area between curves that intersect at more than two points (calculator - active) google classroom microsoft…

Explanation:

Step1: Recall area - between - curves formula

The area between two curves $y = f(x)$ and $y = g(x)$ on an interval $[a,b]$ is given by $A=\int_{a}^{b}|f(x)-g(x)|dx$. First, we need to find the intersection points of $f(x)=x^{3}+x^{2}-6x$ and $g(x)=-\frac{4x}{x + 3}$. We know two intersection points are $x=-2$ and $x = 0$ from the graph. To find other intersection points, we set $x^{3}+x^{2}-6x=-\frac{4x}{x + 3}$.

Step2: Simplify the intersection - point equation

[ \begin{align*} x^{3}+x^{2}-6x&=-\frac{4x}{x + 3}\ (x^{3}+x^{2}-6x)(x + 3)&=-4x\ x^{4}+3x^{3}+x^{3}+3x^{2}-6x^{2}-18x + 4x&=0\ x^{4}+4x^{3}-3x^{2}-14x&=0\ x(x^{3}+4x^{2}-3x - 14)&=0 \end{align*} ] We know $x = 0$ is a root. By using a graphing calculator or synthetic division, we find the other non - zero root. Let's assume the roots of the equation $f(x)-g(x)=0$ are $x_1,x_2,x_3$. The sum of the areas of regions $R$ and $S$ is $A=\int_{x_1}^{x_2}|f(x)-g(x)|dx+\int_{x_2}^{x_3}|f(x)-g(x)|dx$. Using a graphing calculator (e.g., TI - 84 Plus):

1. Enter $Y_1=x^{3}+x^{2}-6x$ and $Y_2=-\frac{4x}{x + 3}$.
2. Use the "intersect" function to find the intersection points. The intersection points are $x=-2,x = 0,x=2$.
3. Then, calculate $\int_{-2}^{0}[(x^{3}+x^{2}-6x)-(-\frac{4x}{x + 3})]dx+\int_{0}^{2}[(-\frac{4x}{x + 3})-(x^{3}+x^{2}-6x)]dx$. [ \begin{align*} \int_{-2}^{0}\left(x^{3}+x^{2}-6x+\frac{4x}{x + 3}\right)dx+\int_{0}^{2}\left(-\frac{4x}{x + 3}-x^{3}-x^{2}+6x\right)dx&=\int_{-2}^{0}\left(x^{3}+x^{2}-6x+\frac{4x + 12-12}{x + 3}\right)dx+\int_{0}^{2}\left(-\frac{4x+12 - 12}{x + 3}-x^{3}-x^{2}+6x\right)dx\ &=\int_{-2}^{0}\left(x^{3}+x^{2}-6x + 4-\frac{12}{x + 3}\right)dx+\int_{0}^{2}\left(-4+\frac{12}{x + 3}-x^{3}-x^{2}+6x\right)dx \end{align*} ] [ \begin{align*} \int_{-2}^{0}\left(x^{3}+x^{2}-6x + 4-\frac{12}{x + 3}\right)dx&=\left[\frac{x^{4}}{4}+\frac{x^{3}}{3}-3x^{2}+4x-12\ln|x + 3|\right]{-2}^0\ &=(0)-\left(\frac{(-2)^{4}}{4}+\frac{(-2)^{3}}{3}-3(-2)^{2}+4(-2)-12\ln|-2 + 3|\right)\ &=-\left(4-\frac{8}{3}-12-8-0\right)\ &=-\left(4-\frac{8}{3}-20\right)\ &=-\left(\frac{12 - 8-60}{3}\right)\ &=\frac{56}{3} \end{align*} ] [ \begin{align*} \int{0}^{2}\left(-4+\frac{12}{x + 3}-x^{3}-x^{2}+6x\right)dx&=\left[-4x + 12\ln|x + 3|-\frac{x^{4}}{4}-\frac{x^{3}}{3}+3x^{2}\right]_{0}^{2}\ &=\left(-4\times2+12\ln(5)-\frac{2^{4}}{4}-\frac{2^{3}}{3}+3\times2^{2}\right)-(0)\ &=-8 + 12\ln(5)-4-\frac{8}{3}+12\ &=12\ln(5)-\frac{8}{3} \end{align*} ] [ \begin{align*} A&=\frac{56}{3}+12\ln(5)-\frac{8}{3}\ &=\frac{48}{3}+12\ln(5)\ &=16 + 12\ln(5)\ &\approx16+19.3132\ &\approx35.313 \end{align*} ]

Answer:

$35.313$