the curves x = -y² + 3y + 11 and x = y² + y - 1 are graphed. which expression represents the area bounded by…

Explanation:

Step1: Find intersection - points

Set $-y^{2}+3y + 11=y^{2}+y - 1$. Rearrange to get $2y^{2}-2y - 12 = 0$, then $y^{2}-y - 6=0$. Factor: $(y - 3)(y + 2)=0$. So $y=-2$ and $y = 3$.

Step2: Determine the upper - and lower - functions

For $-2\leq y\leq3$, the right - hand curve is $x=-y^{2}+3y + 11$ and the left - hand curve is $x=y^{2}+y - 1$. The area $A=\int_{a}^{b}[(right - hand\ function)-(left - hand\ function)]dy$.

Step3: Set up the integral

$A=\int_{-2}^{3}[(-y^{2}+3y + 11)-(y^{2}+y - 1)]dy=\int_{-2}^{3}(-2y^{2}+2y + 12)dy$.

Answer:

$\int_{-2}^{3}(-2y^{2}+2y + 12)dy$ (Option D)