The area of region between the curve `y=abs(x^2-1) and y=1` is :

Options

A. `8/3(sqrt2-1)`
B. `8/3(sqrt2+1)`
C. `4/3(sqrt2-1)`
D. `4/3(sqrt2+1)`

Answer

To find the area of the region between the curve \( y = |x^2 - 1| \) and the line \( y = 1 \), we can follow these steps:

### Step 1: Identify the points of intersection
We need to find the points where the curve \( y = |x^2 - 1| \) intersects the line \( y = 1 \).

1. Set \( |x^2 - 1| = 1 \).
2. This gives us two cases to consider:
- Case 1: \( x^2 - 1 = 1 \) ⇒ \( x^2 = 2 \) ⇒ \( x = \pm \sqrt{2} \)
- Case 2: \( x^2 - 1 = -1 \) ⇒ \( x^2 = 0 \) ⇒ \( x = 0 \)

Thus, the points of intersection are \( x = -\sqrt{2}, 0, \sqrt{2} \).

### Step 2: Determine the area between the curves
The area between the curves can be found by integrating the difference of the two functions from \( -\sqrt{2} \) to \( \sqrt{2} \).

1. For \( x \) in the interval \( [-\sqrt{2}, 0] \):
- Here, \( |x^2 - 1| = 1 - x^2 \) (since \( x^2 < 1 \)).
- The area \( A_1 \) is given by:
\[ A_1 = \int_{-\sqrt{2}}^{0} \left(1 - (1 - x^2)\right) \, dx = \int_{-\sqrt{2}}^{0} x^2 \, dx \]

2. For \( x \) in the interval \( [0, \sqrt{2}] \):
- Here, \( |x^2 - 1| = x^2 - 1 \) (since \( x^2 > 1 \)).
- The area \( A_2 \) is given by:
\[ A_2 = \int_{0}^{\sqrt{2}} \left((x^2 - 1) - 1\right) \, dx = \int_{0}^{\sqrt{2}} (x^2 - 2) \, dx \]

### Step 3: Calculate the integrals
1. Calculate \( A_1 \):
\[ A_1 = \int_{-\sqrt{2}}^{0} x^2 \, dx = \left[\frac{x^3}{3}\right]_{-\sqrt{2}}^{0} = 0 - \left(-\frac{(\sqrt{2})^3}{3}\right) = \frac{2\sqrt{2}}{3} \]

2. Calculate \( A_2 \):
\[ A_2 = \int_{0}^{\sqrt{2}} (x^2 - 2) \, dx = \left[\frac{x^3}{3} - 2x\right]_{0}^{\sqrt{2}} = \left(\frac{(\sqrt{2})^3}{3} - 2\sqrt{2}\right) - (0) = \frac{2\sqrt{2}}{3} - 2\sqrt{2} = \frac{2\sqrt{2}}{3} - \frac{6\sqrt{2}}{3} = -\frac{4\sqrt{2}}{3} \]

### Step 4: Total area
Since the area cannot be negative, we take the absolute value of \( A_2 \):
\[ \text{Total Area} = A_1 + |A_2| = \frac{2\sqrt{2}}{3} + \frac{4\sqrt{2}}{3} = \frac{6\sqrt{2}}{3} = 2\sqrt{2} \]

### Final Answer
The area of the region between the curve \( y = |x^2 - 1| \) and the line \( y = 1 \) is \( 2\sqrt{2} \).