let the region r be the area enclosed by the function f(x)=x^(1/3)+2 and g(x)=(1/4)x + 2. write an integral…

Explanation:

Step1: Find intersection points

Set $f(x)=g(x)$, so $x^{\frac{1}{3}} + 2=\frac{1}{4}x + 2$. Then $x^{\frac{1}{3}}=\frac{1}{4}x$. Let $t = x^{\frac{1}{3}}$, so $t=\frac{1}{4}t^{3}$, which gives $t^{3}-4t = 0$, $t(t^{2}-4)=0$, $t(t - 2)(t + 2)=0$. So $t=0,2,-2$ and $x = 0,8,-8$.

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 $- 8\leqslant x\leqslant0$, $f(x)\geqslant g(x)$ and for $0\leqslant x\leqslant8$, $f(x)\geqslant g(x)$. So the integral in terms of $x$ is $\int_{-8}^{8}(x^{\frac{1}{3}}+2 - (\frac{1}{4}x + 2))dx=\int_{-8}^{8}(x^{\frac{1}{3}}-\frac{1}{4}x)dx$. Here $x_1=-8$, $x_2 = 8$ and the integrand is $x^{\frac{1}{3}}-\frac{1}{4}x$.

Step3: Rewrite functions in terms of y

From $y=x^{\frac{1}{3}}+2$, we get $x=(y - 2)^{3}$. From $y=\frac{1}{4}x + 2$, we get $x = 4(y - 2)$. Find intersection - points in terms of y. When $x=-8$, $y=0$; when $x = 8$, $y=4$. The area between two curves $x = h(y)$ and $x = k(y)$ is $A=\int_{c}^{d}|h(y)-k(y)|dy$. So the integral in terms of $y$ is $\int_{0}^{4}((y - 2)^{3}-4(y - 2))dy$. Here $y_1 = 0$, $y_2=4$ and the integrand is $(y - 2)^{3}-4(y - 2)$.

Answer:

$x_1=-8$ $x_2 = 8$ $\int_{-8}^{8}(x^{\frac{1}{3}}-\frac{1}{4}x)dx$ $y_1 = 0$ $y_2=4$ $\int_{0}^{4}((y - 2)^{3}-4(y - 2))dy$