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The indefinite integral: `int (dw)/w^3`

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good2beme | Student, Undergraduate | Honors

Posted July 9, 2013 at 5:32 PM via web

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The indefinite integral: `int (dw)/w^3`

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted July 9, 2013 at 5:40 PM (Answer #1)

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The integral `int (dw)/w^3` has to be determined.

`int (dw)/w^3`

=> `int w^-3 dw`

=> `(w^(-3+1))/(-3+1)`

=> `w^-2/-2`

=> `-1/(2*w^2)`

The integral `int (dw)/w^3 = -1/(2*w^2) + C`

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted July 9, 2013 at 5:50 PM (Answer #2)

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Determine `int (dw)/w^2` :

`int (dw)/(w^2)=int w^(-2)dw` Use the general power rule:

`=(-1)w^(-1)+C`

`=-1/w +C`

Thus `int (dw)/(w^2)=-1/w + C`

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Check: if we take the derivative we should get the integrand back:

`d/(dw)( -1/w +C)=d/(dw)(-w^(-1))+d/(dw)(C)`

`=w^(-2)+0`

`=1/w^2` as required.

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