 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/ib/paper-4/15f/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Part (i) gives

L (t) = 1/ s^{2}

and

L (t e^{- t}) = 1/ (s + 1)^{2}

f (t) = \int_{0}^{t} (t - v) v e^{- v} d v = t \int_{0}^{t} v e^{- v} d v - \int_{0}^{t} v^{2} e^{- v} d v .

\int_{0}^{t} v e^{- v} d v = 1 - (t + 1) e^{- t}, \int_{0}^{t} v^{2} e^{- v} d v = 2 - (t^{2} + 2 t + 2) e^{- t} .

Substitution gives $f (t) = t - 2 + (t + 2) e^{- t}$ , agreeing with the contour integration result.

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