 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2011/paper-2/18/b/solution

The intersections satisfy

6 x - x^{2} = 2 x + 3

, or

(x - 1) (x - 3) = 0

, so the bounded region has

1 \leq x \leq 3

. Over that interval the parabola is above the line, and their height difference is

(x - 1) (3 - x)

. Thus its area is

A = \int_{1}^{3} (4 x - x^{2} - 3) d x = [2 x^{2} - \frac{x ^{3}}{3} - 3 x]_{1}^{3} = \frac{4}{3} .

Using upper minus lower ensures a positive geometric area.

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