## Thomas' Calculus 13th Edition

a) $av(f+g)=av(f)+av(g)$ b) $av(kf)=k \cdot av(f)$ c) $\int_a^b f(x) dx \leq \int_a^b g(x) dx$
(a) Need to use definition. $av(f)=\dfrac{1}{(b-a)} \int_a^b f(x) dx$ $av(f+g)=\dfrac{1}{(b-a)} \int_a^b f(x) dx+\dfrac{1}{(b-a)} \int_a^b g(x)dx$ or, $av(f+g)=av(f)+av(g)$ (b) Need to use definition. $av(f)=\dfrac{1}{(b-a)} \int_a^b f(x) dx$ $av(kf)=k[\dfrac{1}{(b-a)} \int_a^b f(x) dx]$ $\implies av(kf)=k \cdot av(f)$ c) Need to use definition. $av(f)=\dfrac{1}{(b-a)} \int_a^b f(x) dx$ Here, we have $av(kf) \leq av (g)$ $\implies \dfrac{1}{(b-a)} \int_a^b f(x) dx \leq \dfrac{1}{(b-a)} \int_a^b g(x) dx$ Hence, $\int_a^b f(x) dx \leq \int_a^b g(x) dx$