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Math Help - Area between 2 graphs

  1. #1
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    Area between 2 graphs

    Find the area enclosed between and from to .

    \int_{-2}^{6} [(0.4^2+3) - (x)] dx

    \frac {0.4x^3}{3} + 3x - \frac {x^2}{2}

    If I'm not mistaken, I now have to use the Fb - Fa formula. However, my answer turns out wrong so I'm assuming I made a mistake in the calculations shown above. Can someone please tell me where it is?
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  2. #2
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    Quote Originally Posted by Archduke01 View Post
    Find the area enclosed between and from to .

    \int_{-2}^{6} [(0.4^2+3) - (x)] dx

    \frac {0.4x^3}{3} + 3x - \frac {x^2}{2}

    If I'm not mistaken, I now have to use the Fb - Fa formula. However, my answer turns out wrong so I'm assuming I made a mistake in the calculations shown above. Can someone please tell me where it is?
    your antiderivative is correct ... error in evaluation from x = -2 to x = 6, maybe?
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    \frac {0.4x^3}{3} + 3x - \frac {x^2}{2}

    b = 6
    \frac {0.4*6^3}{3} + 3*6 - \frac {6^2}{2} = 28.8

    a = -2
    <br />
\frac {0.4*(-2)^3}{3} + 3*(-2) - \frac {(-2)^2}{2} = -5.07

    F(b) - F(a) = 28.8 - (-5.07) = 33.86

    I don't understand ... I've triple-checked, and I get the same answer every time. It's wrong.
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    \frac{0.4(-2)^3}{3} + 3(-2) - \frac{(-2)^2}{2}

    \frac{0.4(-8)}{3} - 6 - \frac{4}{2}

    \frac{-3.2}{3} - 6 - 2 = -9.0666...
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