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Homework Help: Reverse diferentiation problem

  1. Aug 9, 2007 #1
    1. The problem statement, all variables and given/known data
    (x^2+3x-1)/x^4 the answer is,(-x^-1)-(3/2)(x^-2)+(1/3)(x^-3)+C


    2. Relevant equations
    Just reverse diferentiation


    3. The attempt at a solution
    =x^2+3x-1+x^-4
    =(x^3)/(3)+(3x^2)/(2)-x+(1/3)(x^-3+C)
    (1/3)(x^3)+(3/2)(x^2)-x+(1/3)(x^-3)+C
    I know i got the answer but somehow it is backwards(exponents mostly). What am I doing wrong?
     
  2. jcsd
  3. Aug 9, 2007 #2
    Look at your signs on the exponents before you intergrate and be careful, that should take care of it.
     
  4. Aug 9, 2007 #3

    malawi_glenn

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    Homework Helper

    Remember in "backwards differentiation":
    i) Exponent +1
    ii) divide with the new exponent.

    [tex](x^2+3x-1)/x^4 = x^{-2} + 3x^{-3} - x^{-4} [/tex]

    the first term:
    [tex]x^{-2} \rightarrow (x^{-2+1})/(-2+1) = -x^{-1}[/tex]
    etc.
     
  5. Aug 9, 2007 #4

    HallsofIvy

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    Perhaps it would help if you told us what the question is!

    Are you trying to integrate (x^2+ 3x-1)/x^4?

    You then write "= x^2+ 3x-1+ x^-4" which is not at all the same thing: you are adding x^-4 rather than dividing by x^4.

    (x^2+ 3x- 1)/x^4= (x^2/x^4)+ 3(x/x^4)- 1/x^4= x^-2+ 3x^-3- x^-4.
    Now integrate.
     
  6. Aug 9, 2007 #5
    Now I know what I did wrong now, I just put x^-4 on top instead of subtracting the lower from the upper which I should have done. Thanks glenn
     
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