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Chapter 8

Q. 8.T.6

(Linearity Property)

If f_{1}, f_{2} ∈ \mathcal{R}(a, b) and k_{1}, k_{2} ∈ \mathbb{R}, then k_{1}f_{1} + k_{2}f_{2} ∈ \mathcal{R}(a, b) and

\int_{a}^{b}{k_{1}f_{1} + k_{2}f_{2}} = k_{1} \int_{a}^{b}{f_{1} + k_{2}} \int_{a}^{b}{f_{2}}.

Step-by-Step

Verified Solution

The result is immediate if k_{1} = k_{2} = 0, so we assume that \left|k_{1}\right| + \left|k_{2}\right| > 0. Let ε > 0 be given. Using Theorem 8.5, we can choose δ_{i} > 0 (i = 1, 2) such that, if P ∈ \mathcal{P} (a, b) satisfies \left\|P\right\| < δ_{i} and α is a mark on P, then

\left|S (f_{i}, P, α) − \int_{a}^{b}{f_{i}}\right| < \frac{ε}{\left|k_{1}\right| + \left|k_{2}\right|},   i = 1, 2.

Put δ = \min\left\{δ_{1}, δ_{2}\right\} and let P ∈ \mathcal{P} (a, b) satisfy \left\|P\right\| < δ. Since

S (k_{1}f_{1} + k_{2}f_{2}, P, α) = k_{1}S (f_{1}, P, α) + k_{2}S (f_{2}, P, α) ,

we clearly have

\left|S \left(k_{1}f_{1} + k_{2}f_{2}, P, α\right) − \left(k_{1} \int_{a}^{b}{f_{1}} + k_{2} \int_{a}^{b}{f_{2}}\right)\right|

≤ \left|k_{1}\right| \left|S (f_{1}, P, α) − \int_{a}^{b}{f_{1}}\right| + \left|k_{2}\right| \left|S (f_{2}, P, α) − \int_{a}^{b} f_{2}\right| < ε.

Theorem 8.5 now yields the desired result.