Solution
= Solution
Normalization requires $1=\int_0^{T_2}At\,dt=AT_2^2/2$. Therefore
$$
\boxed{A=\frac2{T_2^2},\qquad f_2(t)=\begin{cases}2t/T_2^2,&0<t<T_2,\\0,&\text{otherwise}.\end{cases}}
$$
The 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>.