Answer
a) $av(f+g)=av(f)+av(g)$
b) $av(kf)=k \cdot av(f)$
c) $\int_a^b f(x) dx \leq \int_a^b g(x) dx$
Work Step by Step
Apply the definition: $ av(f)=\dfrac{1}{(b-a)} \int_a^b f(x) dx$
a) $av(f+g)=\dfrac{1}{(b-a)} \int_a^b f(x) dx+\dfrac{1}{(b-a)} \int_a^b g(x)dx$
This implies that
$av(f+g)=av(f)+av(g)$
b) $av(kf)=k[\dfrac{1}{(b-a)} \int_a^b f(x) dx]$
This implies that
$av(kf)=k \cdot av(f)$
c) $av(kf) \leq av (g)$
This implies that
$\dfrac{1}{(b-a)} \int_a^b f(x) dx \leq \dfrac{1}{(b-a)} \int_a^b g(x) dx$
Hence, $\int_a^b f(x) dx \leq \int_a^b g(x) dx$
