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Math Help - Intergration Help!

  1. #1
    Member looi76's Avatar
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    Intergration Help!

    Question:
    Find the total area of the region shaded in the following diagram:

    y = 3x^2 - 12x + 9

    Attempt:
    y = 3x^2 - 12x + 9
    a = 3, b = -12, c = 9

    x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

    x = \frac{-(-12) \pm \sqrt{(-12)^2-4\times3\times9}}{2\times3}

    x = \frac{12 \pm \sqrt{36}}{6}

    x = \frac{12 + 6}{6} , x = \frac{12 - 6}{6}
    x = 3 , x = 1

    =3x^2-12x+9

    =\frac{3x^{2+1}}{2+1} - \frac{12x^{1+1}}{1+1} + 9x

    =\frac{3x^{3}}{3} - \frac{12x^{2}}{2} + 9x

    =x^3-6x^2+9x

    = [x^3-6x^2+9x]^3_1

    = [(3)^3-6(3)^2+9(3)] - [(1)^3-6(1)^2+9(1)]

    =-4
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    Try it again, remember that areas can't be negative.
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  3. #3
    GAMMA Mathematics
    colby2152's Avatar
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    Quote Originally Posted by Krizalid View Post
    Try it again, remember that areas can't be negative.
    With these questions, you always want to split the integrals in parts to make sure each segment of area is positive.
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  4. #4
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    Also have in mind that integrals of graphs UNDER the x-axes are always negative. Therefore, if you look for an AREA (and not just for an integral), you have to switch it to positive.
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