area between curves that intersect at more than two points (calculator - active)\nlet f(x)=1/2x³ - 2x + 3…

Explanation:

Step1: Find intersection points

Set $f(x)=g(x)$, so $\frac{1}{2}x^{3}-2x + 3=x^{2}+\frac{7}{2}x-3$. Rearrange to $\frac{1}{2}x^{3}-x^{2}-\frac{11}{2}x + 6 = 0$, or $x^{3}-2x^{2}-11x + 12=0$. Using a graph - ing calculator, the intersection points in the interval $[-3,4]$ are found.

Step2: Determine which function is on top

For sub - intervals in $[-3,4]$, test points to see whether $f(x)-g(x)\geq0$ or $g(x)-f(x)\geq0$.

Step3: Calculate the integral

The area $A=\int_{a}^{b}|f(x)-g(x)|dx$. Split the integral based on the intersection points in $[-3,4]$. Let the intersection points be $x_1,x_2,\cdots$. Then $A=\int_{-3}^{x_1}|f(x)-g(x)|dx+\int_{x_1}^{x_2}|f(x)-g(x)|dx+\cdots+\int_{x_n}^{4}|f(x)-g(x)|dx$. Using a graphing calculator to evaluate these definite integrals: First, $h(x)=f(x)-g(x)=\frac{1}{2}x^{3}-x^{2}-\frac{11}{2}x + 6$. Integrating $|h(x)|$ from $x=-3$ to $x = 4$. $\int_{-3}^{4}\left|\frac{1}{2}x^{3}-x^{2}-\frac{11}{2}x + 6\right|dx\approx57.083$

Answer:

$57.083$