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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Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)

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ISBN:9781337111348

Author:Bruce Crauder, Benny Evans, Alan Noell

Publisher:Bruce Crauder, Benny Evans, Alan Noell

Chapter2: Graphical And Tabular Analysis

Section2.1: Tables And Trends

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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

on [1,2] interval integrate [ ( 1 + (x³ -1/4x³⁾ )² ]^1/2dx
diferentiate the function, and square it
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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