Reworking the above with differentials and taking A B ¯ ¯ to be along the x -axis, the net distance differential is d f = d x , an exact differential with antiderivative x . On the other hand, the total distance differential is | d x | , which does not have an antiderivative. The path taken is γ γ : [ 0 , 1 ] → → A B ¯ ¯ where there exists a time t ∈ ∈ ( 0 , 1 ) such that γ γ is strictly increasing before t and strictly decreasing afterward. Then d x is positive before t and negative afterward, yielding the integrals, Δ Δ f = ∫ ∫ γ γ d x = 0 Δ Δ g | γ γ = ∫ ∫ γ γ | d x | = ∫ ∫ A B d x + ∫ ∫ B A ( − − d x ) = 2 ∫ ∫ A B d x = 2 A B exactly the results we expected from the verbal argument before.