1. **State the problem:** We need to find the area of the region bounded by the curve $y = x^2 - 4x + 3$ and the x-axis. 2. **Find the points where the curve intersects the x-axis:** Set $y = 0$ and solve for $x$: $$x^2 - 4x + 3 = 0$$ 3. **Solve the quadratic equation:** $$x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} = \frac{4 \pm \sqrt{16 - 12}}{2} = \frac{4 \pm 2}{2}$$ So the roots are: $$x = \frac{4 - 2}{2} = 1 \quad \text{and} \quad x = \frac{4 + 2}{2} = 3$$ 4. **Set up the integral for the area:** The curve is above the x-axis between $x=1$ and $x=3$, so the area is: $$\text{Area} = \int_1^3 (x^2 - 4x + 3) \, dx$$ 5. **Calculate the integral:** $$\int (x^2 - 4x + 3) \, dx = \frac{x^3}{3} - 2x^2 + 3x + C$$ 6. **Evaluate the definite integral:** $$\left[ \frac{x^3}{3} - 2x^2 + 3x \right]_1^3 = \left( \frac{27}{3} - 2 \cdot 9 + 9 \right) - \left( \frac{1}{3} - 2 + 3 \right) = (9 - 18 + 9) - \left( \frac{1}{3} + 1 \right) = 0 - \frac{4}{3} = -\frac{4}{3}$$ 7. **Interpret the result:** The integral is negative because the curve is below the x-axis between 1 and 3. The area is the absolute value: $$\text{Area} = \frac{4}{3}$$ **Final answer:** The area of the region bounded by the curve and the x-axis is $\boxed{\frac{4}{3}}$.