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Math Help - Finding area of curve....

  1. #1
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    Finding area of curve....

    Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. (Do this on paper. Your instructor may ask you to turn in this graph.)


    This is the solution I got, but it was marked wrong:

    x^2-2x = x+4

    x^2-2x-x-4 =0
    x^2-3x-4 =0
    (x+1)(x-4), bounds are (-1,4)
    = (int from -1 and 4) x^2-3x-4
    =(x^3/3)-(3x^2/(2))-4x
    when I plug it in, I get 191/6, but it's wrong, don't know what I did wrong..please help!
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  2. #2
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    Quote Originally Posted by Jgirl689 View Post
    Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. (Do this on paper. Your instructor may ask you to turn in this graph.)


    This is the solution I got, but it was marked wrong:

    x^2-2x = x+4

    x^2-2x-x-4 =0
    x^2-3x-4 =0
    (x+1)(x-4), bounds are (-1,4)
    = (int from -1 and 4) x^2-3x-4

    your integrand is incorrect ... should be top function - bottom function.
    \int_{-1}^4 (x+4) - (x^2-2x) \, dx
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