let the region r be the area enclosed by the function f(x)=x³ + 2 and g(x)=4x + 2. write an integral in…

let the region r be the area enclosed by the function f(x)=x³ + 2 and g(x)=4x + 2. write an integral in terms of x and also an integral in terms of y that would represent the area of the region r. if necessary, round limit values to the nearest thousandth. answer attempt 1 out of 3 x1=□ x2=□ ∫x1x2□dx y1=□ y2=□ ∫y1y2□dy

let the region r be the area enclosed by the function f(x)=x³ + 2 and g(x)=4x + 2. write an integral in terms of x and also an integral in terms of y that would represent the area of the region r. if necessary, round limit values to the nearest thousandth. answer attempt 1 out of 3 x1=□ x2=□ ∫x1x2□dx y1=□ y2=□ ∫y1y2□dy

Answer

Explanation:

Step1: Find intersection points

Set $f(x)=g(x)$, so $x^{3}+2 = 4x + 2$. Simplify to $x^{3}-4x=0$, factor as $x(x^{2}-4)=0$, then $x(x - 2)(x + 2)=0$. The solutions are $x=-2,0,2$. The relevant intersection - points for the region shown are $x = 0$ and $x = 2$.

Step2: Determine the upper and lower functions for $x$ - integral

For $x\in[0,2]$, $g(x)=4x + 2$ is above $f(x)=x^{3}+2$. The area in terms of $x$ is given by $\int_{a}^{b}(g(x)-f(x))dx$. Here, $a = 0$, $b = 2$, and $g(x)-f(x)=(4x + 2)-(x^{3}+2)=4x - x^{3}$.

Step3: Express $x$ in terms of $y$

For $y=x^{3}+2$, we have $x=\sqrt[3]{y - 2}$. For $y = 4x+2$, we have $x=\frac{y - 2}{4}$.

Step4: Find the $y$ - values of intersection points

When $x = 0$, $y=f(0)=g(0)=2$. When $x = 2$, $y=f(2)=g(2)=10$.

Step5: Determine the right - hand and left - hand functions for $y$ - integral

For $y\in[2,10]$, $x=\frac{y - 2}{4}$ is to the right of $x=\sqrt[3]{y - 2}$. The area in terms of $y$ is $\int_{c}^{d}(\text{right}-\text{left})dy$, where $c = 2$, $d = 10$, and $\text{right}-\text{left}=\frac{y - 2}{4}-\sqrt[3]{y - 2}$.

Answer:

$x_1 = 0$ $x_2 = 2$ $\int_{0}^{2}(4x - x^{3})dx$ $y_1 = 2$ $y_2 = 10$ $\int_{2}^{10}(\frac{y - 2}{4}-\sqrt[3]{y - 2})dy$