Proving Integral Property for Constant One-Forms in R^3

[tronter]

Given a constant one-form k_1 \ dx + k_2 \ dy + k_3 \ dz in \bold{R}^{3}, and three points \vec{a}, \ \vec{b}, \ \vec{c} in \bold{R}^3, prove that \int_{\vec{a}}^{\vec{c}} k_1 \ dx + k_2 \ dy + k_3 \ dz = \int_{\vec{a}}^{\vec{b}} k_1 \ dx + k_2 \ dy + k_3 \dz + \int_{\vec{b}}^{\vec{c}} k_1 \ dx + k_2 \ dy + k_3 \ dz.

So we want to show that k_{1}(c_1-a_1) + k_2(c_2-a_2) + k_3(c_3-a_3) = k_{1}(b_1-a_1) + k_2(b_2-a_2) + k_3(b_3-a_3) + k_1(c_1-b_1) + k_2(c_2-b_2) + k_3(c_3-a_3).

Doesn't this follow from the transitive property, or the triangle inequality?

 

[HallsofIvy]

Since there is no inequality there, it certainly isn't a "triangle inequality"!

Perhaps if you stated the "transitive property" for integrals.

 

[tronter]

Well I think the key idea is that we are using constant 1-forms. If it were not constant, then we could not use the transitive property.

 

[mrandersdk]

k_{1}(c_1-a_1) + k_2(c_2-a_2) + k_3(c_3-a_3) = k_{1}(b_1-a_1) + k_2(b_2-a_2) + k_3(b_3-a_3) + k_1(c_1-b_1) + k_2(c_2-b_2) + k_3(c_3-b_3) You had an error here in the last parentese

if you collect the terms with same k you get

k_{1}(b_1-a_1) + k_2(b_2-a_2) + k_3(b_3-a_3) + k_1(c_1-b_1) + k_2(c_2-b_2) + k_3(c_3-b_3) =

k_{1}(b_1-a_1 +c_1-b_1) + k_2(b_2-a_2+c_2-b_2)+k_3(b_3-a_3+c_3-b_3)= k_{1}(c_1-a_1) + k_2(c_2-a_2)+k_3(c_3-a_3)

is that what you want?