let the region r be the area enclosed the function f(x)=3x³, the horizontal line y = 2, and the y - axis…

let the region r be the area enclosed the function f(x)=3x³, the horizontal line y = 2, 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)=3x³, the horizontal line y = 2, 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 the intersection - point of $y = 3x^{3}$ and $y = 2$

Set $3x^{3}=2$, then $x^{3}=\frac{2}{3}$, and $x=\sqrt[3]{\frac{2}{3}}\approx0.874$.

Step2: Integral with respect to $x$

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

Step3: Solve for $x$ in terms of $y$

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

Step4: Integral with respect to $y$

The region is bounded by $x=\sqrt[3]{\frac{y}{3}}$ and $x = 0$ for $0\leq y\leq2$. The area formula with respect to $y$ is $\int_{c}^{d}xdy$. Here, $c = 0$, $d = 2$, and $x=\sqrt[3]{\frac{y}{3}}$. So the integral with respect to $y$ is $\int_{0}^{2}\sqrt[3]{\frac{y}{3}}dy$.

Answer:

$x_1 = 0$ $x_2=\sqrt[3]{\frac{2}{3}}\approx0.874$ $\int_{0}^{\sqrt[3]{\frac{2}{3}}}(2 - 3x^{3})dx$ $y_1 = 0$ $y_2 = 2$ $\int_{0}^{2}\sqrt[3]{\frac{y}{3}}dy$