Date: Jan 21, 2005 3:07 PM
Author: ticbol
Subject: Re: calc help

INT.[x(4 -2x^2)^(1/2)]dx

Let u = (4 -2x^2)
So, du = -4x dx
And (4 -2x^2)^(1/2) = u^(1/2)
Then,
INT.[x(4 -2x^2)^(1/2)]dx
= INT.[(4 -2x^2)^(1/2)]x dx
= INT.[(4 -2x^2)^(1/2)](-4/-4)x dx
= (-1/4) INT.[(4 -2x^2)^(1/2)](-4x dx)
= -(1/4)INT.[u^(1/2)]du
= -(1/4)[1/(3/2)*(u^(3/2))] +C
= -(1/4)(2/3)u^(3/2) +C
= -(1/6)u^(3/2) +C
= -(1/6)(4 -2x^2)^(3/2) +C -----answer.


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