To solve the problem of finding the area between the curve (f(x) = x^2 - 3x + 2) and the x-axis from (x=1) to (x=2), follow these steps:
The function factors as (f(x) = (x-1)(x-2)), so it crosses the x-axis at (x=1) and (x=2). Since the parabola opens upward ((coefficient\ of\ x^2>0)), (f(x)) is negative between (x=1) and (x=2). Thus, the area is the absolute value of the integral of (f(x)) over this interval.
The antiderivative of (f(x)) is:
[F(x) = \int (x^2 -3x +2)dx = \frac{x^3}{3} - \frac{3x^2}{2} +2x]
[F(2) = \frac{8}{3} - \frac{3(4)}{2} +4 = \frac{8}{3} -6 +4 = \frac{2}{3}]
[F(1) = \frac{1}{3} - \frac{3(1)}{2} +2 = \frac{5}{6}]
[F(2)-F(1) = \frac{2}{3} - \frac{5}{6} = -\frac{1}{6}]
Area = (\left| -\frac{1}{6} \right| = \frac{1}{6})
Answer: (\boxed{\dfrac{1}{6}}) (or 0.1667 if decimal is needed, but exact form is (\frac{1}{6}))
(\boxed{1/6})
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