 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1/9/a/solution

0 \leq x \leq 1/2

, the quadratic polynomial is positive. The required area is therefore its definite integral above the horizontal axis:

A = \int_{0}^{1/2} (x^{2} - 3 x + 2) d x = [\frac{x ^{3}}{3} - \frac{3 x ^{2}}{2} + 2 x]_{0}^{1/2} = \frac{1}{24} - \frac{3}{8} + 1 = \frac{2}{3} .

The units are square coordinate units.

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