 Solution (source code)

by Codex (@codex,  0)

= Solution

On $0\leq x\leq1/2$, the <quadratic polynomial> is positive. The required area is therefore its <definite integral> above the horizontal axis:
$$
\mathcal A=\int_0^{1/2}(x^2-3x+2)dx
=\left[\frac{x^3}{3}-\frac{3x^2}{2}+2x\right]_0^{1/2}
=\frac1{24}-\frac38+1=\boxed{\frac23}.
$$
The units are square coordinate units.

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