Why an Interior Gap Needs an Explicit Choice
A concentration-time profile is observed only at selected times. If one interior concentration is missing but the samples immediately before and after it are available, the surrounding observations still define an interval across the gap. An area can be calculated over that wider interval, but the calculation necessarily treats the unobserved part of the profile in some way. The resulting AUC is therefore an estimate under an interpolation assumption, not a recovered measurement.
This distinction matters because a single missing sample can affect more than one trapezoid. If the missing value were available, it would divide the interval into two subintervals. Without it, an analyst may bridge directly between the neighboring measured concentrations, or may estimate the missing concentration under a stated model and calculate the two smaller areas. These approaches agree only under particular assumptions.
Write the observed endpoint concentrations as \(C_a=C(t_a)\) and \(C_b=C(t_b)\), with \(t_a<t_i<t_b\). The gap duration is \(\Delta t=t_b-t_a\). Concentrations are assumed to refer to the same analyte and measurement scale, with time and concentration units stated consistently. The AUC has units of concentration multiplied by time, such as mg·h/L.
Linear Interpolation and Its Area
Linear interpolation assumes that concentration changes at a constant rate between the two observed endpoints. Under that assumption, the missing concentration at \(t_i\) is estimated along the straight line joining \((t_a,C_a)\) and \((t_b,C_b)\). This assumption is most natural when a straight concentration-time segment is scientifically plausible over the gap; it is not guaranteed merely because the gap contains one missing sample.
The linear-trapezoidal area across a pair of observations is the time interval multiplied by the average of the endpoint concentrations. Splitting the gap at \(t_i\) and using the interpolated value gives
The equality follows because the interpolated point lies on the same straight line as the endpoints. Thus the two smaller trapezoids have exactly the same combined area as one trapezoid drawn directly between the observed endpoints. Under linear interpolation, estimating the missing concentration is not necessary to calculate the gap AUC, although it may be useful for plotting or for describing the assumed profile.
Worked Example: A Missing Sample on a Rising Segment
Suppose concentrations are observed as 1.0 mg/L at 0 h and 7.0 mg/L at 3 h, but the scheduled sample at 1 h is missing. Assume, for this calculation, that concentration changes linearly across the entire gap. Estimate the missing concentration and the AUC from 0 to 3 h.
Interpolated concentration. The time fraction from 0 h to 1 h is \(1/3\), so
Area using the imputed value. The two intervals each have duration 1 h and 2 h, respectively:
Endpoint check. Direct bridging gives \(3(1.0+7.0)/2=12.0\ \mathrm{mg\cdot h/L}\), the same result. This is an arithmetic check, not evidence that the true unobserved profile was linear.
Log-Linear Interpolation for Positive Concentrations
A different assumption is that the logarithm of concentration changes linearly with time. Equivalently, concentration follows an exponential curve over the gap. This approach requires positive endpoint concentrations. It is commonly considered for a declining segment when exponential decline is a plausible local description; it should not be applied mechanically across a rising-and-falling segment or across a possible peak.
For an exponentially changing concentration \(C(t)=C_a\exp[-k(t-t_a)]\), where \(k\) is constant over the interval, direct integration gives an area equal to \((C_a-C_b)/k\). Since \(\log(C_a/C_b)=k\Delta t\), this is the log-trapezoidal formula. The formula therefore corresponds to an exponential segment, rather than a straight line on the concentration scale.
Worked Example: A Missing Sample on a Declining Segment
Suppose concentration is 6.0 mg/L at 8 h and 3.0 mg/L at 10 h, with the 9 h sample missing. Assume positive concentrations and a single exponential decline across the gap. Calculate the interpolated value and the log-trapezoidal AUC.
Interpolated concentration. Since 9 h is halfway between the endpoints, the log concentration is halfway between \(\log(6.0)\) and \(\log(3.0)\):
Area over the full gap. The log-trapezoidal estimate is
Check by splitting the gap. Using the unrounded interpolated concentration, the first subinterval contributes \(1(6.0-4.2426407)/\log(6.0/4.2426407)\approx5.070\ \mathrm{mg\cdot h/L}\). The second contributes \(1(4.2426407-3.0)/\log(4.2426407/3.0)\approx3.586\ \mathrm{mg\cdot h/L}\). Their sum is approximately \(8.656\ \mathrm{mg\cdot h/L}\), agreeing with the direct calculation. Rounding the intermediate concentration too early can slightly alter the split-area sum.
Why the Two Interpolations Can Disagree
For the same declining endpoints of 6.0 and 3.0 mg/L over 2 h, the linear-trapezoidal estimate is \(2(6.0+3.0)/2=9.000\ \mathrm{mg\cdot h/L}\), whereas the log-trapezoidal estimate is 8.656 mg·h/L. Linear interpolation assumes a straight line in concentration; log-linear interpolation assumes a straight line in log concentration. Neither method can determine which curve actually occurred between two observed samples.
The difference is not a computational error. For a declining exponential curve, concentration is convex upward on the original scale: its second derivative is positive. A straight chord between two points on a convex curve lies above the curve, so the linear trapezoid overestimates the area of that exponential segment. If concentration truly changes linearly, the linear trapezoid is exact instead. The direction and size of discrepancy depend on the curve shape and the endpoints; the two estimates should not be described as interchangeable.
Worked Example: Sensitivity to the Interpolation Choice
Use the same hypothetical concentrations, 6.0 mg/L at 8 h and 3.0 mg/L at 10 h, but compare the two standard assumptions directly.
Linear assumption. The endpoint trapezoid is \(2(6.0+3.0)/2=9.000\ \mathrm{mg\cdot h/L}\). At the midpoint, linear interpolation would give \((6.0+3.0)/2=4.5\ \mathrm{mg/L}\).
Log-linear assumption. The midpoint estimate is \(\sqrt{6.0\times3.0}=4.243\ \mathrm{mg/L}\), and the log-trapezoidal area is 8.656 mg·h/L as calculated above.
Difference. The linear estimate is \(9.000-8.656=0.344\ \mathrm{mg\cdot h/L}\) larger, or about \(0.344/8.656\times100=3.97\%\) of the log-linear estimate. This difference quantifies sensitivity to these two interpolation assumptions; it is not a confidence interval and does not identify the true area.
A comparison is informative when it is scientifically plausible to consider both shapes, especially if a gap is long or its endpoints differ substantially. It is less meaningful to calculate both methods and select whichever result is preferred. The model choice should be consistent with the segment’s direction, the surrounding profile, the measurement scale, and the prespecified analysis convention.
Assumptions, Diagnostics, and Failure Modes
Before calculating an area across a missing sample, check that the gap is genuinely interior and that both bounding samples are usable. A recorded value should not be silently treated as an observed concentration if it is invalid, below quantification, or otherwise handled under a separate data rule. Also confirm the scheduled time and units: a time error changes the interval width and therefore changes the area.
- Linear interpolation is a poor local description when the interval is known or suspected to contain a sharp bend, rapid absorption, a peak, or a steep change in elimination rate. A straight line may smooth over such behavior.
- Log-linear interpolation is not defined at zero or negative concentrations and is generally unsuitable across a segment that changes direction or may contain an unobserved maximum. Positive endpoints alone do not establish exponential behavior.
- A long gap is more consequential than a short gap because there is more time for unobserved curvature or a hidden peak. A small computed difference between interpolation methods does not rule out other shapes.
- Unexpected endpoint patterns require review. For example, a concentration that rises substantially after an apparent decline may reflect a real profile feature, a timing or assay issue, or both. The gap formula alone cannot diagnose the cause.
A missing interior value can also obscure the observed maximum concentration if the scheduled time could have coincided with a peak. Interpolation may produce a plausible concentration for area calculation, but that imputed value is not an observed maximum and should not be presented as one. Likewise, AUC calculated over a gap may be less sensitive to a narrow peak than a peak-based metric, but a gap can still alter the estimated area.
When the missing observation makes the local shape doubtful, appropriate responses include flagging the profile, reporting a sensitivity analysis using defensible alternatives, or applying a prespecified missing-data rule. If the concentration-time information is insufficient to justify an interpolation, the correct conclusion is that the gap AUC is uncertain, not that one formula has recovered the missing concentration.
An Interior Gap Is Not a Truncated Profile
An interior gap has an observed concentration on both sides, so a local interpolation can connect the endpoints. At the last observation, there is no later observed endpoint. If a scheduled final sample is not obtained, the measured profile ends at the last available time; the unobserved tail cannot be estimated by bridging between neighboring samples because there is no right-hand neighbor.
Worked Example: A Profile Ending Before the Planned Last Sample
Suppose hypothetical concentrations are 0 mg/L at 0 h, 8.0 mg/L at 2 h, and 4.0 mg/L at 4 h. A planned 6 h sample is not collected. Calculate the observed linear-trapezoidal AUC through 4 h and state what is not determined.
Observed area from 0 to 2 h: \(2(0+8.0)/2=8.0\ \mathrm{mg\cdot h/L}\).
Observed area from 2 to 4 h: \(2(8.0+4.0)/2=12.0\ \mathrm{mg\cdot h/L}\).
Total through the last observation: \(8.0+12.0=20.0\ \mathrm{mg\cdot h/L}\). This is the estimated AUC from 0 to 4 h under linear trapezoids. The area after 4 h is not calculated from these data: no 6 h concentration is available to define an endpoint for an additional trapezoid. Any estimate beyond 4 h would require a separately justified tail model and must not be mislabeled as an interior-gap interpolation.
The distinction is practical as well as mathematical. Bridging an interior gap estimates area between two measured concentrations; handling truncation requires assumptions about concentration after the final observation. Reporting the last-observation AUC and an extrapolated tail as if they were the same quantity obscures how much of the area is supported by observed data.
Common Mistake
A frequent error is to say, “One sample is missing, so I will insert the average of the two neighboring concentrations and proceed,” without checking the times or the intended interpolation scale. The arithmetic midpoint is correct only when the missing time is halfway between the endpoints and the assumption is linear concentration change. If the time is not halfway, linear interpolation must use the actual time fraction. If an exponential decline is the intended model, the midpoint concentration is the geometric, not arithmetic, mean. A separate mistake is to apply the same bridging logic at the final observation, where the later endpoint does not exist.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- Concentrations are 2.0 mg/L at 1 h and 5.0 mg/L at 5 h; the 2 h sample is missing. Under linear interpolation, what is the imputed concentration at 2 h and the gap AUC from 1 to 5 h?
- For positive concentrations declining from 8.0 mg/L to 2.0 mg/L over 3 h, what log-trapezoidal area results? State the model assumption.
- Why can a linear trapezoid overestimate area when the true segment is an exponential decline?
- A final planned sample is missing, but the previous two concentrations are available. Can those two values be used to bridge the unobserved terminal interval? Explain.
Solutions. (1) The linear time fraction is \((2-1)/(5-1)=1/4\), so \(\widehat C_2=2.0+(1/4)(5.0-2.0)=2.75\ \mathrm{mg/L}\). The gap AUC is \(4(2.0+5.0)/2=14.0\ \mathrm{mg\cdot h/L}\). (2) Assuming a single exponential decline, the area is \(3(8.0-2.0)/\log(8.0/2.0)=18/\log(4)=12.98\ \mathrm{mg\cdot h/L}\). (3) An exponential decline is convex upward on the concentration scale, so its endpoint chord lies above the curve and has greater area. (4) No. The final interval has no later observed endpoint, so it is truncation, not an interior gap; estimating area beyond the last observation requires a separate tail assumption.