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Math Help - Area under a curve

  1. #1
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    Area under a curve

    Hello all!

    Using (3/x-2) - (2/x-1) calculate the area enclosed between the curve
    y= (x+1)/(x-2)(x-1) and the x axis, between x = 3 and x = 5

    I would like to confirm whether the work I have done is correct or not
    and if anyone could give me some guidance

    ∫3/x-2 dx - ∫2/x-1

    3∫1/x-2 dx + 2∫1/x-1 dx

    5
    [3ln|x-2| + 2ln|x-1|]
    3

    (3ln|5-2| + 2ln|5-1|) - (3ln|3-2| + 2ln|3-1|)

    = 4.682 units squared
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  2. #2
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    Quote Originally Posted by Benji123 View Post
    Hello all!

    Using (3/x-2) - (2/x-1) calculate the area enclosed between the curve
    y= (x+1)/(x-2)(x-1) and the x axis, between x = 3 and x = 5

    I would like to confirm whether the work I have done is correct or not
    and if anyone could give me some guidance

    ∫3/x-2 dx - ∫2/x-1

    3∫1/x-2 dx + 2∫1/x-1 dx

    5
    [3ln|x-2| + 2ln|x-1|]
    3

    (3ln|5-2| + 2ln|5-1|) - (3ln|3-2| + 2ln|3-1|)

    = 4.682 units squared
    Looks good!

    \frac{3}{x-2}-\frac{2}{x-1}=\frac{3(x-1)-2(x-2)}{(x-2)(x-1)}=\frac{x+1}{(x-2)(x-1)}


    When applying the chain rule.....

    u=x-2\ \Rightarrow\ \frac{du}{dx}=1\ \Rightarrow\ du=dx

    v=x-1\ \Rightarrow\ dv=dx

    hence all correct,

    or is it?
    Last edited by Archie Meade; July 22nd 2010 at 08:34 AM.
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  3. #3
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    Quote Originally Posted by Benji123 View Post
    Hello all!

    Using (3/x-2) - (2/x-1) calculate the area enclosed between the curve
    y= (x+1)/(x-2)(x-1) and the x axis, between x = 3 and x = 5

    I would like to confirm whether the work I have done is correct or not
    and if anyone could give me some guidance

    ∫3/x-2 dx - ∫2/x-1

    3∫1/x-2 dx + 2∫1/x-1 dx

    5
    [3ln|x-2| + 2ln|x-1|]
    3

    (3ln|5-2| + 2ln|5-1|) - (3ln|3-2| + 2ln|3-1|)

    = 4.682 units squared
    Unfortunately,

    there is one error as you've calculated the integral of \frac{3}{x-2}+\frac{2}{x-1}

    instead of the integral of \frac{3}{x-2}-\frac{2}{x-1}

    so your calculated area should be less than 2.
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  4. #4
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    Therefore,

    ∫3/x-2 dx - ∫2/x-1

    3∫1/x-2 dx - 2∫1/x-1 dx

    5
    [3ln|x-2| - 2ln|x-1|]
    3

    (3ln|5-2| - 2ln|5-1|) - (3ln|3-2| - 2ln|3-1|)

    =0.523248144 - (-1.386294361)

    = 1.910 units squared ?
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  5. #5
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    Yes,
    that's it!
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  6. #6
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    Thanks I really appreciate it and pointing out my error. I don't get why alot of revision notes don't give you answers to the questions so you can check.
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