question let the region r be the area enclosed by the function f(x)=3x^(1/3) and g(x)=(3/2)x. write an…

Explanation:

Step1: Find intersection points

Set $3x^{\frac{1}{3}}=\frac{3}{2}x$. Let $t = x^{\frac{1}{3}}$, then $3t=\frac{3}{2}t^{3}$. Rearranging gives $t^{3}-2t = 0$, factoring out $t$ we get $t(t^{2}-2)=0$. So $t = 0,\pm\sqrt{2}$. Then $x = 0,\pm2\sqrt{2}$. Since the region in the graph is in the first - quadrant, we consider $x = 0$ and $x = 2\sqrt{2}\approx2.828$.

Step2: Integral in terms of x

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x = b$ is $A=\int_{a}^{b}|f(x)-g(x)|dx$. Here, for $0\leq x\leq2\sqrt{2}$, $3x^{\frac{1}{3}}\geq\frac{3}{2}x$, so the integral in terms of $x$ is $\int_{0}^{2\sqrt{2}}(3x^{\frac{1}{3}}-\frac{3}{2}x)dx$.

Step3: Solve for x in terms of y

From $y = 3x^{\frac{1}{3}}$, we get $x=\frac{y^{3}}{27}$; from $y=\frac{3}{2}x$, we get $x=\frac{2y}{3}$.

Step4: Find intersection points in terms of y

Set $\frac{y^{3}}{27}=\frac{2y}{3}$. Rearranging gives $y^{3}-18y = 0$, factoring out $y$ we get $y(y^{2}-18)=0$. So $y = 0,\pm3\sqrt{2}$. Since the region is in the first - quadrant, we consider $y = 0$ and $y = 3\sqrt{2}\approx4.243$.

Step5: Integral in terms of y

The area between two curves $x = h(y)$ and $x = k(y)$ from $y = c$ to $y = d$ is $A=\int_{c}^{d}|h(y)-k(y)|dy$. Here, for $0\leq y\leq3\sqrt{2}$, $\frac{2y}{3}\geq\frac{y^{3}}{27}$, so the integral in terms of $y$ is $\int_{0}^{3\sqrt{2}}(\frac{2y}{3}-\frac{y^{3}}{27})dy$.

Answer:

Integral in terms of $x$: $\int_{0}^{2\sqrt{2}}(3x^{\frac{1}{3}}-\frac{3}{2}x)dx$ Integral in terms of $y$: $\int_{0}^{3\sqrt{2}}(\frac{2y}{3}-\frac{y^{3}}{27})dy$