It's trickier than it looks. Think differently

First, we find the points of intersection of $y^{2} = 4+x$ and $x+2y = 4$ . Equating $x$ on both sides, we get $y^{2}-4 = 4-2y$ . Solving for $y$ gives us $(-4,2)$ .

We can now integrate $x$ with respect to $y$ . Observe that the area between the two equations is the absolute value of their difference.

$A = \int_{-4}^{2} y^{2}-4 dy - \int_{-4}^{2} 4-2y dy \\ = \int_{-4}^{2} y^{2}+2y-8 dy \\ = [\frac{1}{3}y^{3}+y^{2}-8y]_{-4}^{2} \\ = 36$

Hence, the area is $\boxed{36}$ .

$\int_{-4}^{2}(4-2(-2+y)-y^{2})$ = $36$

so area bounded by the lines is $36$

first integrate 1 from y^2-4 to 4-2y with respect to x then integrate the answere with reespect to y from -4 to 2.

double integration dxdy x changes from (y^2-4) to (4-2y) and y changes from (-4) to (2).