Computing Partial AUC From Fitted Concentration Curves

Why Integrate Over a Time Window?

Total AUC summarizes exposure across a full dosing interval or another specified duration. It can, however, conceal when exposure occurs. Two formulations can have similar total AUCs but different early exposure, or their relative exposure can change between an early and a late interval. A partial AUC makes the time window explicit by integrating concentration only between prespecified times.

In a model-based comparison, the integral is calculated from the fitted concentration-time function for each formulation. The integral is exact for that mathematical function when an analytic expression is available; it is not automatically exact for the underlying biological concentration profile. Its relevance therefore depends both on the choice of interval and on whether the fitted curves adequately represent the concentrations over that interval.

Definition: For formulation $j$ with fitted concentration function $C_j(t;\theta_j)$, the partial AUC from time $a$ to time $b$, where $0 \leq a < b$, is $$ \operatorname{pAUC}_j(a,b)=\int_a^b C_j(t;\theta_j)\,dt. $$ If concentration is measured in mg/L and time in hours, the partial AUC has units mg·h/L. The interval $[a,b]$ is part of the definition of the metric; changing either boundary changes the quantity being compared.

The comparison can be expressed as an absolute contrast, $\operatorname{pAUC}_T(a,b)-\operatorname{pAUC}_R(a,b)$, or as a ratio, $\operatorname{pAUC}_T(a,b)/\operatorname{pAUC}_R(a,b)$, where $T$ and $R$ denote test and reference formulations. State which contrast is used and retain the same interval for both formulations. A ratio is undefined if the reference partial AUC is zero.

Assumptions and Calculation

Before integrating, specify the target formulation predictions and interval. The fitted curves must be expressed on a common time scale and concentration scale, and the chosen model must give meaningful predictions throughout $[a,b]$. The integral must exist; for ordinary pharmacokinetic concentration curves, this requires the curve to be integrable on the finite interval. The model and parameter estimates should be appropriate for the population or individual predictions that define the intended estimand. These conditions do not establish that the model is correct: fit and biological plausibility still need assessment.

For a fitted exponential component, the integral has a simple closed form. If $C(t)=A e^{-kt}$, with $k>0$, then

\int_a^b A e^{-kt}\,dt =A\left[\frac{e^{-ka}-e^{-kb}}{k}\right].

Because integration is linear, a curve made of a sum or difference of exponential components can be integrated one component at a time. For instance, for $C(t)=A(e^{-k_1t}-e^{-k_2t})$,

\operatorname{pAUC}(a,b) =A\left\{\frac{e^{-k_1a}-e^{-k_1b}}{k_1} -\frac{e^{-k_2a}-e^{-k_2b}}{k_2}\right\}.

The units follow from the coefficient and the integration variable. If $A$ is in mg/L and $k_1,k_2$ are in h$^{-1}$, each exponential integral contributes hours, so the result is in mg·h/L.

Worked Example: Integrating a Fitted Oral Concentration Curve

Suppose hypothetical fitted curves, in mg/L with time $t$ in hours, are $C_R(t)=10(e^{-0.2t}-e^{-1.0t})$ for the reference and $C_T(t)=10(e^{-0.2t}-e^{-0.5t})$ for the test. Both curves are nonnegative for $t \geq 0$, since each subtracts a faster-decaying exponential from a slower-decaying one. Calculate their partial AUCs over 0–2 hours and 2–8 hours, then compare the ratios.

For any $k>0$, define $G_k(a,b)=(e^{-ka}-e^{-kb})/k$. Then the reference partial AUC is $10[G_{0.2}(a,b)-G_{1.0}(a,b)]$, and the test partial AUC is $10[G_{0.2}(a,b)-G_{0.5}(a,b)]$.

Over 0–2 hours: $G_{0.2}(0,2)=(1-e^{-0.4})/0.2\approx1.648400$, $G_{1.0}(0,2)=1-e^{-2}\approx0.864665$, and $G_{0.5}(0,2)=(1-e^{-1})/0.5\approx1.264241$. Therefore,

\begin{aligned} \operatorname{pAUC}_R(0,2) &=10(1.648400-0.864665) \approx 7.837\ \text{mg·h/L},\\ \operatorname{pAUC}_T(0,2) &=10(1.648400-1.264241) \approx 3.842\ \text{mg·h/L}. \end{aligned}

The test-to-reference ratio over this interval is $3.842/7.837\approx0.4901$. Thus the fitted test curve has about 49.0% of the reference partial AUC over 0–2 hours.

Over 2–8 hours: $G_{0.2}(2,8)=(e^{-0.4}-e^{-1.6})/0.2\approx2.342117$, $G_{1.0}(2,8)=e^{-2}-e^{-8}\approx0.135000$, and $G_{0.5}(2,8)=(e^{-1}-e^{-4})/0.5\approx0.699128$. Substitution gives $\operatorname{pAUC}_R(2,8)=10(2.342117-0.135000)\approx22.071$ mg·h/L and $\operatorname{pAUC}_T(2,8)=10(2.342117-0.699128)\approx16.430$ mg·h/L. The ratio is approximately $16.430/22.071=0.7444$.

The interval-specific ratios differ substantially: about 0.4901 early and 0.7444 later. For comparison, integrating these curves from 0 to infinity gives $10(1/0.2-1/1.0)=40$ mg·h/L for the reference and $10(1/0.2-1/0.5)=30$ mg·h/L for the test, a total-AUC ratio of 0.7500. The total ratio is not a substitute for either partial-AUC ratio; it summarizes exposure over a different time range.

How the Boundaries Shape the Contrast

The dependence on interval is not merely a reporting convention. By the fundamental theorem of calculus, for a continuous fitted curve,

\frac{\partial \operatorname{pAUC}_j(a,b)}{\partial b}=C_j(b), \qquad \frac{\partial \operatorname{pAUC}_j(a,b)}{\partial a}=-C_j(a).

Increasing the upper boundary adds the concentration at the newly included time; increasing the lower boundary removes the concentration at the time excluded. For the absolute test-minus-reference contrast, the change at the upper boundary is governed by $C_T(b)-C_R(b)$. Consequently, if the fitted curves cross, moving a boundary across that region can change the magnitude or direction of the contrast. The ratio can also change as either area accumulates at a different rate.

Worked Example: A Contrast That Changes With the Window

Consider hypothetical fitted post-peak curves $C_R(t)=12e^{-0.4t}+3e^{-0.1t}$ and $C_T(t)=10e^{-0.4t}+5e^{-0.1t}$, in mg/L, with time in hours. Each curve is a positive sum of decaying components and is integrable on the intervals considered. For a component with rate $k$, use $G_k(a,b)=(e^{-ka}-e^{-kb})/k$.

Over 0–2 hours, $G_{0.4}(0,2)=(1-e^{-0.8})/0.4\approx1.376678$, and $G_{0.1}(0,2)=(1-e^{-0.2})/0.1\approx1.812692$. Thus,

\begin{aligned} \operatorname{pAUC}_R(0,2) &=12(1.376678)+3(1.812692) \approx21.958\ \text{mg·h/L},\\ \operatorname{pAUC}_T(0,2) &=10(1.376678)+5(1.812692) \approx22.830\ \text{mg·h/L}. \end{aligned}

The test-to-reference ratio is $22.830/21.958\approx1.040$, a test partial AUC about 4.0% higher over 0–2 hours.

For 2–5 hours, $G_{0.4}(2,5)=(e^{-0.8}-e^{-2})/0.4\approx0.784984$, and $G_{0.1}(2,5)=(e^{-0.2}-e^{-0.5})/0.1\approx2.122001$. Therefore, the reference partial AUC is $12(0.784984)+3(2.122001)\approx15.786$ mg·h/L, while the test partial AUC is $10(0.784984)+5(2.122001)\approx18.460$ mg·h/L. The ratio is $18.460/15.786\approx1.169$.

The later-window ratio is about 1.169 rather than 1.040. The slower component contributes more heavily as time advances, and the two formulations have different coefficients for that component. The fitted curves and their formulation contrast therefore depend on which time window is summarized.

Numerical Integration When There Is No Closed Form

Not every fitted concentration function has a convenient analytic integral. A numerical quadrature rule can approximate the area by combining curve evaluations at selected times. For example, Simpson’s rule with an even number $n$ of equally spaced subintervals, spacing $h=(b-a)/n$, uses weighted endpoint and interior values:

\int_a^b C(t)\,dt \approx \frac{h}{3}\left[C(t_0)+C(t_n) +4\sum_{\substack{i=1\\i\ \mathrm{odd}}}^{n-1}C(t_i) +2\sum_{\substack{i=2\\i\ \mathrm{even}}}^{n-2}C(t_i)\right], \quad t_i=a+ih.

This approximation is for the specified curve, not a method for repairing a misspecified fitted model. Increasing the number of evaluation points generally improves quadrature for a sufficiently smooth curve, but does not resolve uncertainty in fitted parameters or poor model fit.

Worked Example: Simpson Approximation and an Analytic Check

Let a hypothetical fitted curve be $C(t)=6t e^{-0.6t}$ mg/L, with $t$ in hours. Approximate its area from 1 to 3 hours using Simpson’s rule with $n=4$, so $h=0.5$ hours. The five curve values are $C(1)\approx3.293$, $C(1.5)\approx3.659$, $C(2)\approx3.614$, $C(2.5)\approx3.347$, and $C(3)\approx2.975$ mg/L. Substitution gives

\begin{aligned} \operatorname{pAUC}(1,3) &\approx \frac{0.5}{3}\left[ 3.29287+2.97538+ 4(3.65913+3.34695)+2(3.61433)\right]\\ &\approx 6.920\ \text{mg·h/L}. \end{aligned}

An analytic check is available here because $\int 6t e^{-0.6t}\,dt=-(10t+16.6667)e^{-0.6t}$. Evaluating at 1 and 3 gives an area of approximately 6.921 mg·h/L. The close agreement supports the numerical arithmetic for this smooth curve and grid; it does not validate the model against observed concentrations.

Diagnostics, Alternatives, and Limitations

Check that the curve predictions are plausible throughout the chosen interval, not only at the observed sample times. Useful diagnostics include plotting fitted curves alongside concentrations, examining residual patterns over time, checking for negative or implausible predictions, and verifying that the integration limits match the intended time window. For numerical integration, repeat the calculation on a finer grid or compare it with an analytic integral when available. A material change under grid refinement points to inadequate numerical resolution.

The main alternative to integrating a fitted curve is to calculate area directly from observed concentrations, commonly by a trapezoidal rule. Direct numerical area is less dependent on a structural curve model, but it relies on observed values at the relevant times and its accuracy can be sensitive to sampling density and the timing of measurements. Model-based integration can provide a smooth area over an interval, but it depends on the fitted model between observations and may be particularly sensitive where data are sparse or the curve changes rapidly. These approaches answer related, not identical, practical questions; the analysis plan should define the estimand and method before examining the contrast.

A prespecified interval should have a scientific or regulatory rationale. Choosing boundaries after seeing which interval gives the most favorable formulation ratio can exaggerate evidence for a desired conclusion. Also, calculating a plug-in integral from fitted parameter estimates does not itself quantify uncertainty in that area or in its ratio. The resulting point estimate should not be presented as if model uncertainty had disappeared.

Common pitfall: A reader may assume that similar total AUCs imply similar partial AUCs, or interpret one partial-AUC ratio as applying to every interval. This is incorrect: area depends on the curve within the selected window, and formulation curves can differ in shape or cross. Define the interval, integrate each fitted curve over that same interval, and interpret the resulting contrast only for that window.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

For $C(t)=4e^{-0.2t}$ mg/L, calculate the partial AUC from 1 to 3 hours.
In the first worked example, why is the 0–2-hour ratio not interchangeable with the ratio of total AUCs?
For a continuous curve, what is the effect on partial AUC of moving the upper boundary $b$ slightly later?
What does close agreement between a Simpson approximation and an analytic integral establish, and what does it not establish?
Can equal total AUCs guarantee equal partial AUCs over an early interval? Explain.

Solutions. (1) $\operatorname{pAUC}(1,3)=4(e^{-0.2}-e^{-0.6})/0.2=20(0.818731-0.548812)\approx5.398$ mg·h/L. (2) The total ratio integrates over 0 to infinity, whereas the partial ratio uses only 0–2 hours; the formulations have different curve shapes, so their relative areas vary by interval. (3) The added area is approximately $C(b)\Delta b$ for a small increase $\Delta b$, because $\partial\operatorname{pAUC}/\partial b=C(b)$. (4) It supports the numerical quadrature calculation for the specified curve and grid; it does not show that the curve fits observed concentrations or that its parameters are certain. (5) No. Equal total areas constrain the integral over the full range, not the distribution of that area over time, so the early partial areas can differ.