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CalculusQ&A LibraryA 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?Question

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) = x^{3} - 1/4x^{3}

- on [1,2] interval integrate [ ( 1 + (x
^{3}-1/4x^{3)})^{2}]^{1/2 }dx - diferentiate the function, and square it
- simplify the intergrand to [ 1 + x
^{6 }- 1/2 + 1/16x^{6}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?

Step 1

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