find the area of the shaded region.\nf(x)=20x + x^2 - x^3, g(x)=0\nthe area is . (type an integer or a…

Explanation:

Step1: Find intersection points

Set $f(x)=20x + x^{2}-x^{3}=0$. Factor out $x$: $x(20 + x - x^{2})=0$. Then factor the quadratic: $x(5 - x)(4 + x)=0$. The roots are $x=-4,0,5$.

Step2: Set up integral for area

The area $A=\int_{-4}^{0}(20x + x^{2}-x^{3}-0)dx+\int_{0}^{4}(20x + x^{2}-x^{3}-0)dx$.

Step3: Integrate term - by - term

The antiderivative of $20x + x^{2}-x^{3}$ is $F(x)=20\times\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}=10x^{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}$.

Step4: Evaluate definite integrals

For $\int_{-4}^{0}(20x + x^{2}-x^{3})dx=F(0)-F(-4)=0-(10\times(-4)^{2}+\frac{(-4)^{3}}{3}-\frac{(-4)^{4}}{4})=-(160-\frac{64}{3}-64)=- (96-\frac{64}{3})=\frac{64}{3}-96$. For $\int_{0}^{4}(20x + x^{2}-x^{3})dx=F(4)-F(0)=(10\times4^{2}+\frac{4^{3}}{3}-\frac{4^{4}}{4})-0=(160+\frac{64}{3}-64)=96+\frac{64}{3}$.

Step5: Calculate total area

$A = (\frac{64}{3}-96)+(96+\frac{64}{3})=\frac{128}{3}$.

Answer:

$\frac{128}{3}$