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 dx 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?
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 dx 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?
Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
6th Edition
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
Problem 1TU: If a coffee filter is dropped, its velocity after t seconds is given by v(t)=4(10.0003t) feet per...
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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 ]1/2 dx - 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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