Answer
$\frac{13}{6}$
Work Step by Step
Step 1. Graph the functions as shown and we can identify intersections at $(1,1)$, $(4,2)$ and $(5,1)$. The area of the enclosed region by the functions is the integral with respect to $y$ from $1$ to $2$.
Step 2. Rewrite the functions as $x=6-y$ and $x=y^2$
Step 3. Evaluate
$A=\int_{1}^2(6-y-y^2)dy=(6y-\frac{1}{2}y^2-\frac{1}{3}y^3)|_{1}^2=(6(2)-\frac{1}{2}(2)^2-\frac{1}{3}(2)^3)-(6-\frac{1}{2}-\frac{1}{3})=\frac{13}{6}$
