1. **State the problem:** Find the area of the region enclosed by the curve $y=(2x-1)(2x+1)$ and the x-axis. 2. **Rewrite the function:** Expand the product: $$y = (2x-1)(2x+1) = 4x^2 - 1$$ 3. **Find the points where the curve intersects the x-axis:** Set $y=0$: $$4x^2 - 1 = 0$$ $$4x^2 = 1$$ $$x^2 = \frac{1}{4}$$ $$x = \pm \frac{1}{2}$$ 4. **Set up the integral for the area:** The curve is a parabola opening upwards shifted down by 1. The region enclosed between the curve and the x-axis is between $x=-\frac{1}{2}$ and $x=\frac{1}{2}$. Since the curve dips below the x-axis between these points, the area is: $$\text{Area} = \int_{-\frac{1}{2}}^{\frac{1}{2}} |4x^2 - 1| \, dx$$ Because $4x^2 - 1 \leq 0$ in this interval, 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:** $$\left[x - \frac{4x^3}{3}\right]_{-\frac{1}{2}}^{\frac{1}{2}} = \left(\frac{1}{2} - \frac{4(\frac{1}{2})^3}{3}\right) - \left(-\frac{1}{2} - \frac{4(-\frac{1}{2})^3}{3}\right)$$ Calculate each term: $$\frac{4(\frac{1}{2})^3}{3} = \frac{4(\frac{1}{8})}{3} = \frac{\frac{1}{2}}{3} = \frac{1}{6}$$ So the first bracket: $$\frac{1}{2} - \frac{1}{6} = \frac{3}{6} - \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$$ For the second bracket: $$-\frac{1}{2} - \frac{4(-\frac{1}{2})^3}{3} = -\frac{1}{2} - \frac{4(-\frac{1}{8})}{3} = -\frac{1}{2} + \frac{1}{6} = -\frac{3}{6} + \frac{1}{6} = -\frac{2}{6} = -\frac{1}{3}$$ 8. **Subtract:** $$\frac{1}{3} - (-\frac{1}{3}) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$$ **Final answer:** The area enclosed by the curve and the x-axis is $$\boxed{\frac{2}{3}}$$