This is part of solving the correlation function for standard 1D Brownian Motion. I am unsure how to compute the integral, or explain the solution I have. Any help is appreciated. The integral is
$$ \int_0^t\int_0^s\min(\rho,l)\ dld\rho $$
$$ = \int_0^s\int_0^s\min(\rho,l)\ dld\rho + \int_s^t\int_0^s\ \min(\rho,l)\ dld\rho $$
$$ = 2\int_0^sd\rho\int_0^{\rho}\min(\rho,l)\ dl + \int_s^td\rho\int_0^s\min(\rho,l)\ dl $$
$$ = 2\int_0^sd\rho\int_0^{\rho}l\ dl + \int_s^td\rho\int_0^sl\ dl $$
$$ = 2\int_0^sd\rho\int_0^{\rho}l\ dl + \int_s^td\rho\int_0^sl\ dl $$
$$ = \frac{s^3}{3} + \frac{s^2}{2}(t - s) $$
If you have an alternate solution to the first line, that would be great, otherwise an explanation (added intermediate steps) from line 3 I would be so grateful. This solution may be incorrect as well. Thanks.
I've searched for some similar problems, but the limits are the same, and I am not sure how to reason through these problems either.
integrals involving minimum function how to solve double integral of a min function