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July 29, 2014

July 29, 2014

Posted by **COFFEE** on Monday, June 11, 2007 at 11:48pm.

y = ( x^3/6 ) + ( 1/2x )

1/2 <or= x <or= 1

im looking over my notes, but i'm getting stuck.

here's my work so far:

A ( 1 , 2/3 )

B ( 1/2 , 49/48 )

y' = [1/6 (3x^2)] + [1/2 (-1x^-2)]

y' = ( x^2 / 2 ) - ( x^-2 / 2 )

(y')^2 = [( x^2 / 2 ) - ( x^-2 / 2 )]^2

y = (x^4 / 4) - (1/2) + (x^-4 / 4)

Integral: [from 1 to 1/2]

( 1 )

∫ sqrt[1 + (x^4 / 4) - (1/2) + (x^-4 / 4)] dx

( 1/2 )

( 1 )

∫ sqrt[(x^4 / 4) + (1/2) + (x^-4 / 4)] dx

( 1/2 )

..now i'm stuck. hopefully i computed it correctly.. how do i finish this to get a numerical answer? please help! thanks!!

Write the integral as:

Integral of

sqrt[(x^2/2 - x^(-2)/2)^2 + 1] dx

from x = 1/2 to x = 1

substitute x = Exp(t)

Then

x^2/2 - x^(-2)/2 =

[Exp(2t) - Exp(-2t)]/2 = Sinh(2t)

And thus:

[(x^2/2 - x^(-2)/2)^2 + 1 =

Sinh(2t)^(2) + 1 = Cosh(2t)^(2)

And we have:

sqrt[(x^2/2 - x^(-2)/2)^2 + 1] =

Cosh(2t)

Finally, using that:

dx = Exp(t) dt

we can write the integral as:

Integral of Cosh(2t) Exp(t) dt

from t = -Log(2) to 0 =

Integral of 1/2[Exp(3t) + Exp(-t)] dt

from t = -Log(2) to 0 =

95/144

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