1. **Problem Statement:** Find the area between the graph of the function $f(x) = 1 - x^2$ and the x-axis from $x = -1$ to $x = 1$. 2. **Formula:** The area between the curve and the x-axis from $a$ to $b$ is given by the definite integral $$\text{Area} = \int_a^b |f(x)| \, dx.$$ Since $f(x)$ is positive or zero on $[-1,1]$ (because $1 - x^2 \geq 0$ for $|x| \leq 1$), we can integrate without absolute value: $$\text{Area} = \int_{-1}^1 (1 - x^2) \, dx.$$ 3. **Calculate the integral:** $$\int (1 - x^2) \, dx = \int 1 \, dx - \int x^2 \, dx = x - \frac{x^3}{3} + C.$$ 4. **Evaluate definite integral:** $$\int_{-1}^1 (1 - x^2) \, dx = \left[x - \frac{x^3}{3}\right]_{-1}^1 = \left(1 - \frac{1^3}{3}\right) - \left(-1 - \frac{(-1)^3}{3}\right) = \left(1 - \frac{1}{3}\right) - \left(-1 + \frac{1}{3}\right).$$ 5. **Simplify:** $$= \left(\frac{2}{3}\right) - \left(-\frac{2}{3}\right) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3}.$$ 6. **Interpretation:** The area between the curve $f(x) = 1 - x^2$ and the x-axis from $x = -1$ to $x = 1$ is $\frac{4}{3}$ square units.