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Math Help - What am I doing wrong in this problem (area enclosed between curves)?

  1. #1
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    What am I doing wrong in this problem (area enclosed between curves)?

    Find the area of the region enclosed by y=x^2-5x and y=x+7.

    What I did was:

    \displaystyle\int^7_{-1} (x+7)-(x^2-5x) dx

    \frac{-x^3}{3}+\frac{6x^2}{2}+7 from -1 to 7 (sorry, not sure how to type that)

    \left(-\frac{343}{3}+147+7\right) - \left(\frac{1}{3}+3-7\right)

    Area = \frac{130}{3}

    This is what I got, but our computer program tells me that I am wrong. Is there a mistake I'm making?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Dana_Scully View Post
    Find the area of the region enclosed by y=x^2-5x and y=x+7.

    What I did was:

    \displaystyle\int^7_{-1} (x+7)-(x^2-5x) dx

    \frac{-x^3}{3}+\frac{6x^2}{2}+7 from -1 to 7 (sorry, not sure how to type that)

    \left(-\frac{343}{3}+147+7\right) - \left(\frac{1}{3}+3-7\right)

    Area = \frac{130}{3}

    This is what I got, but our computer program tells me that I am wrong. Is there a mistake I'm making?
    a little hickup... the integral of 7 is not 7


    by the way, you can type \Bigg|_{-1}^7 to get \Bigg|_{-1}^7. otherwise, see our LaTeX tutorial and learn how to use \right and \left tags
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  3. #3
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    Just a minor mistake:

    \begin{aligned} \int_{-1}^7 \left(x + 7 - (x^2 - 5x)\right) dx & = \int_{-1}^7 \left( - x^2 + 6x + 7\right) dx \\ & =  \left(- \frac{x^3}{3} + \frac{6x^2}{2} + 7{\color{red}x} \right)\bigg|_{-1}^7 \\ & \ \ \vdots \end{aligned}
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  4. #4
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    Thanks, I had actually written 7x but somehow confused it in my calculations...and thanks for telling me how to write the, uh, integral thing.
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