Answer
$\frac{9}{8}$
Work Step by Step
Step 1. Graph the functions as shown and we can identify intersections at $(\frac{1}{4},-1)$ and $(1,2)$. 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 $x=\frac{1}{4}y^2$ and $x=\frac{1}{4}y+\frac{1}{2}$
Step 3. Evaluate
$A=\int_{-1}^2(\frac{1}{4}y+\frac{1}{2}-\frac{1}{4}y^2)dy=(\frac{1}{8}y^2+\frac{1}{2}y-\frac{1}{12}y^3)|_{-1}^2=(\frac{1}{8}(2)^2+\frac{1}{2}(2)-\frac{1}{12}(2)^3)-(\frac{1}{8}(-1)^2+\frac{1}{2}(-1)-\frac{1}{12}(-1)^3)=\frac{9}{8}$
