 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2007/paper-1/section-b/7/i/solution

The region is

0 \leq y \leq 1

2 y \leq x \leq 2

. Reversing the order describes the same triangle as

0 \leq x \leq 2

0 \leq y \leq x /2

. Since the integrand is continuous on this compact region, Fubini's theorem permits

\int_{0}^{1} \int_{2 y}^{2} e^{x^{2}} d x d y = \int_{0}^{2} \int_{0}^{x /2} e^{x^{2}} d y d x = \frac{1}{2} \int_{0}^{2} x e^{x^{2}} d x = \frac{e ^{4} - 1}{4} .

The new inner integral introduces the factor

x

u = x^{2}

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