Determine the area between the curves $$ y = x^2 $$ and $$ y = 4 - x^2 $$ for the interval where they intersect.

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To determine the area between the curves $$ y = x^2 $$ and $$ y = 4 - x^2 $$ where they intersect, follow these steps:

1. **Find the Points of Intersection:**
   
   Set $$ x^2 = 4 - x^2 $$:
   $$
   2x^2 = 4 \quad \Rightarrow \quad x^2 = 2 \quad \Rightarrow \quad x = \pm\sqrt{2}
   $$
   
   The curves intersect at $$ x = -\sqrt{2} $$ and $$ x = \sqrt{2} $$.

2. **Determine the Upper and Lower Functions:**
   
   For $$ -\sqrt{2} \leq x \leq \sqrt{2} $$:
   - Upper function: $$ y = 4 - x^2 $$
   - Lower function: $$ y = x^2 $$

3. **Set Up the Integral for the Area:**
   
   The area $$ A $$ between the curves is:
   $$
   A = \int_{-\sqrt{2}}^{\sqrt{2}} \left( (4 - x^2) - x^2 \right) dx = \int_{-\sqrt{2}}^{\sqrt{2}} (4 - 2x^2) \, dx
   $$
   
   Since the integrand is even, we can simplify the computation:
   $$
   A = 2 \int_{0}^{\sqrt{2}} (4 - 2x^2) \, dx
   $$

4. **Compute the Integral:**
   
   $$
   \int (4 - 2x^2) \, dx = 4x - \frac{2x^3}{3}
   $$
   
   Evaluate from $$ 0 $$ to $$ \sqrt{2} $$:
   $$
   \left[4\sqrt{2} - \frac{2(\sqrt{2})^3}{3}\right] - \left[0 - 0\right] = 4\sqrt{2} - \frac{4\sqrt{2}}{3} = \frac{12\sqrt{2} - 4\sqrt{2}}{3} = \frac{8\sqrt{2}}{3}
   $$
   
   Multiply by 2 (due to symmetry):
   $$
   A = 2 \times \frac{8\sqrt{2}}{3} = \frac{16\sqrt{2}}{3}
   $$

**Final Answer:**
$$
\boxed{\dfrac{16\,\sqrt{2}}{3}}
$$
The area between the curves $$ y = x^2 $$ and $$ y = 4 - x^2 $$ from $$ x = -\sqrt{2} $$ to $$ x = \sqrt{2} $$ is $$ \frac{16\sqrt{2}}{3} $$.

Determine the area between the curves and for the interval where they intersect.

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