1. **State the problem:** Find the area of the region enclosed by the curves given by the function $y = (2x - 1)(2x + 1)$. 2. **Rewrite the function:** Expand the product to get a standard quadratic form: $$y = (2x - 1)(2x + 1) = 4x^2 - 1$$ 3. **Identify the region enclosed:** The curve is a parabola opening upwards with vertex at $(0, -1)$. To find the enclosed area, we need to find the points where the curve intersects the x-axis (i.e., where $y=0$): $$4x^2 - 1 = 0$$ $$4x^2 = 1$$ $$x^2 = \frac{1}{4}$$ $$x = \pm \frac{1}{2}$$ 4. **Set up the integral:** The area enclosed between the curve and the x-axis from $x = -\frac{1}{2}$ to $x = \frac{1}{2}$ is given by: $$\text{Area} = \int_{-\frac{1}{2}}^{\frac{1}{2}} |4x^2 - 1| \, dx$$ Since $4x^2 - 1 \leq 0$ in this interval (because $4x^2 \leq 1$), the function is negative or zero, so the absolute value is: $$|4x^2 - 1| = 1 - 4x^2$$ 5. **Calculate the integral:** $$\text{Area} = \int_{-\frac{1}{2}}^{\frac{1}{2}} (1 - 4x^2) \, dx$$ 6. **Integrate term-by-term:** $$\int (1) \, dx = x$$ $$\int (-4x^2) \, dx = -\frac{4x^3}{3}$$ 7. **Evaluate the definite integral:** $$\text{Area} = \left[x - \frac{4x^3}{3}\right]_{-\frac{1}{2}}^{\frac{1}{2}} = \left(\frac{1}{2} - \frac{4(\frac{1}{2})^3}{3}\right) - \left(-\frac{1}{2} - \frac{4(-\frac{1}{2})^3}{3}\right)$$ Calculate each term: $$\frac{4(\frac{1}{2})^3}{3} = \frac{4(\frac{1}{8})}{3} = \frac{\frac{1}{2}}{3} = \frac{1}{6}$$ $$\frac{4(-\frac{1}{2})^3}{3} = \frac{4(-\frac{1}{8})}{3} = -\frac{1}{6}$$ So: $$\text{Area} = \left(\frac{1}{2} - \frac{1}{6}\right) - \left(-\frac{1}{2} + \frac{1}{6}\right) = \left(\frac{3}{6} - \frac{1}{6}\right) - \left(-\frac{3}{6} + \frac{1}{6}\right) = \frac{2}{6} - (-\frac{2}{6}) = \frac{2}{6} + \frac{2}{6} = \frac{4}{6} = \frac{2}{3}$$ 8. **Final answer:** The area of the region enclosed by the curve and the x-axis is: $$\boxed{\frac{2}{3}}$$