 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-1/10/solution

The function graphs intersect when

2 x = 4 x e^{- x}

, so

x = 0

x = ln 2

. For

0 < x < ln 2

e^{- x} > 1/2

, making

4 x e^{- x}

the upper graph. The enclosed area is therefore

A = \int_{0}^{l n 2} (4 x e^{- x} - 2 x) d x = [- 4 (x + 1) e^{- x} - x^{2}]_{0}^{l n 2} = 2 - 2 ln 2 - (ln 2)^{2} .

x e^{- x}

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