Why is the average of a downward sloping line, $f(x)$ over the domain $[c,d]$ just $\frac{f(c)+f(d)}{2}$

Question

The result is easy to show (just show that the average of a downward sloping function, $\frac{1}{d-c}\int_c^d (a-bx)dx = \frac{a-bc + a-bd}{2}$),

but I feel like there should be some clean intuition as to why this is the case.

I can't seem to come up with some simple intuition though (my guess is that it has something to do with a linear function accumulating area in a "nice" way)

The result seems kind of strange to me because the average of a function, in integral form, is basically (area)/(width of interval), but (area) does not show up in $\frac{a-bc + a-bd}{2}$

Answer 1

Area expression is hidden: $$\dfrac{f(c)+f(d)}{2} = \dfrac{ \dfrac{1}{2}[f(c)+f(d)](d-c)}{d-c} $$

Lookup area of trapezoid

Answer 2

Using the fact that, as the other answer states, "Area expression is hidden" (see expression in the other answer),

and nothing that integrating a line from $c$ to $d$ gives a triangle, which has area $.5*$base$*$height, in other words it has area $$ .5*(d-c)(f(c)-f(d)) $$ when we divide this by the width of the interval, we get $$ .5(f(c)-f(d)) $$ which is the expression that we get from evaluating the integral (in other words, the integral is giving us an expression of the form area/width)