### Quiz 5.6:

Question:

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 $$x=2$$
$$(x+1)(x-2)(x-2) = 0$$
$$x=-1,2$$

$$\int \limits _{-1}^2 \left ( x^3- \left ( 3x^2-4 \right ) \right ) dx$$
$$= \left [ \frac{x^4}{4} - x^3 + 4x \right ]_{-1}^2=\left ( 4-8+8 \right ) - \left (\frac{1}{4}+1-4 \right ) = \frac{27}{4}$$