Key Concepts and Formulas

• Area Between Curves: The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$, where $f(x) \ge g(x)$ on $[a, b]$, is given by $A = \int_a^b [f(x) - g(x)] dx$.
• Finding Intersection Points: To find where two curves intersect, set their equations equal to each other and solve for $x$.
• Even Functions and Symmetry: If $f(x)$ is an even function (i.e., $f(-x) = f(x)$), then $\int_{-a}^a f(x) dx = 2 \int_0^a f(x) dx$.

Step-by-Step Solution

Step 1: Identify the Curves

We are given the curves $y = x^2 - 1$ and $y = 1 - x^2$. Why: This step simply restates the problem's given information, making it easier to refer to the equations later.

Step 2: Find the Intersection Points

To find the intersection points, we set the two equations equal to each other: $x^2 - 1 = 1 - x^2$ Why: The intersection points determine the limits of integration. At these points, the $y$-values of the two curves are equal.

Step 3: Solve for x

Solving the equation from Step 2: $2x^2 = 2$ $x^2 = 1$ $x = \pm 1$ Thus, the intersection points occur at $x = -1$ and $x = 1$. Why: This step calculates the x-coordinates of the intersection points, which will serve as our limits of integration.

Step 4: Determine the Upper and Lower Curves

We need to determine which curve is above the other in the interval $[-1, 1]$. Let's pick a test point within the interval, say $x = 0$: For $y = x^2 - 1$, when $x = 0$, $y = 0^2 - 1 = -1$. For $y = 1 - x^2$, when $x = 0$, $y = 1 - 0^2 = 1$. Since $1 > -1$, the curve $y = 1 - x^2$ is above the curve $y = x^2 - 1$ in the interval $[-1, 1]$.

Why: This step determines which function will be $f(x)$ and which will be $g(x)$ in our area integral. It's crucial to subtract the lower curve from the upper curve to get a positive area.

Step 5: Set Up the Integral

The area $A$ is given by the integral: $A = \int_{-1}^1 [(1 - x^2) - (x^2 - 1)] dx$ Why: This step applies the area between curves formula with the correct limits of integration and the correct order of subtraction of the functions.

Step 6: Simplify the Integrand

Simplifying the integrand: $A = \int_{-1}^1 (1 - x^2 - x^2 + 1) dx$ $A = \int_{-1}^1 (2 - 2x^2) dx$ Why: Simplifying the integrand makes the integration easier.

Step 7: Use Symmetry

Since $2 - 2x^2$ is an even function, we can use symmetry to simplify the integral: $A = 2 \int_0^1 (2 - 2x^2) dx$ $A = 4 \int_0^1 (1 - x^2) dx$ Why: Using symmetry reduces the range of integration and often simplifies the calculation by making the lower limit 0.

Step 8: Integrate

Integrating the expression: $A = 4 \left[x - \frac{x^3}{3}\right]_0^1$ Why: This step finds the antiderivative of the integrand.

Step 9: Evaluate the Integral

Evaluating the integral at the limits: $A = 4 \left[\left(1 - \frac{1^3}{3}\right) - \left(0 - \frac{0^3}{3}\right)\right]$ $A = 4 \left(1 - \frac{1}{3}\right)$ $A = 4 \left(\frac{2}{3}\right)$ Why: This step applies the Fundamental Theorem of Calculus to find the definite integral's value.

Step 10: Calculate the Area

$A = \frac{8}{3}$ Why: This step performs the final calculation to arrive at the area.

Common Mistakes & Tips

• Always sketch the curves to visualize the region and ensure you're subtracting the correct functions.
• Double-check the intersection points to avoid errors in the limits of integration.
• Remember to use symmetry when applicable to simplify the integral.

Summary

We found the area between the curves $y = x^2 - 1$ and $y = 1 - x^2$ by finding their intersection points, setting up the definite integral of the difference between the upper and lower curves, using symmetry to simplify the calculation, and evaluating the integral. The area is $\frac{8}{3}$ square units.

Final Answer

The final answer is $\boxed{\frac{8}{3}}$, which corresponds to option (A).