The problem describes finding the area of a region between two graphs: $f(x) = x^2 - 3$ and $g(x) = 1$, over the interval $x = -2$ to $x = 2$.

Steps to Solve:

1. Area Between Two Curves Formula: The area between two curves $f(x)$ and $g(x)$ over an interval $[a, b]$ is given by: $A = \int_a^b \left[ g(x) - f(x) \right] \, dx$ In this case, $g(x) = 1$ and $f(x) = x^2 - 3$, with the limits $a = -2$ and $b = 2$.

2. Set up the Integral: $A = \int_{-2}^{2} \left[ 1 - (x^2 - 3) \right] \, dx$ Simplify the expression inside the integral: [ A = \int_{-2}^{2} \left[ 1 - x^2 + 3 \right] , dx = \int_{-2}^{2} \left[ 4 - x^2 \right] , dx ]

3. Evaluate the Integral: We can now break the integral into two simpler parts: $A = \int_{-2}^{2} 4 \, dx - \int_{-2}^{2} x^2 \, dx$

4. Solve Each Integral:

   • The first integral: $\int_{-2}^{2} 4 \, dx = 4 \times (2 - (-2)) = 4 \times 4 = 16$

   • The second integral: $\int_{-2}^{2} x^2 \, dx$ Since $x^2$ is an even function, we can use the symmetry of the interval to simplify this: $\int_{-2}^{2} x^2 \, dx = 2 \times \int_{0}^{2} x^2 \, dx$ Now, evaluate the integral: $\int x^2 \, dx = \frac{x^3}{3}$ So: $\left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{2^3}{3} - 0 = \frac{8}{3}$ Thus: $\int_{-2}^{2} x^2 \, dx = 2 \times \frac{8}{3} = \frac{16}{3}$

5. Final Area: Now, subtract the two results: $A = 16 - \frac{16}{3} = \frac{48}{3} - \frac{16}{3} = \frac{32}{3}$

Thus, the area of the shaded region is $\frac{32}{3}$ square units.

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Would you like further details or have any questions?

Here are some related questions:

1. How do you find the area between curves over different intervals?
2. How does symmetry simplify the integration process?
3. What if the functions were non-polynomial—how would the method change?
4. How do you handle curves that intersect at multiple points?
5. How do odd and even functions affect integral calculations?

Tip: Always check for symmetry in problems involving definite integrals—it can greatly simplify the calculations!