Solving Double Integrals: y = 4x^3 - x^4 & y = 3 - 4x + 4x^2

[haroldholt]

I'm having some trouble with this particular question.

∫∫x dA bound by y = 4x^3 - x^4 and y = 3 - 4x + 4x^2.

All I can think to do is equate the two equations to find where they intercept to give the bounds for the double integral giving 0 = x^4 - 4x^3 + 4x^2 - 4x + 3. But I don't know where to go from here.

Any help would be appreciated.

 

[Pseudo Statistic]

That's pretty nasty.
Maybe you could try Newton's method? Take an initial guess, e.g. x=1...
Actually, it looks like x=1 works.. (1 - 4 + 4 - 4 + 3 = -3 + 4 - 4 + 3 = 0)
So since we have that, divide through by (x-1) to get the other roots.
Have fun. :D

 

[NateTG]

The only possible rational roots are:
$$\pm 1, \pm 3$$
so you could start by checking whether those are intersections.

 

[haroldholt]

What method did you use to find those roots?

 

[NateTG]

http://mathworld.wolfram.com/RationalZeroTheorem.html

Gives you a list of rational zeros to try.

 

[haroldholt]

Cheers mate. Can't say I've ever heard of the rational zero theorem.