CALCULUS
posted by Sarah .
Find the length of the curve:
y=ln(1x^2), 0<_x<_1/2

s = ∫[0,1/2] ds
= ∫[0,1/2] √(1+y'^2) dx
= ∫[0,1/2] √(1+(2x/(x^21))^2) dx
= ∫[0,1/2] (x^2+1)/(x^21) dx
now use partial fractions to get
= ∫[0,1/2] 1 + 1/(x1)  1/(x+1) dx
= (x + lnx1  lnx+1) [0,1/2]
= (1/2 + ln(1/2)  ln(3/2))  (0+00)
= 1/2  ln3
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