# 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?

Question
6 views

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?

check_circle

Step 1

Hey, since there are multiple questions posted, we will answer first question. If you want any specific question to be answered, then please submit...

### Want to see the full answer?

See Solution

#### Want to see this answer and more?

Solutions are written by subject experts who are available 24/7. Questions are typically answered within 1 hour.*

See Solution
*Response times may vary by subject and question.
Tagged in
MathCalculus