area between curves that intersect at more than two points (calculator - active)\nlet f(x)=8 - 2x and…

Explanation:

Step1: Find intersection points

Set $f(x)=g(x)$, so $8 - 2x=x^{3}-7x^{2}+12x$. Rearrange to $x^{3}-7x^{2}+14x - 8 = 0$. We are interested in the interval $[1,4]$.

Step2: Determine which function is on top

We can test values in sub - intervals of $[1,4]$. Let's choose $x = 1.5$ and $x=3$. For $x = 1.5$: $f(1.5)=8-2\times1.5 = 5$, $g(1.5)=(1.5)^{3}-7\times(1.5)^{2}+12\times1.5=3.375 - 15.75+18 = 5.625$, so $g(x)>f(x)$ on some sub - interval. For $x = 3$: $f(3)=8-2\times3 = 2$, $g(3)=3^{3}-7\times3^{2}+12\times3=27 - 63 + 36 = 0$, so $f(x)>g(x)$ on some sub - interval. The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $a$ to $b$ is given by $A=\int_{a}^{b}|f(x)-g(x)|dx=\int_{a}^{c}(g(x)-f(x))dx+\int_{c}^{b}(f(x)-g(x))dx$ (where $c$ is the intersection point in $(a,b)$). Using a graphing calculator, we calculate $\int_{1}^{4}|(x^{3}-7x^{2}+12x)-(8 - 2x)|dx$. $\int_{1}^{4}|x^{3}-7x^{2}+14x - 8|dx$. We know that $x^{3}-7x^{2}+14x - 8=(x - 1)(x - 2)(x - 4)$. $\int_{1}^{2}((x^{3}-7x^{2}+12x)-(8 - 2x))dx+\int_{2}^{4}((8 - 2x)-(x^{3}-7x^{2}+12x))dx$ $=\int_{1}^{2}(x^{3}-7x^{2}+14x - 8)dx+\int_{2}^{4}(-x^{3}+7x^{2}-14x + 8)dx$ For $\int(x^{3}-7x^{2}+14x - 8)dx=\frac{x^{4}}{4}-\frac{7x^{3}}{3}+7x^{2}-8x+C$ $\int_{1}^{2}(\frac{x^{4}}{4}-\frac{7x^{3}}{3}+7x^{2}-8x)\big|{1}^{2}=(\frac{2^{4}}{4}-\frac{7\times2^{3}}{3}+7\times2^{2}-8\times2)-(\frac{1^{4}}{4}-\frac{7\times1^{3}}{3}+7\times1^{2}-8\times1)$ $=(4-\frac{56}{3}+28 - 16)-(\frac{1}{4}-\frac{7}{3}+7 - 8)$ For $\int(-x^{3}+7x^{2}-14x + 8)dx=-\frac{x^{4}}{4}+\frac{7x^{3}}{3}-7x^{2}+8x+C$ $\int{2}^{4}(-\frac{x^{4}}{4}+\frac{7x^{3}}{3}-7x^{2}+8x)\big|{2}^{4}=(-\frac{4^{4}}{4}+\frac{7\times4^{3}}{3}-7\times4^{2}+8\times4)-(-\frac{2^{4}}{4}+\frac{7\times2^{3}}{3}-7\times2^{2}+8\times2)$ After calculating with a calculator: $\int{1}^{2}(x^{3}-7x^{2}+14x - 8)dx+\int_{2}^{4}(-x^{3}+7x^{2}-14x + 8)dx\approx5.083$

Answer:

$5.083$