Stand a fence along a winding path, as tall as the function is high. Its area is the scalar line integral.
With f = 1 the fence is one unit tall everywhere, so its area is just the path's length.
Bend the path and watch the fence follow.
Walk a path across the ground and, at every step, put up a fence panel as tall as the function is above you. The total area of that fence is the scalar line integral ∫ f ds.
This is a different animal from the work integral. There you asked how much a vector field pushed you along, and the direction you walked mattered enormously — reverse it and the answer flipped sign. Here you're adding up a plain number times a length of path. Walk it backwards and nothing changes at all; the fence is the same fence.
The ds is doing real work, and the third readout shows what happens without it. Integrating f dt weights each point by how much parameter you spent there instead of how much ground you covered — so a stretch traced slowly gets counted far too heavily. Since ds = |r′(t)| dt, the speed is exactly the correction, and the two numbers only agree when the curve happens to be traced at unit speed.
Set f = 1 and the fence is a unit-height ribbon, so its area is the path's length — the arclength formula falling out as the simplest possible case. Divide the area by the length instead and you get the honest average of f along the path, which is what "average" ought to mean when the steps aren't evenly spaced.