 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/16/d/solution

The expected value is zero, so the variance is

E [X^{2}]

Var (X) = \frac{2}{N} \int_{0}^{2} x^{2} (1 + e^{- x}) d x .

The polynomial term contributes

8/3

. Two applications of integration by parts give

\int x^{2} e^{- x} d x = - (x^{2} + 2 x + 2) e^{- x} + C,

so its definite value on

[0, 2]

2 - 10 e^{- 2}

. Therefore

Var (X) = \frac{28/3 - 20 e ^{- 2}}{6 - 2 e ^{- 2}} .

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