Integral Theorem

**manjohn12** · March 15th 2009, 01:29 PM

Theorem. Let $f$ and $g$ be real valued functions that are Riemann integrable on $[a,b]$ . Then the following statements hold:

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

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

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

Outline of Proof. The basic idea is that we basically look at $\mathcal{R}(f+g,P), \ \mathcal{R}(kf, P),$ and $\mathcal{R}(f-g, P)$ and show that they are in fact equal to those right hand sides?

**ThePerfectHacker** · March 15th 2009, 03:41 PM

Originally Posted by manjohn12

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

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

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

EDIT: Fixed

**manjohn12** · March 15th 2009, 03:42 PM

Originally Posted by ThePerfectHacker

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

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

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

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

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