= Solution
This is the <Brownian covariance kernel>, so its integral operator is a <covariance operator>. To obtain its <eigendecomposition>, suppose $Uf=\lambda f$ with $\lambda\ne0$. Splitting the integral at $s$ gives
$$
\lambda f(s)=\int_0^s t f(t)dt+s\int_s^1f(t)dt.
$$
Differentiation yields $\lambda f'(s)=\int_s^1f(t)dt$ and $\lambda f''(s)=-f(s)$, with boundary conditions $f(0)=0$ and $f'(1)=0$. Hence the normalized eigenpairs are
$$
\phi_k(t)=\sqrt2\sin\!\left((k-\tfrac12)\pi t\right),
\qquad
\lambda_k=\frac1{(k-\tfrac12)^2\pi^2},
\qquad k\geq1.
$$
The eigenvalues are positive and summable, consistently with positivity and the <trace-class operator> property of a covariance operator.
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