1. **State the problem:** Find the area bounded by the curve $y = 1 - x^2$ and the line $y = 0$. 2. **Identify the region:** The parabola $y = 1 - x^2$ opens downward and intersects the x-axis where $y=0$. 3. **Find intersection points:** Solve $1 - x^2 = 0$ to find $x$ values. $$1 - x^2 = 0 \implies x^2 = 1 \implies x = \pm 1$$ 4. **Set up the integral:** The area bounded is between $x = -1$ and $x = 1$, above the x-axis. $$\text{Area} = \int_{-1}^{1} (1 - x^2) \, dx$$ 5. **Calculate the integral:** $$\int (1 - x^2) \, dx = \int 1 \, dx - \int x^2 \, dx = x - \frac{x^3}{3} + C$$ 6. **Evaluate definite integral:** $$\text{Area} = \left[x - \frac{x^3}{3}\right]_{-1}^{1} = \left(1 - \frac{1^3}{3}\right) - \left(-1 - \frac{(-1)^3}{3}\right)$$ $$= \left(1 - \frac{1}{3}\right) - \left(-1 + \frac{1}{3}\right) = \frac{2}{3} - \left(-\frac{2}{3}\right) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3}$$ 7. **Final answer:** The area bounded by the parabola and the x-axis is $\boxed{\frac{4}{3}}$.