be real-valued functions that are Riemann integrable on $[a,b]$. Then the following hold:

1. If $f(x) \geq 0$ for all $x \in [a,b]$ then $\int_{a}^{b} f \geq 0$.

2. If $f(x) \leq g(x)$ for all $x \in [a,b]$ then $\int_{a}^{b} f \leq \int_{a}^{b} g$.

3. If $m \leq f(x) \leq M$ for all $x \in [a,b]$, then $m(b-a) \leq \int_{a}^{b} f \leq M(b-a)$.
If you prove #1 then #2 follows immediately by considering $f-g$ which in consequence #3 follows immediately.

For #1 assume that $I < 0$ we know that $|\mathcal{R}(f,P) - I| < \epsilon$ for $||P||<\delta$.
But, $|\mathcal{R}(f,P) - I| = \mathcal{R}(f,P) - I$ since $I < 0$.
Thus, $|\mathcal{R}(f,P) - 0| = \mathcal{R}(f,P) < \mathcal{R}(f,P) - I < \epsilon$.

We have a contradition because we have shown the integral of $f$ is $0$ and $I$ but the integral is well-defined and so this forces $I = 0$ which is where we get a contradiction.