 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-310/1/c/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

a (t_{0}) = 1

H (a) = H_{0} Ω_{r, 0} a^{- 4} + Ω_{m, 0} a^{- 3} + Ω_{Λ, 0},

where

Ω_{r, 0} + Ω_{m, 0} + Ω_{Λ, 0} = 1

. Since

d t = d a / (a H)

t_{0} = \frac{1}{H _{0}} \int_{0}^{1} \frac{d a}{a Ω _{r, 0} a ^{- 4} + Ω _{m, 0} a ^{- 3} + Ω _{Λ, 0}} .

t_{0} = \frac{2}{3 H _{0}} .

H_{0} ≃ 70 km s^{- 1} Mp c^{- 1}

, for which the Hubble time is

H_{0}^{- 1} ≃ 14.0 Gyr

, gives

t_{0} ≃ 9.3 Gyr .

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