determine the area of the shaded region in the figure.\nthe area of the shaded region is \n(type an exact…

Explanation:

Step1: Find intersection points

Set $y^{2}-12 = y$, then $y^{2}-y - 12=0$. Factoring gives $(y - 4)(y+3)=0$. So $y=-3$ and $y = 4$.

Step2: Set up integral for area

The right - hand function is $x = y$ and the left - hand function is $x=y^{2}-12$. The area $A$ between two curves $x = f(y)$ and $x = g(y)$ from $y = a$ to $y = b$ is $A=\int_{a}^{b}[f(y)-g(y)]dy$. Here, $f(y)=y$, $g(y)=y^{2}-12$, $a=-3$ and $b = 4$. So $A=\int_{-3}^{4}(y-(y^{2}-12))dy=\int_{-3}^{4}(y - y^{2}+12)dy$.

Step3: Integrate term - by - term

$\int_{-3}^{4}(y - y^{2}+12)dy=\left[\frac{y^{2}}{2}-\frac{y^{3}}{3}+12y\right]_{-3}^{4}$.

Step4: Evaluate the definite integral

First, substitute $y = 4$: $\frac{4^{2}}{2}-\frac{4^{3}}{3}+12\times4=\frac{16}{2}-\frac{64}{3}+48=8-\frac{64}{3}+48=\frac{24 - 64+144}{3}=\frac{104}{3}$. Then substitute $y=-3$: $\frac{(-3)^{2}}{2}-\frac{(-3)^{3}}{3}+12\times(-3)=\frac{9}{2}+9 - 36=\frac{9 + 18-72}{2}=-\frac{45}{2}$. Subtract the two results: $\frac{104}{3}-\left(-\frac{45}{2}\right)=\frac{104\times2+45\times3}{6}=\frac{208 + 135}{6}=\frac{343}{6}$.

Answer:

$\frac{343}{6}$