Posted by david on Wednesday, February 4, 2009 at 3:03am.
As a practical matter, you will have to find the graphs. You have a problem of "negative" areas if your ytop becomes ybottom on your integration, so the total a area integrated becomes less (postive and negative areas combined.). So I would graph them (graphing calcs have a few nice advantages).
Just looking at the equations, where y1-y1 ( a probable crossing point)
x^3-14x+45=-x^3+14x-45 or
2x(x^2-14x+45)=0 which after factoring, give you three roots
2x(x-9)(x-5)=0. So knowing nothing else, you could look at from -inf to zero, zero to 5, 5 to 9, and 9 to inf.
But at - inf,the curves go to y +- unbounded,so that does bound an areas, and the same is true on the upper end. So your area of interest is
zero to 5, 5 to 9
Without graphing, you might be wondering which one is y top?
WEll, it wont matter if you integrate both as absolutes, then add them
area= abs INT ytop-ybottom dx from x=0 to 5 PLUS abs INT ytop-ybottom dx from x=5 to 9.
Which is ytop and y bottom? Machts nichts.
But it is nicer, and more clear, if you have a graph first.
factor the first into
y = x(x-5)(x-9)
so you have a cubic which crosses the x-axis at 0, 5, and 9
your second equation is simply the reflection of that in the x-axis.
So the closed regions are between 0 and 5 and between 5 and 9, with the bottom half of each equal to its matching top half.
This is an integration problem in calculus
I would then take
2[integral of](x^3–14x^2+45x)dx from 0 to 5 + 2[integral of](-x^3+14x^2-45x)dx from 5 to 9
I hope you can take it from there.