Find the area enclosed by \( x^3-y=0 \) and \( 3x^2-y=4 \)

Solution:

First we need to find our points of intersection.

\( x^3 = 3x^2-4 \)
\( x^3-3x^2+4=0 \)
This has possible rational zeros of \( \pm 1, \pm 2 , \pm 4 \)
Testing these, we see we have \( x = -1 \) and \( x=2 \)
\( (x+1)(x-2)(x-2) = 0 \)
\( x=-1,2 \)