 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2003/paper-2/5/b/iii/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

E [X] = k \int_{0}^{\infty} t e^{- k t} d t

t e^{- k t}

tending to zero, gives

E [X] = \int_{0}^{\infty} e^{- k t} d t = \frac{1}{k} .

 New to topics? Read the docs here!