1. **State the problem:** We need to find the area of the region bounded by the curve $y = x^2 - 1$ and the x-axis. 2. **Identify the points of intersection:** The region is bounded where the curve intersects the x-axis, i.e., where $y=0$. Solve $x^2 - 1 = 0$: $$x^2 = 1$$ $$x = \pm 1$$ 3. **Set up the integral:** The area between the curve and the x-axis from $x=-1$ to $x=1$ is given by $$\text{Area} = \int_{-1}^{1} |x^2 - 1| \, dx$$ Since $x^2 - 1 \leq 0$ between $-1$ and $1$, the curve is below the x-axis, so the absolute value is $$|x^2 - 1| = 1 - x^2$$ 4. **Calculate the integral:** $$\text{Area} = \int_{-1}^{1} (1 - x^2) \, dx = \left[ x - \frac{x^3}{3} \right]_{-1}^{1}$$ Evaluate at the bounds: $$\left(1 - \frac{1}{3}\right) - \left(-1 + \frac{-1}{3}\right) = \left(\frac{2}{3}\right) - \left(-\frac{4}{3}\right) = \frac{2}{3} + \frac{4}{3} = 2$$ 5. **Final answer:** The area of the region bounded by the curve and the x-axis is $2$ square units.