this question is designed to be answered with a calculator. region r is enclosed by the graphs of x = 2y²…

Answer:

C. 9

Explanation:

Step1: Find intersection points

Set $2y^{2}-4 = 2y$. Rearrange to $2y^{2}-2y - 4=0$, then $y^{2}-y - 2 = 0$. Factor: $(y - 2)(y+1)=0$. So $y=-1,2$.

Step2: Set up integral for area

Use the formula for the area between two curves with respect to $y$: $A=\int_{a}^{b}(right - left)dy$. Here, $x_1 = 2y$ and $x_2=2y^{2}-4$, $a=-1$, $b = 2$. So $A=\int_{-1}^{2}(2y-(2y^{2}-4))dy$.

Step3: Integrate

$A=\int_{-1}^{2}(2y - 2y^{2}+4)dy=\left[y^{2}-\frac{2}{3}y^{3}+4y\right]_{-1}^{2}$.

Step4: Evaluate definite - integral

$A=(2^{2}-\frac{2}{3}(2)^{3}+4\times2)-((-1)^{2}-\frac{2}{3}(-1)^{3}+4\times(-1))$. $A=(4-\frac{16}{3}+8)-(1 + \frac{2}{3}-4)$. $A=(12-\frac{16}{3})-(\frac{3 + 2}{3}-4)$. $A=\frac{36-16}{3}-(\frac{5}{3}-4)$. $A=\frac{20}{3}-\frac{5 - 12}{3}$. $A=\frac{20}{3}+\frac{7}{3}=\frac{27}{3}=9$.