= Solution
After <bath cooling followed by insulated equilibration>, the remaining mean <temperature> is conserved: integrating the insulated <heat equation> gives
$$
\frac{d}{d\tau}\int_0^L U(x,\tau)\,dx=\kappa[U_x(L,\tau)-U_x(0,\tau)]=0.
$$
For constant <density>, specific <heat capacity> and cross-section, this is <conservation of energy>. Its <Fourier cosine series> has a constant mode equal to the initial <integral average>; every nonconstant mode decays. Therefore the eventual uniform <temperature> is
$$
T_f=\frac1L\int_0^LT(x,t_f)\,dx.
$$
For each mode with an <odd integer> index in part (c), $L^{-1}\int_0^L\sin((2j+1)\pi x/L)dx=2/[(2j+1)\pi]$. Integrating the bath-cooled series yields
$$
\boxed{T_f=\frac{8T_0}{\pi^2}\sum_{j=0}^{\infty}\frac{1}{(2j+1)^2}\exp\left[-\frac{(2j+1)^2\pi^2\kappa t_f}{L^2}\right]}.
$$
For $t_f>0$, <uniform convergence> justifies termwise <integration>; the $t_f=0$ result also follows by dominated convergence of the summable positive coefficients. As a check, $\sum_{j\geq0}(2j+1)^{-2}=\pi^2/8$ gives $T_f=T_0$ for zero bath time, and $T_f\to0$ as $t_f\to\infty$. More bath time strictly lowers the retained average <temperature> when $T_0>0$ and $\kappa>0$.
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