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Math Help - Chk Answer - Find area of bounded region (integrals)

  1. #1
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    Chk Answer - Find area of bounded region (integrals)

    Q. Sketch the region bounded by the graphs of: y=9-x^2, y=5-3x and then find it's area.

    A. A = \int_{-1}^{4}(9-x^2)-3x dx = \left[-\frac{1}{6}x(x(2x+9)-54\right]_{-1}^{4}

    = -\frac{1}{6}4(4(2*4+9)-54) - (-\frac{1}{6})-1(-1(2*(-1)+9)-54)

    = -\frac{56}{6} - \left[-\frac{61}{6}\right] = -\frac{5}{6}
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  2. #2
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    You have some mistakes in your integration. You shouldn't have negative area.

    \int_{-1}^{4}\left[(9-x^{2})-(5-3x)\right]dx
    Last edited by galactus; November 24th 2008 at 06:39 AM.
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  3. #3
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    I see what started me down the wrong path. How is this?

    <br />
\int_{-1}^{4}\left[(9-x^{2})-(5-3x)\right]dx = \left[ \frac{1}{6}x(9-2x)+24\right]_{-1}^{4}

    =\left[\frac{1}{6}4(9-2(4))+24\right]  -  \left[\frac{1}{6}(-1)(9-2(-1))+24\right]
    = \left[\frac{4}{6}+24\right]-\left[(-\frac{11}{6})+24\right]=\frac{15}{6}
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  4. #4
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    No, not quite. Try for 125/6
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  5. #5
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    Ok, I'm stumped, can you show me your calculations so I can compare and see what I did wrong?
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