In pharmacokinetics, it is assumed that the area under a curve is the same even if a substance is dosed gradually. This applies when "first order kinetics" is used. Is this mathematically true?
I graphed two examples, and used a simple app to estimate the area under the curve, and it is close to identical. But would be good to get a mathematical point of view.
I'll take the governing equation for first-order kinetics to be
$$\frac{dN}{dt}=-KN+\frac{dN_{in}}{dt} \tag{1}$$
where $N(t)$ is the amount of drug in the system at time $t$, $K$ is the elimination constant, and $N_{in}(t)$ is the amount of drug infused up to time $t$. I'll assume $N(0)=N_{in}(0)=0$ (no drug is present or has been infused by time $t=0$) and that that $N_{in}(t)=0$ for $t\geq T$ (the drug infusion has ceased by time $T$). As such, for $t\geq T$ we have $$\frac{dN}{dt}=-KN$$ and therefore $N(t)=Ae^{-kt}$ which vanishes as $t\to\infty$.
With these assumptions, suppose we rearrange eq. $(1)$ as $$N(t)=\frac{1}{K}\frac{dN_{in}}{dt}-\frac{1}{K}\frac{dN}{dt}=\frac{1}{K}\frac{d}{dt}(N_{in}-N).$$ Then the area under the $N(t)$ curve is obtained by integrating from $0$ to $\infty$, and the fundamental theorem of calculus yields
\begin{align} \int_0^\infty N(t)\,dt &=\frac{1}{K}\int_0^\infty \frac{d}{dt}(N-N_{in})\,dt\\ &=\frac{1}{K}[N_{in}-N]_0^\infty \\ &= \frac{1}{K}N_{in}(\infty). \end{align} Note that this only depends on the final value of $N_{in}(t)$ and not the infusion schedule.