Deriving AUC And Cmax From Model Predictions

From a Fitted Curve to an Exposure Metric

A fitted concentration-time function contains more information than the concentrations recorded at a few sampling times. It describes a curve between observations, so exposure and peak concentration can be calculated from the curve itself. The calculation is straightforward in principle, but its result depends on the function, the time interval, and the way the maximum is found. AUC over a finite interval, AUC extrapolated to infinity, and a peak found among observed samples are not interchangeable quantities.

This tutorial focuses on extracting AUC and Cmax from a fitted concentration-time function. The function may be a structural pharmacokinetic prediction evaluated at estimated parameters; the calculations below concern the resulting curve, not how those parameters were estimated. The function and its parameter values should be stated, as should the dosing occasion or profile being summarized and the time interval over which the metric is defined.

Definition: Let \(C(t)\) be a predicted concentration at time \(t\). Model-derived AUC over \([t_0,t_1]\) is \(\int_{t_0}^{t_1} C(t)\,dt\). Model-derived Cmax over that interval is \(\max_{t_0\leq t\leq t_1} C(t)\). AUC has concentration-times-time units; Cmax has concentration units. Both are defined relative to the stated curve and interval.

For example, AUC from zero to a specified last time describes exposure only over that interval. AUC from zero to infinity requires a function defined and integrable over the whole nonnegative time axis. Cmax over a planned observation window may differ from the maximum over all \(t\geq0\). Thus, time limits are part of the metric, not a minor computational detail.

Analytic Integration and Peak Conditions

When an antiderivative is available, integration gives AUC directly. For a differentiable curve, an interior time \(t^*\) that maximizes concentration must satisfy \(C'(t^*)=0\), unless the derivative does not exist there. Finding stationary points is not sufficient by itself: evaluate the curve at all candidates and at both interval boundaries, then select the largest value. If the function is piecewise-defined, include its breakpoints as candidates as well.

The first-order oral absorption curve gives a useful derivation. Suppose, for \(t\geq0\), \(C(t)=A(e^{-k_e t}-e^{-k_a t})\), where \(A>0\), \(k_a>k_e>0\), and the rate constants have units of inverse time. Differentiating gives:

C'(t)=A(-k_e e^{-k_e t}+k_a e^{-k_a t}).

At an interior peak, setting the derivative to zero and rearranging yields \(k_a e^{-k_a t}=k_e e^{-k_e t}\). Taking logarithms gives \(\ln(k_a/k_e)=(k_a-k_e)t\). Therefore:

t_{\max}=\frac{\ln(k_a/k_e)}{k_a-k_e}.

This stationary point is the global maximum on \(t\geq0\) for this curve: concentration is zero at \(t=0\), rises initially because \(C'(0)=A(k_a-k_e)>0\), and eventually approaches zero as \(t\) tends to infinity. This conclusion relies on the stated positive parameter constraints and the single-peak form; it should not be carried over automatically to more complex curves.

Worked Example: AUC and Cmax for a Single-Peak Curve

Consider the hypothetical fitted curve \(C(t)=10(e^{-0.2t}-e^{-1.2t})\) mg/L for \(t\geq0\), with time in hours. Calculate AUC from zero to 12 hours and the curve's Cmax.

AUC. Integrating each exponential gives:

\begin{aligned} \operatorname{AUC}_{0-12} &=10\left[\frac{1-e^{-0.2(12)}}{0.2}-\frac{1-e^{-1.2(12)}}{1.2}\right]\\ &=10\left[\frac{1-e^{-2.4}}{0.2}-\frac{1-e^{-14.4}}{1.2}\right]\\ &\approx10(4.5464-0.8333)=37.13\ \text{mg}\cdot\text{h/L}. \end{aligned}

Cmax. Here \(k_a=1.2\) h\(^{-1}\) and \(k_e=0.2\) h\(^{-1}\). The peak time is \(\ln(1.2/0.2)/(1.2-0.2)=\ln(6)\approx1.792\) h. Substitution gives \(C(1.792)=10(e^{-0.3584}-e^{-2.1504})\approx10(0.698793-0.116438)=5.824\) mg/L. This time lies within the interval from zero to 12 hours. The derivative changes from positive to negative there, and the endpoints have lower concentrations, so this is also the maximum on that finite interval.

If the intended interval were zero to infinity, the AUC would instead be \(10(1/0.2-1/1.2)=41.67\) mg·h/L. That is a different metric, not a correction to the finite-interval result.

Multiple Peaks: Search the Whole Domain

With a sum of components, delayed input, or another multiphasic curve, there may be several local maxima. The derivative can have several roots, and a local maximum at one time need not be the global Cmax. A reliable procedure is to identify all stationary points within the interval, evaluate concentration at each, and compare those values with the endpoints and any piecewise boundaries. Numerical optimization can help locate candidates, but a numerical search should be checked for missed peaks, especially if peaks are narrow or close together.

AUC for a sum of integrable components is the sum of their integrals over the same interval. For a component that begins only after a delay, integrate it over the part of the interval where it is defined and nonzero. In contrast, Cmax of a sum is generally not the sum of the component maxima: their peaks may occur at different times, and the total curve must be maximized directly.

Worked Example: Comparing Two Peaks in a Delayed-Input Curve

Suppose a hypothetical fitted profile, in mg/L for time \(t\) in hours, is \(C(t)=12te^{-t}\) for \(0\leq t<4\), and \(C(t)=12te^{-t}+24(t-4)e^{-2(t-4)}\) for \(4\leq t\leq8\). The second component represents a delayed contribution that is zero at its onset. Find AUC from zero to 8 hours and identify the global Cmax.

AUC. The first component is integrated over the full interval. For the second component, set \(u=t-4\), so it contributes only over \(0\leq u\leq4\):

\begin{aligned} \operatorname{AUC}_{0-8} &=12\int_0^8 te^{-t}\,dt+24\int_0^4 ue^{-2u}\,du\\ &=12[1-9e^{-8}]+6[1-9e^{-8}]\\ &=18[1-9e^{-8}]\\ &\approx18(1-0.003019)=17.95\ \text{mg}\cdot\text{h/L}. \end{aligned}

The antiderivatives used are \(\int te^{-t}dt=-(t+1)e^{-t}\) and \(\int ue^{-2u}du=-(u/2+1/4)e^{-2u}\). The two contributions are approximately 11.96 and 5.982 mg·h/L, respectively.

Cmax. Before hour 4, the curve \(12te^{-t}\) has its maximum at \(t=1\), where \(C(1)=12/e\approx4.415\) mg/L. After hour 4, differentiate the full expression:

C'(t)=12e^{-t}(1-t)+24e^{-2(t-4)}[1-2(t-4)].

Solving \(C'(t)=0\) numerically on the later segment gives a local maximum at about \(t=4.47\) h, with concentration about 5.02 mg/L. Checking the endpoints and the earlier local maximum shows this later peak is the global maximum on zero to 8 hours. The numerical root is approximate; it should be reported with suitable precision rather than as an exact analytic value. The calculation illustrates why reporting only the first peak or only the largest component peak can fail.

Numerical Integration and Observed-Sample Summaries

For a fitted function without a convenient antiderivative, numerical quadrature approximates its integral. One common method is the composite trapezoidal rule: divide the interval into subintervals and add the areas of trapezoids whose heights are fitted concentrations at the grid points. A finer grid can improve approximation, but accuracy also depends on how well the curve is resolved, whether the integrand is smooth, and whether the numerical integration settings are adequate. Numerical integration does not remove uncertainty or bias in the fitted curve.

A trapezoidal AUC calculated from measured concentrations is different in construction: it connects observed samples by straight lines. A model-derived AUC integrates the fitted curve between and beyond those sample times according to the model. Similarly, the maximum observed concentration is the largest recorded sample, whereas model-derived Cmax is the maximum of the fitted curve over its specified interval. The two may be close with dense sampling and a suitable model, but they need not coincide.

Worked Example: Model-Derived Metrics Versus Sample-Based Summaries

For comparison, use the hypothetical curve \(C(t)=10(e^{-0.2t}-e^{-1.2t})\) mg/L and sample it at 0, 1, 2, 4, and 8 hours. The predicted concentrations, rounded to three decimals, are 0.000, 5.177, 5.796, 4.411, and 2.018 mg/L.

The largest sampled value is 5.796 mg/L at 2 hours. In contrast, the fitted curve's peak is 5.824 mg/L at approximately 1.792 hours, as derived above. The discrete observations do not include the fitted peak time.

Applying the trapezoidal rule to these samples gives:

\begin{aligned} \operatorname{AUC}_{\text{sample}} &=\frac{0+5.177}{2}(1)+\frac{5.177+5.796}{2}(1)\\ &\quad+\frac{5.796+4.411}{2}(2)+\frac{4.411+2.018}{2}(4)\\ &\approx31.14\ \text{mg}\cdot\text{h/L}. \end{aligned}

The model-derived AUC over the same zero-to-8-hour interval is:

\begin{aligned} \operatorname{AUC}_{0-8} &=10\left[\frac{1-e^{-0.2(8)}}{0.2}-\frac{1-e^{-1.2(8)}}{1.2}\right]\\ &\approx10(3.9905-0.8333)=31.57\ \text{mg}\cdot\text{h/L}. \end{aligned}

The difference, about 0.43 mg·h/L, reflects the difference between integrating the smooth fitted function and joining the specified samples with straight lines. Neither arithmetic result implies that one method is universally preferable: the appropriate metric depends on the analysis definition. The sample-based calculation is also affected by sampling density and placement.

Assumptions, Diagnostics, and Limitations

These calculations presume that the concentration-time function is defined over the requested interval, has meaningful units and parameter values, and represents the profile whose metric is intended. For analytic integration, the integral must exist; for Cmax, the maximum must be attained or a suitable supremum must be specified. In usual finite intervals, a continuous curve has a maximum. For an unbounded interval, inspect the limiting behavior rather than assuming a maximum occurs at a finite time.

Before reporting results, check that the interval matches the estimand, units are consistent, and predicted concentrations are physically and scientifically plausible for the intended application. For Cmax, confirm that the optimization covered the full interval and that competing local peaks and boundaries were evaluated. For AUC, verify integration limits and, for numerical work, check convergence by refining the grid or comparing with an analytic result where possible.

Warning signs include a reported peak outside the allowed interval, a numerical optimizer that returns different peaks from different starting values, unexpectedly large AUC changes under minor grid refinement, or substantial negative concentrations when the model is intended to predict nonnegative concentrations. Such findings call for checking the implementation, the model's domain and parameter constraints, and whether the fitted function is suitable for deriving the metric. A precise integral of a poorly fitting or inappropriate function is not necessarily a useful exposure estimate.

Model-derived metrics use the model to represent the curve between observed times; sample-based summaries make fewer assumptions about the curve's shape but are sensitive to the sampling schedule and interpolation convention. Neither approach automatically accounts for uncertainty in fitted parameters. The calculated value is conditional on the specified fitted function and its parameter values. This tutorial does not determine how such uncertainty should enter a bioequivalence decision; it establishes how the curve-level metrics themselves are obtained.

Key takeaway: State the fitted function, interval, and units before calculating AUC or Cmax. Integrate over exactly that interval; for Cmax, compare all candidate peaks and boundaries. Do not label a maximum among observed samples as the model-derived Cmax.

Check Your Understanding

Answer each question, then compare with the concise solutions below.

For \(C(t)=6e^{-0.3t}\) mg/L on zero to 5 hours, write the AUC formula with limits and give the units.
Why is finding one root of \(C'(t)=0\) not enough to establish Cmax for a curve with repeated peaks?
For the single-peak curve in the first worked example, is its AUC from zero to infinity the same as its AUC from zero to 12 hours? Explain.
What is the distinction between the largest sampled concentration and model-derived Cmax?
Name one useful diagnostic for numerical AUC and one for numerical Cmax.

Solutions. (1) \(\int_0^5 6e^{-0.3t}dt=6(1-e^{-1.5})/0.3\approx15.54\), in mg·h/L. (2) Other stationary points or interval boundaries may have higher concentrations, so all candidates must be compared. (3) No: the infinite-interval integral includes the positive tail after hour 12; for that curve the values are 41.67 and 37.13 mg·h/L, respectively. (4) The sampled maximum is the largest recorded value at a sampling time; model-derived Cmax maximizes the fitted curve over its stated domain and can occur between samples. (5) Refine the integration grid and compare results for AUC; for Cmax, check stationary points, boundaries, and whether numerical optimization finds the same global peak under reasonable search settings.