Key Concepts and Formulas

• Area between curves: If $f(x) \ge g(x)$ on $[a,b]$, then the area between the curves $y=f(x)$ and $y=g(x)$ from $x=a$ to $x=b$ is given by $\int_a^b [f(x) - g(x)] dx$.
• Absolute value function: $|x| = x$ if $x \ge 0$ and $|x| = -x$ if $x < 0$.
• Symmetry: If the region is symmetric about an axis, the area can be calculated for one half and then doubled.

Step-by-Step Solution

Step 1: Understand the curves and their intersection points.

We are given the curves $y = |x^2 - 1|$ and $y = 1$. To find the area bounded by these curves, we first need to find their intersection points. We also need to understand the behavior of the absolute value function. The function $y = |x^2 - 1|$ can be rewritten as: $y = \begin{cases} x^2 - 1 & \text{if } x^2 \ge 1 \text{ (i.e., } x \le -1 \text{ or } x \ge 1) \\ 1 - x^2 & \text{if } x^2 < 1 \text{ (i.e., } -1 < x < 1) \end{cases}$

Step 2: Find the intersection points.

We need to solve $|x^2 - 1| = 1$. This gives us two cases:

Case 1: $x^2 - 1 = 1 \Rightarrow x^2 = 2 \Rightarrow x = \pm \sqrt{2}$. Case 2: $1 - x^2 = 1 \Rightarrow x^2 = 0 \Rightarrow x = 0$.

So, the intersection points are $x = -\sqrt{2}, 0, \sqrt{2}$.

Step 3: Set up the integral.

The region is symmetric about the y-axis, so we can calculate the area in the first quadrant (from $x=0$ to $x=\sqrt{2}$) and then double it. In the interval $[0, 1]$, we have $|x^2 - 1| = 1 - x^2$, and in the interval $[1, \sqrt{2}]$, we have $|x^2 - 1| = x^2 - 1$. The area in the first quadrant can be divided into two parts:

• From $x = 0$ to $x = 1$, the upper curve is $y = 1$ and the lower curve is $y = 1 - x^2$.
• From $x = 1$ to $x = \sqrt{2}$, the upper curve is $y = 1$ and the lower curve is $y = x^2 - 1$.

The area in the first quadrant is: $A_{1} = \int_{0}^{1} [1 - (1 - x^2)] dx + \int_{1}^{\sqrt{2}} [1 - (x^2 - 1)] dx$ $A_{1} = \int_{0}^{1} x^2 dx + \int_{1}^{\sqrt{2}} (2 - x^2) dx$

Step 4: Evaluate the integrals.

$A_{1} = \left[ \frac{x^3}{3} \right]_{0}^{1} + \left[ 2x - \frac{x^3}{3} \right]_{1}^{\sqrt{2}}$ $A_{1} = \left( \frac{1}{3} - 0 \right) + \left( 2\sqrt{2} - \frac{2\sqrt{2}}{3} \right) - \left( 2 - \frac{1}{3} \right)$ $A_{1} = \frac{1}{3} + 2\sqrt{2} - \frac{2\sqrt{2}}{3} - 2 + \frac{1}{3}$ $A_{1} = \frac{2}{3} + \frac{4\sqrt{2}}{3} - 2 = \frac{2 + 4\sqrt{2} - 6}{3} = \frac{4\sqrt{2} - 4}{3} = \frac{4(\sqrt{2} - 1)}{3}$

Since the region is symmetric about the y-axis, the total area is twice the area in the first quadrant: $A = 2A_{1} = 2 \cdot \frac{4(\sqrt{2} - 1)}{3} = \frac{8(\sqrt{2} - 1)}{3}$

Step 5: Check for errors. The correct answer is $\frac{8}{3}(\sqrt{2}-1)$. The given correct answer is $\frac{2}{3}(\sqrt{2}+1)$.

Let's recalculate the area.

Area = $2 \left( \int_0^1 (1 - (1-x^2)) dx + \int_1^{\sqrt{2}} (1 - (x^2 - 1)) dx \right)$ Area = $2 \left( \int_0^1 x^2 dx + \int_1^{\sqrt{2}} (2 - x^2) dx \right)$ Area = $2 \left( \left[ \frac{x^3}{3} \right]_0^1 + \left[ 2x - \frac{x^3}{3} \right]_1^{\sqrt{2}} \right)$ Area = $2 \left( \frac{1}{3} + (2\sqrt{2} - \frac{2\sqrt{2}}{3}) - (2 - \frac{1}{3}) \right)$ Area = $2 \left( \frac{1}{3} + \frac{4\sqrt{2}}{3} - 2 + \frac{1}{3} \right)$ Area = $2 \left( \frac{2}{3} + \frac{4\sqrt{2}}{3} - \frac{6}{3} \right) = 2\left( \frac{4\sqrt{2} - 4}{3} \right) = \frac{8(\sqrt{2} - 1)}{3}$

It seems there's an error in the problem statement or the given correct answer. The area we found is $\frac{8}{3}(\sqrt{2}-1)$.

Common Mistakes & Tips

• Remember to consider the absolute value function and split the integral accordingly.
• Always find the intersection points of the curves to determine the limits of integration.
• Don't forget to multiply by 2 if you're using symmetry.

Summary

We found the area bounded by the curves $y = |x^2 - 1|$ and $y = 1$ by finding the intersection points, setting up the definite integral, and evaluating it. We used symmetry to simplify the calculations. The area is $\frac{8(\sqrt{2}-1)}{3}$.

Final Answer

The final answer is $\boxed{\frac{8}{3}(\sqrt{2}-1)}$, which corresponds to option (D).