For this Neumann problem, the steady state temperature of the thin pipe will be the average value of the function $f(x)$. We are given the diffusion equation, two boundary conditions, and an initial condition:
$$u_t=ku_{xx}$$ $$ u'(0,t)=u'(\ell,t)=0$$$$u(x,0)=f(x)$$
With both ends of the pipe insulated, any temperature minima/maxima along the pipe will even out by diffusion, leaving a temperature distribution at a certain constant value regardless of location along the pipe. In an insulated pipe problem like this one, the entire area under the curve of the initial condition function $f(x)$represents the total heat energy within the pipe. Since no heat is escaping or being added, the area under the curve will remain constant, but will spread out evenly at some constant value. (Note the differences between this case and the Dirichlet boundary conditions case, where heat leaks from one or more ends of the pipe until the temperatures along the pipe stabilize at a different constant value - zero.) The steady state temperature will equalize at the constant value mathematically described by the average value of the function $f(x)$ over the interval $[0,\ell]$ (the pipe's length). In general, the average value of a function over the interval $[a,b]$ is given by the following equation:
$$f_{ave}(x)=\frac{1}{b-a}\int_a^b f(x)\ dx$$
In this case, the steady state temperature $u(x)$ in terms of $f$ will be:
$$u(x)=f_{ave}(x)=\frac{1}{\ell-0}\int_0^\ell f(x)\ dx=\frac{1}{\ell}\int_0^\ell f(x)\ dx$$
The graph of the steady state solution will be a horizontal line over the interval from $0$ to $\ell$, similar to the following, where the "y-value" of the line is $u(x)$. (Of course, the actual value of $u(x)$ will depend on the function described by $f$, as well as the length $\ell$ of the thin pipe.) I couldn't quite get a Mathematica plot to cooperate, but here is the general idea.
In the specific case where $\ell=1$ and $f(x)={x^2}(2-x)^2$, the steady state formula is a relatively straightforward calculation, analogous to the general case above. Using the average value equation and substituting values for $\ell$ and $f(x)$,
$$u(x)=\frac{1}{\ell}\int_0^\ell f(x)\ dx =1\int_0^1 {x^2}(2-x)^2\ dx.$$
Distributing terms,
$$u(x)=\int_0^1 {x^2}(4-4x+x^2)\ dx=\int_0^1 4x^2-4x^3+x^4\ dx.$$
Completing the integration gives:
$$u(x)=\left.\left(\frac43{x^3}-\frac44{x^4}+\frac15{x^5}\right)\right|_0^1$$ $$u(x)=\left(\frac43-1+\frac15\right)-0=\frac{12}{15}-\frac{5}{15}+\frac{3}{15}=\frac{8}{15}.$$
Therefore, the steady state temperature for this problem as $t\rightarrow\infty$ is $u(x)=\frac{8}{15}$. Assuming that $k=1$, the solution evolves as follows.
Mathematica code:
System evolution:
