 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2009/paper-2/18/a/solution

The integrand is continuous on the compact rectangle, so Fubini's theorem allows integrating in

y

first. For

x \neq = 0

\int_{0}^{1} x e^{x y} d y = e^{x} - 1

; the same formula holds by continuity at

x = 0

. Hence

\int_{0}^{1} \int_{0}^{2} x e^{x y} d x d y = \int_{0}^{2} (e^{x} - 1) d x = e^{2} - 3.

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