Answer
The area of the region is $27/4$.
Work Step by Step
$x^3-y=0$ and $3x^2-y=4$
- For the curve $x^3-y=0$, we can rewrite into $$y=x^3$$
- For the curve $3x^2-y=4$, we can rewrite into $$y=3x^2-4$$
1) Draft the graph:
The graph is drafted below. We would use integration with respect to $x$ in this exercise.
2) Find limits of integration:
To find the limits of integration, we will find the intersection points.
$$x^3-3x^2-4=0$$ $$x=-1\hspace{1cm}\text{or}\hspace{1cm}x=2$$
Therefore, the upper limit is $2$ and the lower one is $-1$.
3) Find the area:
The region from $x=-1$ to $x=2$ is bounded above by $y=x^3$ and below by $y=3x^2-4$. So according to definition, the area of the region is $$A=\int^2_{-1}(x^3-3x^2+4)dx$$ $$A=\frac{x^4}{4}-x^3+4x\Big]^2_{-1}$$ $$A=(4-8+8)-\Big(\frac{1}{4}+1-4\Big)$$ $$A=4-\Big(-\frac{11}{4}\Big)=\frac{27}{4}$$
