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 intersections with the x-axis:** - For $y=(x-1)^2=0$, $x=1$ - For $y=(x-2)^2=0$, $x=2$ 4. **Find intersections between the two curves:** Set $(x-1)^2 = (x-2)^2$ $$ (x-1)^2 - (x-2)^2 = 0 $$ Use difference of squares: $$ [(x-1)-(x-2)] \cdot [(x-1)+(x-2)] = 0 $$ $$ (1) \cdot (2x - 3) = 0 $$ So, $2x - 3 = 0 \Rightarrow x = \frac{3}{2} = 1.5$ 5. **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. 6. **Set up the integral for the area enclosed between the curves and the x-axis:** - From $x=1$ to $x=1.5$, area is under $y=(x-1)^2$ above x-axis. - From $x=1.5$ to $x=2$, area is under $y=(x-2)^2$ above x-axis. 7. **Calculate the area:** $$ A = \int_1^{1.5} (x-1)^2 \, dx + \int_{1.5}^2 (x-2)^2 \, dx $$ 8. **Compute each integral:** $$ \int (x-a)^2 dx = \frac{(x-a)^3}{3} + C $$ - First integral: $$ \int_1^{1.5} (x-1)^2 dx = \left[ \frac{(x-1)^3}{3} \right]_1^{1.5} = \frac{(0.5)^3}{3} - 0 = \frac{0.125}{3} = \frac{1}{24} $$ - Second integral: $$ \int_{1.5}^2 (x-2)^2 dx = \left[ \frac{(x-2)^3}{3} \right]_{1.5}^2 = 0 - \frac{(-0.5)^3}{3} = - \frac{-0.125}{3} = \frac{1}{24} $$ 9. **Sum the areas:** $$ A = \frac{1}{24} + \frac{1}{24} = \frac{1}{12} $$ **Final answer:** The area enclosed by the graphs and the x-axis is $\boxed{\frac{1}{12}}$.