Integral Order

**manjohn12** · March 15th 2009, 02:02 PM

Theorem. Let $a,b,M$ and $m$ be real numbers with $a<b$ 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 $\int_{a}^{b} g-f \geq 0$ ?

**ThePerfectHacker** · March 15th 2009, 03:49 PM

Originally Posted by manjohn12

Theorem. Let $a,b,M$ and $m$ be real numbers with $a<b$ 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.

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