question let the region r be the area enclosed by the function f(x)=x³ + 1 and g(x)=3x + 1. write an…

Explanation:

Step1: Find intersection points

Set $x^{3}+1 = 3x + 1$. Then $x^{3}-3x=0$, factoring gives $x(x^{2} - 3)=0$. So $x = 0,x=\sqrt{3},x =-\sqrt{3}$.

Step2: Integral in terms of x

The area between two curves $y = f(x)$ and $y = g(x)$ is $A=\int_{a}^{b}|f(x)-g(x)|dx$. Here, for $-\sqrt{3}\leq x\leq0$, $x^{3}+1\leq3x + 1$ and for $0\leq x\leq\sqrt{3}$, $x^{3}+1\geq3x + 1$. So the integral in terms of $x$ is $\int_{-\sqrt{3}}^{0}((3x + 1)-(x^{3}+1))dx+\int_{0}^{\sqrt{3}}((x^{3}+1)-(3x + 1))dx=\int_{-\sqrt{3}}^{0}(3x - x^{3})dx+\int_{0}^{\sqrt{3}}(x^{3}-3x)dx$.

Step3: Solve for x in terms of y

From $y=x^{3}+1$, we have $x=\sqrt[3]{y - 1}$. From $y = 3x+1$, we have $x=\frac{y - 1}{3}$.

Step4: Find y - values of intersection points

When $x = 0,y=1$; when $x=\pm\sqrt{3},y=3\sqrt{3}+1$ and $y=- 3\sqrt{3}+1$. The integral in terms of $y$ is $\int_{-3\sqrt{3}+1}^{3\sqrt{3}+1}\left|\sqrt[3]{y - 1}-\frac{y - 1}{3}\right|dy$.

Answer:

Integral in terms of $x$: $\int_{-\sqrt{3}}^{0}(3x - x^{3})dx+\int_{0}^{\sqrt{3}}(x^{3}-3x)dx$; Integral in terms of $y$: $\int_{-3\sqrt{3}+1}^{3\sqrt{3}+1}\left|\sqrt[3]{y - 1}-\frac{y - 1}{3}\right|dy$