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Math Help - Definite Integration

  1. #1
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    Definite Integration

    find the total area between the region and the the x-axis.
    y = -x^2 - 2x when -3 <= x <= 2 (which i took it as [-3,2])

    This was my work:

    I integrated the equation:

    -(1/3)x^3 - x^2 [-3,2]

    I plugged in the numbers and got(plugged in 2 then -3):

    -(8/3) - 4 - (9 - 9)... which would equal -20/3, but the answer said +28/3.
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  2. #2
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    Quote Originally Posted by driver327 View Post
    find the total area between the region and the the x-axis.
    y = -x^2 - 2x when -3 <= x <= 2 (which i took it as [-3,2])

    This was my work:

    I integrated the equation:

    -(1/3)x^3 - x^2 [-3,2]

    I plugged in the numbers and got(plugged in 2 then -3):

    -(8/3) - 4 - (9 - 9)... which would equal -20/3, but the answer said +28/3.
    While the "area beneath a graph", for a graph that is always above the x-axis is the integral itself, that is NOT the "area between the graph and the x-axis" for a graph that is both above and below the x-axis. The area between a graph, y= f(x), and the x-axis is \int |f(x)|dx since area is always positive.

    This graph is below the x-axis for -3< x< -2, above the x-axis for -3< x< 0 and below the x-axis again for x> 0. To get the total area between it and the x-axis, you need to do it in three parts. For -3< x< 2, the area is \int_{-3}^{-2} |y| dx= \int_{x= -3}^{-2} x^2+ 2x dx. For -2< x< 0, the area is \int_{-2}^0 |y|dx= \int_{-2}^0 -x^2- 2x dx. For 0< x< 2, the area is \int_{x=0}^2 |y| dx= \int_{x=0}^2 x^2+ 2x dx. The total area is the sum of those three integrals.
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