Spoiler for 2 approachesGo to the full post
Draw a line parallel to the 2 horizontal lines and midway between them. Let it cut the curve at O=f(x0). Notice that the curve is always below the middle line, to the left of O and always above it, to the right of O (monotonic increase)
Assume the big dot is at the left of O at point f(x1). if it moves to the right to point f(x2) (x2<x0) its clear that area gained is smaller than the area lost as segment x1-x2 is below the middle line. Hence we can keep moving to the right till we reach x0 as the total area will keep reducing. After point x0, if we keep moving to the right, the situation will reserve.
Hence the big dot lies where a line equidistant and parallel to the 2 horizontal lines cuts f(x).
to find minimum area S = Int(f(x))x1->x + ( (x2-x)*( f(x2) - f(x1) ) - ( Int(f(x))x->x2) ) where Int(f(x))a->b denotes integral of f(x) from x=a to x=b. (How does 1 write integrals in here?)
Differentiating S wrt x and equating to 0 results in f(x) = (f(x1)+f(x2))/2
Same as in the first approach