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

let the region r be the area enclosed by the function f(x)=2x^1 and g(x)=2/3x. 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=□ ∫x1^x2 □dx y1=□ y2=□ ∫y1^y2 □dy

let the region r be the area enclosed by the function f(x)=2x^1 and g(x)=2/3x. 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=□ ∫x1^x2 □dx y1=□ y2=□ ∫y1^y2 □dy

Answer

Explanation:

Step1: Find intersection points

Set $2x^{\frac{1}{3}}=\frac{2}{3}x$. Let $t = x^{\frac{1}{3}}$, then $2t=\frac{2}{3}t^{3}$. Rearranging gives $t^{3}-3t = 0$, factoring out $t$ we get $t(t^{2}-3)=0$. So $t = 0,t=\sqrt{3},t =-\sqrt{3}$. Then $x = 0,x = 3\sqrt{3},x=- 3\sqrt{3}$. Since the region is in the first - quadrant as shown in the graph, we consider $x_1 = 0,x_2=3\sqrt{3}\approx5.196$.

Step2: Integral in terms of x

The upper - function is $y_1 = 2x^{\frac{1}{3}}$ and the lower - function is $y_2=\frac{2}{3}x$. The area $A_x=\int_{x_1}^{x_2}(2x^{\frac{1}{3}}-\frac{2}{3}x)dx=\int_{0}^{5.196}(2x^{\frac{1}{3}}-\frac{2}{3}x)dx$.

Step3: Inverse of the functions

For $y = 2x^{\frac{1}{3}}$, we can solve for $x$: $x=\frac{y^{3}}{8}$. For $y=\frac{2}{3}x$, we have $x=\frac{3}{2}y$.

Step4: Find y - limits

When $x = 0,y = 0$; when $x = 3\sqrt{3},y = 2\sqrt{3}\approx3.464$.

Step5: Integral in terms of y

The right - function is $x_1=\frac{3}{2}y$ and the left - function is $x_2=\frac{y^{3}}{8}$. The area $A_y=\int_{y_1}^{y_2}(\frac{3}{2}y-\frac{y^{3}}{8})dy=\int_{0}^{3.464}(\frac{3}{2}y-\frac{y^{3}}{8})dy$.

Answer:

$x_1 = 0$ $x_2\approx5.196$ $\int_{0}^{5.196}(2x^{\frac{1}{3}}-\frac{2}{3}x)dx$ $y_1 = 0$ $y_2\approx3.464$ $\int_{0}^{3.464}(\frac{3}{2}y-\frac{y^{3}}{8})dy$