𝑦=𝑥2−3and𝑦=1
 This figure is has two graphs. They are the functions f(x) = x^2-3and g(x)=1. In between these graphs is a shaded region, bounded above by g(x) and below by f(x). The shaded area is between x=-2 and x=2.

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!

The query submitted by the user: 𝑦 = 𝑥² − 3 and 𝑦 = 1. This figure has two graphs, f(x) = x² - 3 and g(x) = 1, with a shaded region between them, bounded by g(x) above and f(x) below, over the interval x = -2 to x = 2.

Learn how to find the area between the curves f(x) = x² - 3 and g(x) = 1 over the interval x = -2 to x = 2. This problem covers integral calculus concepts like definite integrals and the area between curves, making it ideal for students in Grades 11-12 or early college-level calculus.

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