Subdividing the Interior of a Rectifiable Curve 
385 
where Zk{k = 1 ^ n) are ordered points on the arc of the 
curve C, z,, being the same as Z, and ij, is any point on the arc 
Zk-iZ],, while ^Zj, is the chord — 
Let K be a curve 
x = ip (t), y = I(t) (to ^t^T) 
for which and f are not only continuous but except at a 
finite number of points have continuous derivatives. We now 
establish our theorem by showing first that along a curve K, for 
example a polygon, which satisfies these continuity conditions, 
we have 
r f (z) dz = F (Z) -F (zo). 
JzoK 
We then complete the proof by using Jordan’s^ method of prov- 
ing that the value of the integral along a polygon K inscribed in 
the arc C approaches the value of the integral along C, where C 
is any rectifiable curve, from which we are able to show that 
[ fiz)dz = F (Z)-F (zo). 
JzoC 
Taking up the first step in the proof we may write ^ iv 
and for one of the intervals along the curve K we may write 
[6] / CCk) ^^k = [u ih, rih) + iv (ft, r,k) ] (AXt + iAlJ,) = 
M (ft, nk) Axt - « (ft, Vk) ^Vk + iv (ft, »t) Axt + iu {^k,Vk) ^Vk- 
If now we let tj, be the value of t for which, by the mean value 
theorem, 
^Xk = F Ckt 
where A4 is the ^interval corresponding to AXyt, and let Cu.-qf) 
be the point on K corresponding to tj,, then we have for the first 
term of [6] 
u = u[(p{tC, I {TF)]FFk) 
We now sum these expressions for k^lion and then pass to the 
limit as At approaches zero. This gives 
11 
A3 = 0 k=\ 
d)] F^l) dt. 
^ Loc. cit. 
