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Proof of integral property

  1. Mar 5, 2008 #1
    Given a constant one-form [tex] k_1 \ dx + k_2 \ dy + k_3 \ dz [/tex] in [tex] \bold{R}^{3} [/tex], and three points [tex] \vec{a}, \ \vec{b}, \ \vec{c} [/tex] in [tex] \bold{R}^3 [/tex], prove that [tex] \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 [/tex].

    So we want to show that [tex] 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) [/tex].

    Doesn't this follow from the transitive property, or the triangle inequality?
     
  2. jcsd
  3. Mar 5, 2008 #2

    HallsofIvy

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

    Perhaps if you stated the "transitive property" for integrals.
     
  4. Mar 5, 2008 #3
    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.
     
  5. Mar 6, 2008 #4
    [tex] 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) [/tex] You had an error here in the last parentese

    if you collect the terms with same k you get

    [tex] 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) = [/tex]

    [tex]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)[/tex]

    is that what you want?
     
    Last edited: Mar 6, 2008
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