Answer
See explanation
Work Step by Step
$f_{ave}[a,b]$ is the average value of $f$ on the interval $[a,b]$, and as such we define:
$f_{ave}[a,b] = \frac{1}{b-a}\int_a^b f(x) dx$
Split the integral at $c$:
$\int_a^b f(x) dx=\int_a^c f(x) dx+\int_c^b f(x) dx$
So
$f_{ave}[a,b]=\frac{1}{b-a}\left(\int_a^c f(x) dx+\int_c^b f(x) dx\right)$
We also have:
$\int_a^c f(x)dx=(c-a)f_{ave}[a,c]$
$\int_b^c f(x)dx=(b-c)f_{ave}[b,c]$
Substitute:
$f_{ave}[a,b]=\frac{1}{b-a}\left((c-a)f_{ave}[a,c]+(b-c)f_{ave}[c,b]\right)$
Rearrange:
$f_{ave}[a,b]=\frac{c-a}{b-a}f_{ave}[a,c]+\frac{b-c}{b-a}f_{ave}[c,b]$
