# Proof of integral property

1. Mar 5, 2008

### 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?

2. Mar 5, 2008

### HallsofIvy

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

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

3. Mar 5, 2008

### 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.

4. Mar 6, 2008

### 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?

Last edited: Mar 6, 2008
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook