Monday

September 22, 2014

September 22, 2014

Posted by **Sarah** on Friday, November 16, 2012 at 2:11am.

y=ln(1-x^2), 0<_x<_1/2

- CALCULUS -
**Steve**, Friday, November 16, 2012 at 11:12ams = ∫[0,1/2] ds

= ∫[0,1/2] √(1+y'^2) dx

= ∫[0,1/2] √(1+(2x/(x^2-1))^2) dx

= ∫[0,1/2] (x^2+1)/(x^2-1) dx

now use partial fractions to get

= ∫[0,1/2] 1 + 1/(x-1) - 1/(x+1) dx

= (x + ln|x-1| - ln|x+1|) [0,1/2]

= (1/2 + ln(1/2) - ln(3/2)) - (0+0-0)

= 1/2 - ln3

**Answer this Question**

**Related Questions**

Calculus - A curve is defined by the parametric equations: x = t2 – t and y = t3...

12th Grade Calculus - 1. a.) Find an equation for the line perpendicular to the ...

Calculus - Find a curve through the point (1,1) whose length integral is given ...

calculus - Consider a plane curve which is described in polar coordinates (r...

Calculus - Consider a plane curve which is described in polar coordinates (r...

calculus - Consider a plane curve which is described in polar coordinates (r...

Calculus - Find the curve's unit tangent vector. Also, find the length of the ...

calculus - 1. Given the curve a. Find an expression for the slope of the curve ...

calculus 2 - .Find the length of the curve y=e^x+1/4 e^−x between x=0 and ...

calculus 2 - 1.Find the length of the curve y=ex+1e−x between x=0 and x=1.