Definite Integration

**driver327** · January 24th 2010, 05:10 AM

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.

**HallsofIvy** · January 24th 2010, 05:44 AM

Originally Posted by driver327

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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