 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-1/14/c/solution

Here

a = 0

cos x

[- h, h]

, where

h = π /6

. Hence

E [X] = 0

by oddness, and the variance is just

E [X^{2}]

. Two applications of integration by parts give

\int x^{2} cos x d x = x^{2} sin x + 2 x cos x - 2 sin x + C .

Thus

Var (X) = 2 [h^{2} sin h + 2 h cos h - 2 sin h] = \frac{π ^{2}}{36} + \frac{π 3}{3} - 2 .

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