- #1
rizardon
- 20
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While working on an integration problem I found that I will arrive at two different solutions depending on how I approach it.
I'm finding the arc length of y=ln(1-x2) on the interval [0,0.5]
The formula for finding the arc length is ∫sqrt[1+[f'(x)]2]dx
So f'(x) = -2x / ( 1-x2 )
Here I first simplify this to 2x / ( x2 - 1 ) and squaring gives
4x2 / ( x2 -1 )2
Working from here I end up integrating from 0 to 0.5
∫ [1 + 1/(x-1) - 1/(x+1)] dx = 0.5 - ln3
On the other hand if I leave f'(x) as it is without simplifying, when I squared f'(x) I get
4x2 / ( 1-x2 )2
and end up integrating from 0 to 0.5
∫ [1 + 1/(1+x) + 1/(1-x)] dx = -0.5 - ln3
Should both have the same solution or is this simply a possible effect from squaring numbers?
Thank you
I'm finding the arc length of y=ln(1-x2) on the interval [0,0.5]
The formula for finding the arc length is ∫sqrt[1+[f'(x)]2]dx
So f'(x) = -2x / ( 1-x2 )
Here I first simplify this to 2x / ( x2 - 1 ) and squaring gives
4x2 / ( x2 -1 )2
Working from here I end up integrating from 0 to 0.5
∫ [1 + 1/(x-1) - 1/(x+1)] dx = 0.5 - ln3
On the other hand if I leave f'(x) as it is without simplifying, when I squared f'(x) I get
4x2 / ( 1-x2 )2
and end up integrating from 0 to 0.5
∫ [1 + 1/(1+x) + 1/(1-x)] dx = -0.5 - ln3
Should both have the same solution or is this simply a possible effect from squaring numbers?
Thank you