Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 14 / b / Solution

Normalization requires $1={\int }_0^{T_2}Atdt=A{T}_2^2/2$. ThereforeThe graph rises linearly from zero to the left-hand limit $2/T_2$, then drops to zero beyond $T_2$. Its enclosed triangular area is one. The combined sketch in (f) uses dimensionless time and shows this density together with the cumulative probabilities.