While elaborating on the solution for the Green's function of a mechanics problem involving disks moving on an interface, I came across the following integral equation for the unknown function $f(s)$: $$ \int_0^\rho \left( \frac{s}{\rho} \right)^3 f(s) \, \mathrm{d}s +\int_\rho^1 f(s) \, \mathrm{d}s = 1 \, , \tag{1} $$ wherein $0 \le \rho \le 1$. This equation needs to be satisfied for every value of $\rho$.
In particular, for $\rho = 0$, we have the mean $$ \int_0^1 f(s) \, \mathrm{d}s = 1 \, . $$
What I tried is to express the unknown function in terms of a Taylor series and evaluate the corresponding equation (1). By identifying term by term, I am unable to obtain a solution that matches all the coefficients.
Any hints on how to proceed with such an integral equation?
The equation simplifies to
\begin{align*} \int_{0}^{1} (s\wedge \rho)^3 f(s) \, \mathrm{d}s = \rho^3, \qquad \rho \in [0, 1]. \end{align*}
Differentiating both sides with respect to $\rho$,
\begin{align*} \int_{0}^{1} 3 (s\wedge \rho)^2 \mathbf{1}_{\{\rho \leq s\}} f(s) \, \mathrm{d}s = 3\rho^2, \qquad \rho \in (0, 1). \end{align*}
This equation simplifies to
\begin{align*} \int_{\rho}^{1} f(s) \, \mathrm{d}s = 0, \qquad \rho \in (0, 1). \end{align*}
Letting $\rho \to 0^+$ yields the equality $\int_{0}^{1} f(s) \, \mathrm{d}s = 0$, which contradicts $\int_{0}^{1} f(s) \, \mathrm{d}s = 1$.
Addendum. The equation has a "distributional solution" $f(x) = \delta(x - 1^-)$:
$$ \int_{0}^{1} (s\wedge\rho)^3 f(s) \, \mathrm{d}s = \int_{0}^{1} (s\wedge\rho)^3 \delta(x - 1^-) \, \mathrm{d}s = \lim_{s\to 1^-} (s\wedge\rho)^3 = \rho^3 $$
for all $\rho \in [0, 1]$.
(The minus-sign superscript in $\delta(x-1^-)$ is introduced in order to indicate that the distributional solution is interpreted as the unit mass placed "infinitesimally left to the point $1$".)