let the region r be the area enclosed the function f(x)=2x³, the horizontal line y = 1, and the y - axis…

let the region r be the area enclosed the function f(x)=2x³, the horizontal line y = 1, and the y - axis. 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 = ∫x1x2 dx y1 = y2 = ∫y1y2 dy

let the region r be the area enclosed the function f(x)=2x³, the horizontal line y = 1, and the y - axis. 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 = ∫x1x2 dx y1 = y2 = ∫y1y2 dy

Answer

Explanation:

Step1: Find intersection point

Set $2x^{3}=1$, then $x = \frac{1}{\sqrt[3]{2}}\approx0.794$.

Step2: Integral in terms of x

The region is bounded by $y = 2x^{3}$, $y = 1$, and $x=0$. For $0\leq x\leq\frac{1}{\sqrt[3]{2}}$, the upper - curve is $y = 1$ and the lower - curve is $y = 2x^{3}$. The area formula is $\int_{a}^{b}(y_{upper}-y_{lower})dx$. So the integral in terms of $x$ is $\int_{0}^{\frac{1}{\sqrt[3]{2}}}(1 - 2x^{3})dx$.

Step3: Solve for x in terms of y

From $y = 2x^{3}$, we get $x=\sqrt[3]{\frac{y}{2}}$.

Step4: Integral in terms of y

The region is bounded by $x=\sqrt[3]{\frac{y}{2}}$, $x = 0$, and $y = 0$ to $y = 1$. The area formula is $\int_{c}^{d}(x_{right}-x_{left})dy$. So the integral in terms of $y$ is $\int_{0}^{1}\sqrt[3]{\frac{y}{2}}dy$.

Answer:

$x_1 = 0$, $x_2=0.794$, $\int_{0}^{0.794}(1 - 2x^{3})dx$ $y_1 = 0$, $y_2 = 1$, $\int_{0}^{1}\sqrt[3]{\frac{y}{2}}dy$