Figure 2 shows the new axes represented by \(T\). It goes in the opposite direction of \(t\). For \(t=t_1\) we are in the case \(T\ <\ \Delta t\). Therefore the solution is given by (\ref{eq5}). However, because \(t_1\) is fixed, the extremes of the integral vary just between 0 and \(T_1\). We can repeat the procedure for any \(T_1\) in this interval. Upon the knowledge of \(f\), the s-function is known in advance and we have just to pick its values at \(T_1\) (and the values of the precipitation at \(t_2-T_2\)) to know the discharge, as shown in Figure 4 below.