find the area of the shaded region.\nf(x)=6x - x^{2}-x^{3}, g(x)=0\nthe area is (type an integer or a…

Explanation:

Step1: Find intersection points

Set $f(x)=6x - x^{2}-x^{3}=0$. Factor out $x$: $x(6 - x - x^{2})=0$, then $x(3 + x)(2 - x)=0$. The solutions are $x = 0$, $x=-3$ and $x = 2$.

Step2: Set up integral for area

The area $A$ between the curves $y = f(x)$ and $y = g(x)=0$ is given by $A=\int_{a}^{b}|f(x)-g(x)|dx$. Here, we need to split the integral based on the sign of $f(x)$ in the intervals determined by the intersection - points. In the interval $[-3,0]$, $f(x)\leq0$ and in the interval $[0,2]$, $f(x)\geq0$. So $A=\int_{-3}^{0}-(6x - x^{2}-x^{3})dx+\int_{0}^{2}(6x - x^{2}-x^{3})dx$.

Step3: Integrate term - by - term

For $\int(6x - x^{2}-x^{3})dx=6\times\frac{x^{2}}{2}-\frac{x^{3}}{3}-\frac{x^{4}}{4}=3x^{2}-\frac{x^{3}}{3}-\frac{x^{4}}{4}+C$. For $\int_{-3}^{0}-(6x - x^{2}-x^{3})dx=-\left[3x^{2}-\frac{x^{3}}{3}-\frac{x^{4}}{4}\right]{-3}^{0}=-\left(0-(3\times(-3)^{2}-\frac{(-3)^{3}}{3}-\frac{(-3)^{4}}{4})\right)=-(0-(27 + 9-\frac{81}{4}))=\frac{81}{4}-36=\frac{81 - 144}{4}=-\frac{63}{4}$. For $\int{0}^{2}(6x - x^{2}-x^{3})dx=\left[3x^{2}-\frac{x^{3}}{3}-\frac{x^{4}}{4}\right]_{0}^{2}=(3\times2^{2}-\frac{2^{3}}{3}-\frac{2^{4}}{4})-(0)=(12-\frac{8}{3}-4)=8-\frac{8}{3}=\frac{24 - 8}{3}=\frac{16}{3}$.

Step4: Calculate total area

$A=\frac{63}{4}+\frac{16}{3}=\frac{189 + 64}{12}=\frac{253}{12}$.

Answer:

$\frac{253}{12}$