1. **State the problem:** Find the area enclosed by the graphs of \(y = x^2 - 2x + 1\), \(y = x^2 - 4x + 4\), and the x-axis. 2. **Rewrite the functions:** \(y = x^2 - 2x + 1 = (x-1)^2\) \(y = x^2 - 4x + 4 = (x-2)^2\) 3. **Find the points where each curve intersects the x-axis:** For \(y = (x-1)^2 = 0\), \(x=1\). For \(y = (x-2)^2 = 0\), \(x=2\). 4. **Find the points where the two curves intersect:** Set \((x-1)^2 = (x-2)^2\): \(x-1 = \pm (x-2)\) Case 1: \(x-1 = x-2\) gives no solution. Case 2: \(x-1 = -(x-2)\) gives \(x-1 = -x + 2\) \(\Rightarrow 2x = 3\) \(\Rightarrow x = \frac{3}{2}\). 5. **Evaluate the functions at \(x=\frac{3}{2}\):** \(y = (\frac{3}{2} - 1)^2 = (\frac{1}{2})^2 = \frac{1}{4}\). 6. **Determine which curve is on top between \(x=1\) and \(x=2\):** At \(x=1.25\), \((1.25-1)^2 = 0.25^2 = 0.0625\), \((1.25-2)^2 = (-0.75)^2 = 0.5625\). So \(y = (x-1)^2\) is below \(y = (x-2)^2\) in this interval. 7. **Find the area enclosed between the two curves from \(x=1\) to \(x=2\):** \[\text{Area} = \int_1^2 \left[(x-2)^2 - (x-1)^2\right] dx\] 8. **Simplify the integrand:** \[(x-2)^2 - (x-1)^2 = (x^2 - 4x + 4) - (x^2 - 2x + 1) = -4x + 4 + 2x - 1 = -2x + 3\] 9. **Integrate:** \[\int_1^2 (-2x + 3) dx = \left[-x^2 + 3x\right]_1^2 = \left(-4 + 6\right) - \left(-1 + 3\right) = 2 - 2 = 0\] 10. **Interpretation:** The integral is zero, meaning the area between the curves from 1 to 2 is zero because the curves touch at \(x=1\) and \(x=2\) and cross at \(x=\frac{3}{2}\). 11. **Find the area enclosed by each curve and the x-axis separately:** - For \(y = (x-1)^2\), area between \(x=0\) and \(x=2\) is: \[\int_0^2 (x-1)^2 dx = \int_0^2 (x^2 - 2x + 1) dx = \left[\frac{x^3}{3} - x^2 + x\right]_0^2 = \left(\frac{8}{3} - 4 + 2\right) - 0 = \frac{8}{3} - 2 = \frac{2}{3}\] - For \(y = (x-2)^2\), area between \(x=0\) and \(x=4\) is: \[\int_0^4 (x-2)^2 dx = \int_0^4 (x^2 - 4x + 4) dx = \left[\frac{x^3}{3} - 2x^2 + 4x\right]_0^4 = \left(\frac{64}{3} - 32 + 16\right) - 0 = \frac{64}{3} - 16 = \frac{16}{3}\] 12. **Final answer:** The area enclosed by the graphs and the x-axis is the sum of these areas: \[\frac{2}{3} + \frac{16}{3} = \frac{18}{3} = 6\] Thus, the total enclosed area is \(6\).