# Integral Theorem

• Mar 15th 2009, 01:29 PM
manjohn12
Integral Theorem
Theorem. Let $\displaystyle f$ and $\displaystyle g$ be real valued functions that are Riemann integrable on $\displaystyle [a,b]$. Then the following statements hold:

1. The function $\displaystyle f+g$ is Riemann integrable on $\displaystyle [a,b]$ and $\displaystyle \int_{a}^{b} (f+g) = \int_{a}^{b} f + \int_{a}^{b} g$.

2. The function $\displaystyle kf$ is Riemann integrable on the interval $\displaystyle [a,b]$ and $\displaystyle \int_{a}^{b} kf = k \int_{a}^{b} f$.

3. The function $\displaystyle f-g$ is Riemann integrable on $\displaystyle [a,b]$ and $\displaystyle \int_{a}^{b} (f-g) = \int_{a}^{b} f - \int_{a}^{b} g$.

Outline of Proof. The basic idea is that we basically look at $\displaystyle \mathcal{R}(f+g,P), \ \mathcal{R}(kf, P),$ and $\displaystyle \mathcal{R}(f-g, P)$ and show that they are in fact equal to those right hand sides?
• Mar 15th 2009, 03:41 PM
ThePerfectHacker
Quote:

Originally Posted by manjohn12
1. The function $\displaystyle f+g$ is Riemann integrable on $\displaystyle [a,b]$ and $\displaystyle \int_{a}^{b} (f+g) = \int_{a}^{b} f + \int_{a}^{b} g$.

Let $\displaystyle I_1 = \smallint_a^b f \text{ and }I_2 = \smallint_a^b g$. For $\displaystyle \epsilon > 0$ there exists $\displaystyle \delta > 0$ so that $\displaystyle | \mathcal{R}(f,P) - I_1| < \frac{\epsilon}{2}$ and $\displaystyle |\mathcal{R}(g,P) - I_2| < \frac{\epsilon}{2}$ as long as $\displaystyle ||P|| < \delta$.

Now, $\displaystyle \mathcal{R}(f+g,P) = \mathcal{R}(f,P) + \mathcal{R}(g,P)$.
Thus, $\displaystyle |\mathcal{R}(f+g,P) - I_1 - I_2| \leq |\mathcal{R}(f,P) - I_1| + | \mathcal{R}(f,P) - I_2| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$.

EDIT: Fixed
• Mar 15th 2009, 03:42 PM
manjohn12
Quote:

Originally Posted by ThePerfectHacker
Let $\displaystyle I_1 = \smallint_a^b f \text{ and }I_2 = \smallint_a^b g$. For $\displaystyle \epsilon > 0$ there exists $\displaystyle \delta > 0$ so that $\displaystyle | \mathcal{R}(f,P) - I_1| < \frac{\epsilon}{2}$ and $\displaystyle |\mathcal{R}(g,P) - I_2| < \frac{\epsilon}{2}$ as long as $\displaystyle ||P|| < \delta$.

Now, $\displaystyle \mathcal{R}(f+g,P) = \mathcal{R}(f,P) + \mathcal{R}(f,P)$.
Thus, $\displaystyle |\mathcal{R}(f+g,P) - I_1 - I_2| \leq |\mathcal{R}(f,P) - I_1| + | \mathcal{R}(f,P) - I_2| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$.

$\displaystyle \mathcal{R}(f+g,P) = \mathcal{R}(f,P) + \mathcal{R}(g,P)$.

Thus, $\displaystyle |\mathcal{R}(f+g,P) - I_1 - I_2| \leq |\mathcal{R}(f,P) - I_1| + | \mathcal{R}(g,P) - I_2| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$.