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July 6, 2015

July 6, 2015

Posted by **david** on Wednesday, February 4, 2009 at 3:03am.

and y=–x^3+14x^2–45x

i am having trouble how the graphs look like and how to do this...

- Calc -
**bobpursley**, Wednesday, February 4, 2009 at 8:28amAs 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.

- Calc -
**Reiny**, Wednesday, February 4, 2009 at 8:31amfactor 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.