Posted by Joel on November 06, 2002 at 18:48:17:
In Reply to: areas between curves posted by mfrizza on November 06, 2002 at 17:03:59:
: does this appear correct?
: find the area between the following functions...
: y = x^3-x
: y = 3x
: first find points of intersection...3x=x^3-x ---> 0 = x^3-4x ---> 0 = x(x^2-4) ---> 0 = x(x+2)(x-2) so points of intersection are (0,0) (-2,-6) (2,6)
: Then I graph...the areas between the curves are clear...
: According to the "Integrals of Symmetric Functions", since both functions are odd, they will net out to zero...which means my net area between the curves is also zero....
: A = S f(x) - g(x)
: Is this correct?
I don't think so. That argument simply defies reason. (Of course, so do some of the other arguments we have been taught in calc, but that just goes too far for me.)
That symmetry rule tells you that the definite integral of an odd function individually from -a to a is zero. But to calculate areas BETWEEN TWO functions, my eyes tell me that those two lines are marking out an area that is measurable and it isn't zero. That's like saying that a circle drawn around the origin has zero area, but if you slide it up r units along the y axis, area magically appears. When we have any other two curves that mark out areas that overlap the axes, we add the areas, essentially taking the "absolute values" of the areas. It makes no sense to do it differently just because the curves are symmetrical. What would you do if one curve was odd and the other was even, or not symmetrical at all?
Anyway, enough of my diatribe. I say you should go ahead and compute 2 areas:
the integral from -2 to 0 of x^3 - 4x
and the integral from 0 to 2 of 2x - x^3
and then add them together for the total area