let the region r be the area enclosed by the function f(x)=x^(1/3) and g(x)=(1/4)x. write an integral in…

Explanation:

Step1: Find intersection points

Set $x^{\frac{1}{3}}=\frac{1}{4}x$. Let $t = x^{\frac{1}{3}}$, then $t=\frac{1}{4}t^{3}$, $t^{3}-4t = 0$, $t(t^{2}-4)=0$, $t(t - 2)(t + 2)=0$. So $t=0,2,-2$ and $x = 0,8,- 8$. We consider the non - negative part of the graph (from the visual), so the intersection points are $x = 0$ and $x = 8$.

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 $0\leq x\leq8$, $x^{\frac{1}{3}}\geq\frac{1}{4}x$. So the integral in terms of $x$ is $\int_{0}^{8}(x^{\frac{1}{3}}-\frac{1}{4}x)dx$.

Step3: Rewrite functions in terms of $y$

If $y = x^{\frac{1}{3}}$, then $x=y^{3}$; if $y=\frac{1}{4}x$, then $x = 4y$.

Step4: Find intersection points in terms of $y$

Set $y^{3}=4y$, $y^{3}-4y=0$, $y(y^{2}-4)=0$, $y(y - 2)(y + 2)=0$. For the non - negative part, $y = 0$ and $y = 2$.

Step5: Integral in terms of $y$

The area between two curves $x = h(y)$ and $x = k(y)$ is $A=\int_{c}^{d}|h(y)-k(y)|dy$. Here, for $0\leq y\leq2$, $4y\geq y^{3}$. So the integral in terms of $y$ is $\int_{0}^{2}(4y - y^{3})dy$.

Answer:

Integral in terms of $x$: $\int_{0}^{8}(x^{\frac{1}{3}}-\frac{1}{4}x)dx$ Integral in terms of $y$: $\int_{0}^{2}(4y - y^{3})dy$