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 \( x=2 \)

\( (x+1)(x-2)(x-2) = 0 \)

\( x=-1,2 \)

\( \int \limits _{-1}^2 \left ( x^3- \left ( 3x^2-4 \right ) \right ) dx \)

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

\( x^3-3x^2+4=0 \)

This has possible rational zeros of \( \pm 1, \pm 2 , \pm 4 \)

Testing these, we see we have \( x = -1 \) and \( x=2 \)

\( (x+1)(x-2)(x-2) = 0 \)

\( x=-1,2 \)

\( \int \limits _{-1}^2 \left ( x^3- \left ( 3x^2-4 \right ) \right ) dx \)

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