Given curves

$y = x^2 - 1 \, , \quad y = 1 - x^2$

Find points of intersection

At intersection, $x^2 - 1 = 1 - x^2$ $2x^2 = 2$ $x^2 = 1$ $x = \pm 1$

So the curves intersect at: $(-1,0) \text{ and } (1,0)$

Determine the upper and lower curve

For $x \in [-1,1]$, $1 - x^2 \ge x^2 - 1$

Hence,

• Upper curve: $y = 1 - x^2$
• Lower curve: $y = x^2 - 1$

Set up the area integral

Area enclosed is $A = \int_{-1}^{1} \left[(1-x^2) - (x^2-1)\right] dx$

Simplify the integrand: $A = \int_{-1}^{1} (2 - 2x^2)\,dx$ $A = 2\int_{-1}^{1} (1-x^2)\,dx$

Evaluate the integral

$A = \int_{-1}^{1} (2 - 2x^2)\,dx$ $= \left[2x - \frac{2x^3}{3}\right]_{-1}^{1}$

At $x=1$: $2(1) - \frac{2(1)^3}{3} = 2 - \frac{2}{3} = \frac{4}{3}$

At $x=-1$: $2(-1) - \frac{2(-1)^3}{3} = -2 + \frac{2}{3} = -\frac{4}{3}$

Therefore, $A = \frac{4}{3} - \left(-\frac{4}{3}\right) = \frac{8}{3}$

Compare with options

$\frac{8}{3}$ matches Option A.

Verification with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.