A number of solutions of length of curve functions I have run across are solved to a point and then instructed to "simplify the integrand before going further" as in the following example: f(y) = x3 - 1/4x3  on [1,2] interval integrate  [ ( 1 + (x3 -1/4x3) )2 ]1/2 dxdiferentiate the function, and square itsimplify the intergrand to [ 1 + x6 - 1/2 + 1/16x6 dx ]1/2I am absolutly puzzled as to where the middle term, - 1/2, comes from! Would you please give me a step-wise understanding of this feature?

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Asked Sep 21, 2019
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A number of solutions of length of curve functions I have run across are solved to a point and then instructed to "simplify the integrand before going further" as in the following example: f(y) = x3 - 1/4x3  

  1. on [1,2] interval integrate  [ ( 1 + (x3 -1/4x3) )2 ]1/2 dx
  2. diferentiate the function, and square it
  3. simplify the intergrand to [ 1 + x6 - 1/2 + 1/16x6 dx ]1/2

I am absolutly puzzled as to where the middle term, - 1/2, comes from! Would you please give me a step-wise understanding of this feature? 

 

 

 

 

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1 1 dx 4x3 1+ dx 4x3 2 1 -dx 42 x |dx xdx - 1 -3+1 1 х + 4 431

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