To solve the problem of finding the area bounded by (y = x^2 - 4x + 3) and (y = 0), follow these steps:

Step 1: Find intersection points with the x-axis

Set (y = 0):
[x^2 - 4x + 3 = 0]
Factor: ((x-1)(x-3) = 0)
Intersection points: (x=1) and (x=3).

Step 2: Determine the sign of the parabola between (x=1) and (x=3)

Test (x=2):
[y = (2)^2 -4(2)+3 = -1]
The parabola is below the x-axis here.

Step 3: Calculate the area

Area = integral of (upper function - lower function) from 1 to 3:
[Area = \int{1}^{3} [0 - (x^2 -4x +3)] dx = \int{1}^{3} (-x^2 +4x -3) dx]

Compute the antiderivative:
[-\frac{x^3}{3} + 2x^2 -3x]

Evaluate at bounds:
At (x=3): (-\frac{27}{3} + 2(9) -9 = -9 +18 -9 = 0)
At (x=1): (-\frac{1}{3} + 2(1) -3 = -\frac{4}{3})

Area = (0 - \left(-\frac{4}{3}\right) = \frac{4}{3})

Answer: A (\boxed{4/3})

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