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## 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

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

## 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.