question find the area bounded by y = x^3 + 2x and y = x^2 + 8x. submit your answer in fractional form…

Explanation:

Step1: Find intersection points

Set $x^{3}+2x=x^{2}+8x$. Rearrange to $x^{3}-x^{2}-6x = 0$. Factor out $x$: $x(x^{2}-x - 6)=0$. Factor the quadratic: $x(x - 3)(x+2)=0$. The solutions are $x=-2,0,3$.

Step2: Determine which function is on top

Let $f(x)=x^{3}+2x$ and $g(x)=x^{2}+8x$. For $-2<x<0$, $f(x)-g(x)=x^{3}-x^{2}-6x=x(x - 3)(x + 2)>0$, so $y=x^{3}+2x$ is on top. For $0<x<3$, $g(x)-f(x)=-x^{3}+x^{2}+6x=-x(x - 3)(x + 2)>0$, so $y=x^{2}+8x$ is on top.

Step3: Calculate the area using integral

The area $A=\int_{-2}^{0}[(x^{3}+2x)-(x^{2}+8x)]dx+\int_{0}^{3}[(x^{2}+8x)-(x^{3}+2x)]dx$. First integral: $\int_{-2}^{0}(x^{3}-x^{2}-6x)dx=\left[\frac{x^{4}}{4}-\frac{x^{3}}{3}-3x^{2}\right]{-2}^0$ $=0-\left(\frac{(-2)^{4}}{4}-\frac{(-2)^{3}}{3}-3(-2)^{2}\right)$ $=-\left(4+\frac{8}{3}-12\right)=\frac{16}{3}$. Second integral: $\int{0}^{3}( - x^{3}+x^{2}+6x)dx=\left[-\frac{x^{4}}{4}+\frac{x^{3}}{3}+3x^{2}\right]_{0}^{3}$ $=-\frac{3^{4}}{4}+\frac{3^{3}}{3}+3\times3^{2}$ $=-\frac{81}{4}+9 + 27=\frac{-81 + 36+108}{4}=\frac{63}{4}$.

Step4: Sum the two - part areas

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

Answer:

$\frac{253}{12}$