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Integration Question

  1. Mar 31, 2005 #1
    I am having a little trouble on this problem for calculus..

    Curve has equation y = x³ - 6x² + 9x + 16

    I am asked to find [tex]\int_{-1}^{2} (x^3 - 6x^2 + 9x + 16) dx[/tex]
    I get the answer to be -3/4 but the book gives the answer as 47.25. I do not know how the book got this. The way I found it was
    [tex][\frac{x^4}{4} - \frac{6x^3}{3} + \frac{9x^2}{2}]_{-1}^{2}[/tex] and then putting x=2 and solving it then putting x=-1 and solving it then subtract the answer from x=-1 from x=2 so i found [tex](4-16+18)-(\frac{1}{4}--2+\frac{9}{2}) = \frac{-3}{4}[/tex]

    It is important that I get this answer correct because It is needed for the next question of finding the area under a curve, could it just be that the book has an incorrect answer as it has known to be in the past or am i going about this completely the wrong way.
    Thanks for any help you guys can provide
     
  2. jcsd
  3. Mar 31, 2005 #2
    Your primitive function is wrong. You need an extra term of 16x in there.
     
  4. Mar 31, 2005 #3

    dextercioby

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    The integral of any constant A is

    [tex] \int A \ dx = Ax+\mathcal{C} [/itex]

    ,where [itex] \mathcal{C} [/itex] is an integration constant...

    Daniel.
     
  5. Mar 31, 2005 #4
    Thanks for the help I see where I was going wrong there, I was a little confused about the constant thing thanks for clearing it up
     
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