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Finding the antiderivative

  1. May 19, 2007 #1
    I need help with finding the antiderivative of
    1. [tex]y=\frac{\ x^2-47x+33}{10}[/tex]
    and
    2. [tex]y=-.03(x-6)^3+x[/tex]
    My attempts:
    1. [tex]y=.1[(x^3/3)-(47x^2/2)+33x)][/tex]
    2. [tex]y=-.03x-.18^3+x[/tex]
    [tex]y=(\frac{\-.03x^2}{2}-.18x)^3+(x^2/2)[/tex]
     
    Last edited: May 19, 2007
  2. jcsd
  3. May 19, 2007 #2
    Oh I guess you edited it since I last saw it. The first integral is correct now. For the second just try a substitution.
     
    Last edited: May 19, 2007
  4. May 19, 2007 #3

    Gib Z

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    Homework Helper

    Come on guys there are easy ones!!

    1. [tex]\int \frac{\ x^2-47x+33}{10} dx = \frac{1}{10} \int x^2 - 47x + 33 dx = \frac{1}{10} ( \int x^2 dx - \int 47x dx + \int 33 dx)[/tex]

    Please tell me you know the reverse power rule?

    Edit: Sorry I just saw your attempt. It is correct.

    2. Your attempt at the 2nd one doesnt seem correct.

    [tex]\int \frac{-3}{100}(x-6)^3 + x dx = \frac{-3}{100} \int (x-6)^3 dx + \int x dx[/tex].

    If you really need it, for that first integral do substitution u= x-6
     
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