help with finding the arclength over an interval [1,2] when using integration. when y=(x^4)/8+1/(4x^2)?

Question

I have the derivative=1/(2x^3)-1/(2x^3)
How is 1+the derivative^2=(1/(2x^3)+1/(2x^3))^2

Or give me an alternative to find the integal when the answer is 33/16

How is 1+the derivative^2=(1/(2x^3)+1/(2x^3))^2

it is not a typo on the website i get help it show this. If you want to see go here
http://calcchat.tdlc.com/free_solutions/main.html
then choose the third ed chapter 6 section 4 question 7
I don't understand how they got it that is why I asked.

Answer 1

You have a typo in your derivative. It should be (1/2)x^3-(1/2x^3).
Take your derivative and square it to get
(1/4)x^6 -(1/2) + 1/(4x^6).
Now add 1 and notice that the middle term simply changes sign. This means that
sqrt[1+(dy/dx)^2]=(1/2)x^3+ 1/(2x^3).
This is the simplification that makes the integration possible. It does not always happen (although most exercises will do this).

Now find the anti-derivative of this function,
(x^4)/8 -1/(4x^2).
Plug in x=2 and subtract what you get when you plug in x=1. This will give the arc length.

Answer 2

I = ∫ [ x^(4) / 8 + (1/4)x^(-2) ] dx

I = [ x^(5) / 40 - x(^(-1)) / 4 ]

I = [x^(5) / 40 - 1/(4x)] between limits of 1 and 2

I = [32/40 - (1/8)] - [1/40 - 1/4]

I = 27/40 + 9/40 = 36/40 = 9 /10

Also if f(x) = (1/8) x^4 + (1/4)x^(-2)

f `(x) = (1/2) x³ - (1/2) x^(-3)

Have carried out an integration and a differentiation and hope these may be of use to you in answering the question. My apologies if not!

Answer 3

y=(x^4)/8+1/(4x^2)
y' = 2x^3 - 8x/16x^4 = 2x^3 - 1/2x^3
1+y'^2 = 1 + (2x^3 - 1/2x^3)^2 = 1+4x^6 -4 +1/4x^6
1 +y'^2 = -3+ 4x^6+1/4x^6
s = integ 1 to 2 [sqrt(-3 +4x^6 +1/4x^6) dx]
s= integ 1 to 2 [1/2x^3sqrt(16x^12 -12x^6+1) dx]