1. ## Integral Order

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

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

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

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

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

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

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

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

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

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

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