let f(x)=8 - 2x and g(x)=x³ - 7x² + 12x. find the sum of the areas enclosed by the graphs of f and g between…

Explanation:

Step1: Determine which function is greater

First, find the difference $h(x)=f(x)-g(x)=8 - 2x-(x^{3}-7x^{2}+12x)=-x^{3}+7x^{2}-14x + 8$. Evaluate $h(x)$ at some points in the interval $[1,4]$. $h(1)=8-2-(1 - 7+12)=0$, $h(2)=8-4-(8 - 28+24)=4 - 4=0$, $h(3)=8-6-(27 - 63+36)=2 - 0 = 2$, $h(4)=8-8-(64 - 112+48)=0$. We need to split the integral based on where the functions cross. The functions cross at $x = 1,x = 2,x = 4$.

Step2: Set up the integral for the area

The area $A=\int_{1}^{2}[(x^{3}-7x^{2}+12x)-(8 - 2x)]dx+\int_{2}^{4}[(8 - 2x)-(x^{3}-7x^{2}+12x)]dx$. Simplify the integrands: For the first - integral, $(x^{3}-7x^{2}+12x)-(8 - 2x)=x^{3}-7x^{2}+14x - 8$. For the second - integral, $(8 - 2x)-(x^{3}-7x^{2}+12x)=-x^{3}+7x^{2}-14x + 8$.

Step3: Integrate term - by - term

The antiderivative of $x^{3}-7x^{2}+14x - 8$ is $\frac{x^{4}}{4}-\frac{7x^{3}}{3}+7x^{2}-8x$. $\int_{1}^{2}(x^{3}-7x^{2}+14x - 8)dx=\left[\frac{x^{4}}{4}-\frac{7x^{3}}{3}+7x^{2}-8x\right]{1}^{2}$ $=\left(\frac{2^{4}}{4}-\frac{7\times2^{3}}{3}+7\times2^{2}-8\times2\right)-\left(\frac{1^{4}}{4}-\frac{7\times1^{3}}{3}+7\times1^{2}-8\times1\right)$ $=(4-\frac{56}{3}+28 - 16)-(\frac{1}{4}-\frac{7}{3}+7 - 8)$ $=(16-\frac{56}{3})-(\frac{3 - 28 + 84 - 96}{12})$ $=\frac{48 - 56}{3}-\frac{-37}{12}=-\frac{8}{3}+\frac{37}{12}=\frac{-32 + 37}{12}=\frac{5}{12}$. The antiderivative of $-x^{3}+7x^{2}-14x + 8$ is $-\frac{x^{4}}{4}+\frac{7x^{3}}{3}-7x^{2}+8x$. $\int{2}^{4}(-x^{3}+7x^{2}-14x + 8)dx=\left[-\frac{x^{4}}{4}+\frac{7x^{3}}{3}-7x^{2}+8x\right]_{2}^{4}$ $=\left(-\frac{4^{4}}{4}+\frac{7\times4^{3}}{3}-7\times4^{2}+8\times4\right)-\left(-\frac{2^{4}}{4}+\frac{7\times2^{3}}{3}-7\times2^{2}+8\times2\right)$ $=(-64+\frac{448}{3}-112 + 32)-(-4+\frac{56}{3}-28 + 16)$ $=(-144+\frac{448}{3})-(-16+\frac{56}{3})$ $=\frac{-432 + 448}{3}-\frac{-48+56}{3}=\frac{16}{3}-\frac{8}{3}=\frac{8}{3}$.

Step4: Calculate the total area

$A=\frac{5}{12}+\frac{8}{3}=\frac{5 + 32}{12}=\frac{37}{12}\approx3.083$.

Answer:

$3.083$