1. Understand the region

We need the area of

$$\{(x,y): x^2 \le y \le |x^2-4|,\; y\ge 1\}.$$

This means:

• lower bound is effectively the larger of $x^2$ and $1$,
• upper bound is $|x^2-4|$.

So for a given $x$, the vertical strip exists only when

$$|x^2-4| \ge x^2 \quad \text{and} \quad |x^2-4|\ge 1.$$

But since $y\ge x^2$, if $x^2\ge 1$ then automatically $y\ge 1$.

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2. Break the modulus

Since

$$|x^2-4|= \begin{cases} 4-x^2, & x^2<4,\\ x^2-4, & x^2\ge 4, \end{cases}$$

we examine both cases.

Case 1: $x^2<4$

Then upper curve is

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

The inequalities become

$$x^2 \le y \le 4-x^2, \quad y\ge 1.$$

For such a strip to exist, we need

$$x^2 \le 4-x^2 \implies 2x^2\le 4 \implies x^2\le 2.$$

So

$$-\sqrt2 \le x \le \sqrt2.$$

Now incorporate $y\ge 1$:

• if $|x|\le 1$, then $x^2\le 1$, so lower limit becomes $1$,
• if $1\le |x|\le \sqrt2$, then $x^2\ge 1$, so lower limit remains $x^2$.

Thus the area is

$$A=\int_{-\sqrt2}^{-1}[(4-x^2)-x^2]dx+\int_{-1}^{1}[(4-x^2)-1]dx+\int_{1}^{\sqrt2}[(4-x^2)-x^2]dx.$$

Using symmetry,

$$A=2\int_1^{\sqrt2}(4-2x^2)dx+\int_{-1}^1(3-x^2)dx.$$

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3. Compute the integrals

First integral

$$2\int_1^{\sqrt2}(4-2x^2)\,dx =2\left[4x-\frac{2x^3}{3}\right]_1^{\sqrt2}.$$

Now,

$$\left(4\sqrt2-\frac{2(\sqrt2)^3}{3}\right)-\left(4-\frac23\right) =\left(4\sqrt2-\frac{4\sqrt2}{3}\right)-\frac{10}{3} =\frac{8\sqrt2}{3}-\frac{10}{3}.$$

Multiplying by $2$,

$$2\int_1^{\sqrt2}(4-2x^2)dx =\frac{16\sqrt2}{3}-\frac{20}{3}.$$

Second integral

$$\int_{-1}^1(3-x^2)dx =\left[3x-\frac{x^3}{3}\right]_{-1}^{1} =\left(3-\frac13\right)-\left(-3+\frac13\right) =\frac{16}{3}.$$

So total area is

$$A=\left(\frac{16\sqrt2}{3}-\frac{20}{3}\right)+\frac{16}{3} =\frac{16\sqrt2-4}{3} =\frac{4}{3}(4\sqrt2-1).$$

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4. Match with options

$$\frac{4}{3}(4\sqrt2-1)$$

corresponds to Option C.

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5. Comparison with stored answer

Stored correct answer: C.

Our derived answer is also C, so they agree.