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    hi ive done the first part of this question but am not quite sure how to approach the second part

    a) find the value of a such that a translation in the direction of the x axis transforms the curve with the equation y=(x^2 + ax -1)/x^3 into the curve with equation y=(x^2 -2)/(x-1)^3

    i found a=2

    b)(the bit i cant do) hence find the exact value of intergral between 2 and 3 of (x^2 -2)/(x-1)^3

    thanks alot for any pointers

    sarah
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    I'm not sure... But I'm guessing that you can split the fraction into

    x^2/(x-1)^3 - 2/(x-1)^3
    and then integrate as normal...
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    Use the substitution u= x-1... then follow integration by substitution!
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    You know that the area under the graph of y = (x^2 - 2)/(x - 1)^3 on the interval [2, 3] is the same as the area under the graph of y = (x^2 + 2x -1)/x^3 but on the interval [1, 2].

    So:
    int(2,3) (x^2 - 2)/(x - 1)^3 dx = int(1,2) (x^2 + 2x -1)/x^3 dx
    = int(1,2) 1/x + 2/x^2 - 1/x^3 dx
    = [ ln(x) - 2/x + 1/(2x^2) ] (1,2)
    = ln(2) + (5/8)
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    thanks alot!
 
 
 
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Updated: April 11, 2006
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