The area of the region $$\{x, y : x^2 \leq y \leq x^2 - 4, y \geq 1\}$$ is

The region is bounded below by the upward‐opening parabola $$y = x^2$$ and above by the downward‐opening parabola $$y = 4 - x^2$$, with the additional condition $$y \ge 1$$. Hence every vertical slice of the region satisfies
$$x^2 \le y \le 4 - x^2,\; y \ge 1.$$

The first task is to decide, for each value of $$x$$, which curve (or the line $$y = 1$$) acts as the lower boundary.

1. Intersection of the two parabolas:
Set $$x^2 = 4 - x^2 \;\; \Rightarrow \;\; 2x^2 = 4 \;\; \Rightarrow \;\; x^2 = 2 \;\; \Rightarrow \;\; x = \pm\sqrt{2},$$
with the common $$y$$-value $$y = 2$$. Thus the curves meet at $$(-\sqrt{2},\,2)$$ and $$(\sqrt{2},\,2)$$.

2. Where does $$y = 1$$ intersect the upper parabola $$y = 4 - x^2$$?
Put $$4 - x^2 = 1 \;\; \Rightarrow \;\; x^2 = 3 \;\; \Rightarrow \;\; x = \pm\sqrt{3}.$$
Hence $$y = 1$$ cuts the region only for $$|x| \le \sqrt{3}$$.

3. Compare $$x^2$$ with the line $$y = 1$$:
If $$|x| \lt 1$$ then $$x^2 \lt 1$$, so the lower boundary is $$y = 1$$ (because $$1 \gt x^2$$).
If $$1 \le |x| \le \sqrt{2}$$ then $$x^2 \ge 1$$, so the lower boundary is $$y = x^2$$ (because now $$x^2 \gt 1$$).
If $$|x| \gt \sqrt{2}$$, the curves $$y = x^2$$ and $$y = 4 - x^2$$ have already crossed (at $$|x| = \sqrt{2}$$), so no region exists beyond that point.

Therefore the region is confined to $$|x| \le \sqrt{2}$$, and we must treat two sub-intervals:

Lower curve is $$y = 1$$, upper curve is $$y = 4 - x^2$$, so the vertical thickness is
$$\bigl(4 - x^2\bigr) - 1 = 3 - x^2.$$

Lower curve is $$y = x^2$$, upper curve is $$y = 4 - x^2$$, giving thickness
$$\bigl(4 - x^2\bigr) - x^2 = 4 - 2x^2.$$

The region is symmetric about the $$y$$-axis, so we calculate the area in the first quadrant and double it:

Area $$\begin{aligned} A &= 2\Biggl[\int_{0}^{1} \bigl(3 - x^2\bigr)\,dx \;+\; \int_{1}^{\sqrt{2}} \bigl(4 - 2x^2\bigr)\,dx\Biggr]. \end{aligned}$$

Evaluate the integrals one by one.

First integral: $$\int_{0}^{1} (3 - x^2)\,dx \;=\; \Bigl(3x - \frac{x^{3}}{3}\Bigr)\Bigl|_{0}^{1} = 3 - \frac{1}{3} = \frac{8}{3}.$$

Second integral: $$\int_{1}^{\sqrt{2}} (4 - 2x^2)\,dx = \Bigl(4x - \frac{2x^{3}}{3}\Bigr)\Bigl|_{1}^{\sqrt{2}} = \Bigl(4\sqrt{2} - \frac{2(\sqrt{2})^{3}}{3}\Bigr) - \Bigl(4 - \frac{2}{3}\Bigr)$$ $$= \Bigl(4\sqrt{2} - \frac{4\sqrt{2}}{3}\Bigr) - \frac{10}{3} = \frac{12\sqrt{2} - 4\sqrt{2} - 10}{3} = \frac{8\sqrt{2} - 10}{3}.$$

Add the two results inside the brackets: $$\frac{8}{3} + \frac{8\sqrt{2} - 10}{3} = \frac{8\sqrt{2} - 2}{3} = \frac{2\bigl(4\sqrt{2} - 1\bigr)}{3}.$$

Finally, multiply by the factor $$2$$ for symmetry:

$$A = 2 \times \frac{2\bigl(4\sqrt{2} - 1\bigr)}{3} = \frac{4}{3}\bigl(4\sqrt{2} - 1\bigr).$$

Hence the area of the given region is $$\displaystyle \frac{4}{3}\bigl(4\sqrt{2} - 1\bigr).$$

That matches Option A.