Worked AUC and Partial-AUC Calculation

From Concentration Samples to Exposure Metrics

A concentration–time profile is observed only at selected sampling times, but AUC summarizes exposure across an interval. Calculating it therefore requires a rule for estimating the concentration between samples. The same small dataset can also provide the observed maximum concentration, \(C_{\max}\), and the time at which that sampled maximum occurred, \(T_{\max}\). These calculations are straightforward when time points, concentration units, and interval boundaries are handled consistently.

This tutorial uses the linear trapezoidal rule for the worked calculations. It treats the concentration as changing linearly between adjacent observations. The resulting AUC is an estimate based on the sampled profile, not a direct measurement of every concentration between samples. No terminal extrapolation beyond the last quantifiable time is included.

Definition: Let \(t_1,\ldots,t_n\) be ordered sampling times and \(C_1,\ldots,C_n\) their concentrations, with \(t_1<\cdots<t_n\). The AUC from the first included time to the last quantifiable time is the sum of the estimated areas between successive included samples. The observed \(C_{\max}\) is the largest sampled concentration, and the observed \(T_{\max}\) is the sampling time or times at which that largest value occurs. AUC has units of concentration multiplied by time; \(C_{\max}\) has concentration units; \(T_{\max}\) has time units.

The Linear Trapezoidal Calculation

For a pair of adjacent observations at \(t_i\) and \(t_{i+1}\), write \(\Delta t_i=t_{i+1}-t_i\). The linear trapezoidal estimate for that interval is the average of its two endpoint concentrations multiplied by the interval duration:

A_i=\frac{C_i+C_{i+1}}{2}(t_{i+1}-t_i).

Adding the interval areas gives the AUC over the included times. If the first included time is dose time \(0\) and the last quantifiable sample is \(t_{\text{last}}\), this is often denoted \(AUC_{0-t_{\text{last}}}\). The time-zero concentration used in a calculation must follow the study’s prespecified convention; it should not be silently inferred from an unreported value. The examples below explicitly provide the starting value.

The rule has a direct justification. Under linear interpolation, concentration at elapsed time \(u\) after \(t_i\) is \(C_i+(C_{i+1}-C_i)u/\Delta t_i\). Integrating this expression from \(u=0\) to \(u=\Delta t_i\) gives:

\begin{aligned} \int_0^{\Delta t_i}\left(C_i+\frac{C_{i+1}-C_i}{\Delta t_i}u\right)\,du &=C_i\Delta t_i+\frac{C_{i+1}-C_i}{2\Delta t_i}(\Delta t_i)^2\\ &=\frac{C_i+C_{i+1}}{2}\Delta t_i. \end{aligned}

Thus, the trapezoidal formula is exact for the piecewise-linear curve defined by the observations. Whether that curve closely represents the actual concentration profile depends on the sampling schedule and the profile’s shape.

Worked Example: AUC to the Last Quantifiable Time

Consider a hypothetical profile with concentrations in mg/L at times in hours: \((0,0), (1,4), (2,8), (4,5), (6,2)\). Suppose the last quantifiable sample is at 6 h and the provided time-zero value of 0 mg/L is the prespecified starting value. Calculate AUC from 0 to 6 h, \(C_{\max}\), and \(T_{\max}\).

For each interval, multiply its duration by the mean of its two endpoint concentrations:

0 to 1 h: \(A_1=(0+4)(1-0)/2=2\) mg·h/L.

1 to 2 h: \(A_2=(4+8)(2-1)/2=6\) mg·h/L.

2 to 4 h: \(A_3=(8+5)(4-2)/2=13\) mg·h/L.

4 to 6 h: \(A_4=(5+2)(6-4)/2=7\) mg·h/L.

Therefore, \(AUC_{0-6}=2+6+13+7=28\) mg·h/L. The largest sampled concentration is 8 mg/L, observed at 2 h, so \(C_{\max}=8\) mg/L and \(T_{\max}=2\) h. This \(C_{\max}\) is the observed maximum; these samples do not establish that the true peak between sampling times was exactly 8 mg/L.

Partial AUC and Boundaries Between Samples

A partial AUC is the area over a specified time window \([a,b]\) within the observed profile. If a boundary is not a sampling time, first estimate its concentration from the adjacent observations using linear interpolation. For a boundary \(t^*\) between \(t_i\) and \(t_{i+1}\), the interpolated value is:

C(t^*)=C_i+\frac{C_{i+1}-C_i}{t_{i+1}-t_i}(t^*-t_i).

Then include the boundary points and all observed points strictly inside the window, in time order, and sum the trapezoid areas between consecutive points. This is equivalent to integrating the same piecewise-linear curve only from \(a\) to \(b\). The requested window must lie within the available observations for this calculation; estimating beyond the observed range requires an additional extrapolation model and is not the same operation.

Worked Example: Partial AUC With an Interpolated Endpoint

A hypothetical profile has times \(0, 1, 3, 4,\) and \(6\) h, and concentrations \(0, 2, 8, 6,\) and \(2\) mg/L, respectively. Calculate the linear-trapezoidal partial AUC from 1 to 5 h.

The start at 1 h is an observed sample with concentration 2 mg/L. The endpoint at 5 h lies between 4 and 6 h. Linear interpolation gives:

\(C(5)=6+\{(2-6)/(6-4)\}(5-4)=6-2=4\) mg/L.

The points within the requested window are now \((1,2), (3,8), (4,6), (5,4)\), with time in hours and concentration in mg/L. The component areas are:

1 to 3 h: \((2+8)(3-1)/2=10\) mg·h/L.

3 to 4 h: \((8+6)(4-3)/2=7\) mg·h/L.

4 to 5 h: \((6+4)(5-4)/2=5\) mg·h/L.

Thus, the partial AUC from 1 to 5 h is \(10+7+5=22\) mg·h/L. The concentration at 5 h is an interpolated value, not a measured sample, and the final area depends on the stated linear-interpolation assumption.

Observed Peak and a Partial-AUC Check

The observed \(C_{\max}\) is found by comparing the concentrations in the dataset, not by finding the largest trapezoid or by fitting an unrequested peak model. \(T_{\max}\) is the time associated with the observed maximum. If two or more samples share that maximum, report the tied observed times or apply a prespecified reporting convention; the data do not identify a unique sampled peak time in that case.

Worked Example: Tied Observed Peak and Partial Window

For a hypothetical profile, times are \(0, 0.5, 1.5, 3,\) and \(5\) h and concentrations are \(0, 6, 12, 12,\) and \(3\) mg/L. Calculate AUC from 0 to 5 h, the observed \(C_{\max}\) and \(T_{\max}\), and partial AUC from 0.75 to 4 h.

The full-profile interval areas are \((0+6)(0.5)/2=1.5\), \((6+12)(1)/2=9\), \((12+12)(1.5)/2=18\), and \((12+3)(2)/2=15\) mg·h/L. Their sum is \(AUC_{0-5}=43.5\) mg·h/L.

The maximum sampled concentration is 12 mg/L, occurring at both 1.5 and 3 h. Therefore \(C_{\max}=12\) mg/L and the observed \(T_{\max}\) values are 1.5 and 3 h. Reporting only one of these times would conceal the tie unless a reporting rule specified which to select.

For the partial window, interpolate at both boundaries. At 0.75 h, between 0.5 h (6 mg/L) and 1.5 h (12 mg/L), \(C(0.75)=6+(12-6)(0.25/1)=7.5\) mg/L. At 4 h, between 3 h (12 mg/L) and 5 h (3 mg/L), \(C(4)=12+(3-12)(1/2)=7.5\) mg/L.

The partial-window points are \((0.75,7.5), (1.5,12), (3,12), (4,7.5)\). Their areas are \((7.5+12)(0.75)/2=7.3125\), \((12+12)(1.5)/2=18\), and \((12+7.5)(1)/2=9.75\) mg·h/L. Consequently, the partial AUC is \(7.3125+18+9.75=35.0625\), or 35.06 mg·h/L to four significant figures.

Assumptions, Diagnostics, and Alternative Rules

These calculations require correctly ordered times, concentrations associated with the correct times, consistent concentration units, and a declared rule for the interval between samples. The linear trapezoidal rule additionally approximates each interval by a straight line. It does not assume that the whole pharmacokinetic profile is linear, nor does it prove that a straight line is a good approximation within every interval.

Before accepting a calculation, check that the last included sample is indeed the last quantifiable time for the stated endpoint, that every interval duration is positive, and that each interval area uses the matching pair of concentrations and times. Confirm that partial-AUC boundaries are inside the observed time range and that interpolation uses the two samples that bracket each boundary. A useful consistency check is that, under the same integration rule, a partial AUC within a full profile cannot exceed the total AUC over that profile when concentrations are nonnegative. The partial-window value should also agree with summing only the relevant subintervals of the full piecewise-linear curve.

Some analyses use a linear-up/log-down rule instead. It uses linear trapezoids for increasing concentrations and, for a decreasing interval with positive unequal endpoint concentrations, a logarithmic trapezoid:

A_i=(t_{i+1}-t_i)\frac{C_i-C_{i+1}}{\ln(C_i/C_{i+1})}.

This rule can represent a log-linear decline differently from a straight-line decline. It is undefined in this form when an endpoint is zero, and equal endpoint concentrations require the limiting value, which is the ordinary rectangle area. The integration method should be chosen and applied as specified for the analysis, rather than selected after seeing which result is preferred. The numerical examples here use linear trapezoids throughout.

A sparse schedule can miss a peak or conceal curvature between samples. Consequently, observed \(C_{\max}\) and \(T_{\max}\) depend on when samples were collected, and a trapezoidal AUC is also sensitive to the profile’s shape between samples. Interpolation gives a reproducible estimate under a stated rule; it does not recover unknown concentrations exactly. These summaries also do not by themselves establish bioequivalence or clinical similarity.

Common Mistakes

Using the concentration at one endpoint as the whole interval. That gives a rectangle, not the linear trapezoid. Use the mean of the two endpoint concentrations multiplied by elapsed time.
Using the wrong time width. Concentrations at 1 and 3 h are separated by 2 h, not 1 h. Subtract the times for every interval.
Treating an interpolated boundary value as an observed sample. Label it as interpolated and use the stated rule; do not imply that it was measured.
Calling the observed peak the true continuous-time peak. The maximum in the table is \(C_{\max}\) by the observed-data definition. A peak between sampling times may have been missed.
Mixing units or integration methods. Convert time units before combining interval areas, and do not combine linear and logarithmic trapezoids without a specified method.

Key takeaway: State the included time range and integration rule, calculate every interval with its own time width, interpolate partial-window boundaries when needed, and report units. \(C_{\max}\) and \(T_{\max}\) describe the sampled maximum, not necessarily the unobserved continuous profile’s peak.

Check Your Understanding

Try each calculation first. The solutions use the linear trapezoidal rule and give units explicitly.

A hypothetical profile has \((t,C)\) values \((0,0), (2,4), (5,4)\), where time is in hours and concentration in mg/L. Calculate AUC from 0 to 5 h.
For samples at 1 h with 3 mg/L and 3 h with 7 mg/L, find the linearly interpolated concentration at 2 h.
A sampled profile has concentrations 5, 9, and 9 mg/L at 1, 2, and 4 h. State observed \(C_{\max}\) and all observed \(T_{\max}\) values.
Explain why a partial AUC endpoint between samples is not an additional observed concentration.

Solutions. (1) From 0 to 2 h, area \(=(0+4)(2)/2=4\) mg·h/L. From 2 to 5 h, area \(=(4+4)(3)/2=12\) mg·h/L. Total AUC is \(4+12=16\) mg·h/L. (2) The interpolation fraction is \((2-1)/(3-1)=1/2\), so \(C(2)=3+(7-3)(1/2)=5\) mg/L. (3) The maximum sampled concentration is 9 mg/L at both 2 and 4 h; thus \(C_{\max}=9\) mg/L and the observed \(T_{\max}\) values are 2 and 4 h. (4) The boundary value is calculated from bracketing observations using an interpolation rule. It estimates the piecewise curve at that time but was not measured by the assay.