Answer
$\frac{243}{8}$
Work Step by Step
Step 1. Graph the functions as shown and we can identify intersections at $(3,-4)$ and $(\frac{21}{4},5)$. The area of the enclosed region by the functions is the integral with respect to $y$ from $-4$ to $5$.
Step 2. Rewrite the functions as $x=\frac{1}{4}y^2-1$ and $x=\frac{1}{4}y+4$
Step 3. Evaluate
$A=\int_{-4}^5(\frac{1}{4}y+4-(\frac{1}{4}y^2-1))dy=\int_{-4}^5(\frac{1}{4}y+5-\frac{1}{4}y^2)dy=
(\frac{1}{8}y^2+5y-\frac{1}{12}y^3)|_{-4}^5=(\frac{1}{8}(5)^2+5(5)-\frac{1}{12}(5)^3)-(\frac{1}{8}(-4)^2+5(-4)-\frac{1}{12}(-4)^3)=43-\frac{101}{8}=\frac{243}{8}$
