1. Write the curves in convenient form

The given curves are:

$y^2+4x=4$ $y-2x=2$

Rewrite them as $x$ in terms of $y$:

• Parabola: $4x=4-y^2 \implies x=1-\frac{y^2}{4}$

• Line: $y-2x=2 \implies 2x=y-2 \implies x=\frac{y-2}{2}$

2. Find points of intersection

At intersection,

$1-\frac{y^2}{4}=\frac{y-2}{2}$

Multiply by $4$:

$4-y^2=2y-4$

$8-2y-y^2=0$

$y^2+2y-8=0$

Factor:

$(y+4)(y-2)=0$

So,

$y=-4,\quad y=2$

Now find corresponding $x$ values:

• For $y=2$: $x=\frac{2-2}{2}=0$

• For $y=-4$: $x=\frac{-4-2}{2}=-3$

Thus intersection points are:

$(0,2),\; (-3,-4)$

3. Decide which curve is on the right

For area using horizontal strips, compare $x$-values:

$x_{\text{parabola}}=1-\frac{y^2}{4}, \qquad x_{\text{line}}=\frac{y-2}{2}$

Take a test value, say $y=0$:

$x_{\text{parabola}}=1, \qquad x_{\text{line}}=-1$

So the parabola lies to the right of the line.

Hence area is

$A=\int_{-4}^{2}\left(x_{\text{right}}-x_{\text{left}}\right)dy$

$A=\int_{-4}^{2}\left(1-\frac{y^2}{4}-\frac{y-2}{2}\right)dy$

Simplify the integrand:

$1-\frac{y^2}{4}-\frac{y}{2}+1=2-\frac{y}{2}-\frac{y^2}{4}$

So,

$A=\int_{-4}^{2}\left(2-\frac{y}{2}-\frac{y^2}{4}\right)dy$

4. Integrate