What the Observed AUC Covers
An AUC value is meaningful only when its time interval is clear. The observed AUC from zero to the last quantifiable time summarizes the concentration-time profile from the first dose time through the final sample with a quantifiable concentration. It does not, by itself, estimate the area after that sample. Keeping that boundary explicit prevents an observed exposure measure from being confused with an estimate that includes a modeled terminal tail.
Here, time zero means the time of the first dose in a single-dose profile. The last quantifiable time, \(t_{\mathrm{last}}\), is the latest scheduled or collected post-dose time at which concentration is quantifiable. The area is calculated from the time-concentration observations within the interval \([0,t_{\mathrm{last}}]\), using trapezoidal calculations between adjacent observations.
The units are concentration multiplied by time—for example, \(\mathrm{mg\cdot h/L}\) when concentration is in \(\mathrm{mg/L}\) and time is in hours. AUC is a summary of exposure over the stated interval, not a concentration at one time. As emphasized in earlier material on AUC as a numerical integral, the calculated area approximates the area under the underlying concentration-time curve; its accuracy depends in part on the sampling schedule and the interpolation rule.
Starting at the Dose Time
For a single-dose study in which the participant is drug-free at dosing, the concentration at the first dose time is zero. If the first post-dose sample is taken at time \(t_1\), the initial interval from \(0\) to \(t_1\) must be included. Under the usual linear interpolation convention, its area is the trapezoid with endpoints \(C(0)=0\) and \(C(t_1)\). Omitting this interval would instead calculate area from the first sample, not from dose time.
This zero-concentration convention is appropriate only when the profile begins with no drug in the body at the defined dose time. If a predose concentration is present—for example, because of a prior dose in a multiple-dose profile—one must use the concentration and interval definition appropriate to that design. Setting a nonzero predose concentration to zero without justification changes the area. Likewise, if the study has no sample at time zero, the dose time and the first observation time must still be distinguished: the interval from zero to the first observation requires a defensible starting concentration and interpolation rule.
Summing the Interval Areas
Let the observations within the interval be at times \(t_0,t_1,\ldots,t_m\), with concentrations \(C_0,C_1,\ldots,C_m\), where \(t_0=0\), \(t_m=t_{\mathrm{last}}\), and \(C_i=C(t_i)\). For an interval with increasing or equal positive concentrations, use the linear trapezoid. For a declining interval with two positive concentrations, the log trapezoid represents exponential change between those endpoints. These are the conventions developed in the earlier material on switching between linear and log trapezoids. A declining interval ending at zero cannot use a log trapezoid; the logarithm is undefined, so the linear trapezoid is used.
The total observed area is the sum of the interval areas:
This sum follows from the additivity of integrals over adjacent, nonoverlapping intervals. Each \(A_i\) estimates the area between two consecutive observations, and the intervals together cover the full range from zero to \(t_{\mathrm{last}}\), with no gaps or overlap. The sum does not assert that the concentration follows the chosen interpolation rule exactly between samples; it is the numerical estimate defined by that rule.
Worked Example: A Profile With Rising and Declining Segments
Consider hypothetical single-dose concentrations measured in \(\mathrm{mg/L}\) at the following times in hours: \(t=0,1,2,4,6\), with concentrations \(0,0.80,1.40,0.70,0.25\). The participant is drug-free at dose time, so \(C(0)=0\). Calculate \(AUC_{0-t_{\mathrm{last}}}\), where \(t_{\mathrm{last}}=6\) h.
From 0 to 1 h, concentration rises from zero to \(0.80\), so use the linear trapezoid:
Initial interval: \(A_{0-1}=(1-0)(0+0.80)/2=0.400\ \mathrm{mg\cdot h/L}\).
From 1 to 2 h, concentration rises from \(0.80\) to \(1.40\), again using the linear trapezoid:
Rising interval: \(A_{1-2}=(2-1)(0.80+1.40)/2=1.10\ \mathrm{mg\cdot h/L}\).
From 2 to 4 h, concentration declines from \(1.40\) to \(0.70\), both positive, so use the log trapezoid:
First declining interval: \(A_{2-4}=2(1.40-0.70)/\ln(1.40/0.70)=1.40/\ln(2)\approx2.020\ \mathrm{mg\cdot h/L}\).
From 4 to 6 h, concentration declines from \(0.70\) to \(0.25\):
Second declining interval: \(A_{4-6}=2(0.70-0.25)/\ln(0.70/0.25)=0.900/\ln(2.8)\approx0.8741\ \mathrm{mg\cdot h/L}\).
Therefore, \(AUC_{0-6}=0.400+1.10+2.020+0.8741=4.394\ \mathrm{mg\cdot h/L}\), to four significant figures. This estimate covers 0 through 6 h; it makes no estimate of exposure after 6 h.
Finding the Last Quantifiable Time
In practice, one or more late samples may be reported as below the lower limit of quantification (BLQ). A BLQ result is not a measured concentration of zero. It indicates that the concentration could not be quantified by the assay under its reporting rules. For \(AUC_{0-t_{\mathrm{last}}}\), identify the last quantifiable concentration and end the observed-area calculation there. Do not append a trapezoid to a later BLQ result as though its concentration were known to be zero.
This rule makes \(t_{\mathrm{last}}\) a property of the observed, quantifiable profile, not necessarily the time of the final sample collected. Study protocols and analysis plans may specify how BLQ observations are handled for other purposes. Such handling does not change the meaning of this particular observed-interval AUC: it ends at the last quantifiable time.
Worked Example: Ending at the Last Quantifiable Sample
A hypothetical profile has samples at 0, 1, 3, 5, and 8 h. The concentrations are \(0,2.00,1.00,0.40\ \mathrm{mg/L}\), followed by a BLQ result at 8 h. Find the observed AUC from dose time to the last quantifiable time.
The 5 h concentration is quantifiable and the 8 h result is BLQ, so \(t_{\mathrm{last}}=5\) h. The calculation includes the observations through 5 h only.
From 0 to 1 h, use a linear trapezoid: \(A_{0-1}=1(0+2.00)/2=1.00\ \mathrm{mg\cdot h/L}\).
From 1 to 3 h, use a log trapezoid: \(A_{1-3}=2(2.00-1.00)/\ln(2.00/1.00)=2/\ln(2)\approx2.885\ \mathrm{mg\cdot h/L}\).
From 3 to 5 h, use a log trapezoid: \(A_{3-5}=2(1.00-0.40)/\ln(1.00/0.40)=1.20/\ln(2.5)\approx1.310\ \mathrm{mg\cdot h/L}\).
Thus \(AUC_{0-5}=1.00+2.885+1.310=5.195\ \mathrm{mg\cdot h/L}\). The BLQ result at 8 h is not treated as zero and does not contribute a 5-to-8 h trapezoid. The calculated value is the observed area through 5 h, not the area through 8 h or through infinity.
Observed Area Is Not Terminal Extrapolation
The distinction between observed area and extrapolated area is substantive, not just a difference in notation. \(AUC_{0-t_{\mathrm{last}}}\) uses observed concentrations through \(t_{\mathrm{last}}\) and the specified interpolation between observations. An estimate extended beyond \(t_{\mathrm{last}}\) requires an additional model for the unobserved terminal decline. That additional calculation is not part of the observed area defined here. The later material on extrapolating AUC to infinity addresses how such a tail is estimated.
Consequently, two profiles can have the same \(AUC_{0-t_{\mathrm{last}}}\) but different unobserved tails, or the same terminal sample and different areas earlier in time. The observed AUC should be reported with its endpoint and units, rather than described simply as “total exposure” when the post-sampling tail has not been included.
Worked Example: Clock Time, Elapsed Time, and the Area Boundary
Suppose a hypothetical single dose is administered at 14:00. The first post-dose sample is collected at 16:00 with concentration \(1.00\ \mathrm{mg/L}\), followed by samples at 18:00 with \(0.50\ \mathrm{mg/L}\) and 21:00 with \(0.25\ \mathrm{mg/L}\). Assume the participant was drug-free at 14:00. Calculate the observed area through the final quantifiable sample.
Expressing the times as elapsed hours after dosing gives observations \((0,0),(2,1.00),(4,0.50),(7,0.25)\). From 0 to 2 h, use a linear trapezoid: \(A_{0-2}=2(0+1.00)/2=1.00\ \mathrm{mg\cdot h/L}\).
From 2 to 4 h, use a log trapezoid: \(A_{2-4}=2(1.00-0.50)/\ln(1.00/0.50)=1/\ln(2)\approx1.443\ \mathrm{mg\cdot h/L}\).
From 4 to 7 h, use a log trapezoid: \(A_{4-7}=3(0.50-0.25)/\ln(0.50/0.25)=0.75/\ln(2)\approx1.082\ \mathrm{mg\cdot h/L}\).
The sum is \(AUC_{0-7}=1.00+1.443+1.082=3.525\ \mathrm{mg\cdot h/L}\). Using clock times instead would give the same interval lengths and area if the dose time were correctly used as the origin. The result ends at 21:00, the last quantifiable sample; it does not include an estimated area after 21:00.
Assumptions, Diagnostics, and Common Pitfalls
The calculation requires a defined dose-time origin, correctly paired sampling times and concentrations, and an identified last quantifiable time. The endpoint concentration at time zero must also be appropriate to the dosing design. For each interval, the interpolation rule must match the observed direction and values: a log trapezoid requires two positive concentrations and a declining segment; a linear trapezoid accommodates rising or equal values and a zero endpoint.
Before summing, check that times are in increasing order, that time units are consistent, and that each concentration is attached to the correct sampling time. Confirm that the first area starts at dose time rather than at the first post-dose sample. Check the endpoint against the BLQ pattern and verify that no interval after the last quantifiable sample has been included. Finally, confirm that the output units are concentration multiplied by time.
Another error is to omit the dose-time-to-first-sample interval. That omission changes the estimand: the resulting sum begins at the first observation, not at \(t=0\). A different mistake is to use a log trapezoid for a rising segment or one with a zero endpoint. The logarithmic formula is not appropriate in either case. The linear trapezoid is used for rising or equal endpoints and whenever a declining interval ends at zero.
The principal alternative to the hybrid trapezoidal convention is to use linear trapezoids throughout. This is straightforward and avoids a log calculation, but for a positive declining segment it assumes linear change between endpoints and can give a different area from exponential interpolation. The hybrid convention uses the log trapezoid for positive declines and the linear trapezoid for rises, ties, or a zero endpoint. Neither calculation reveals the true between-sample curve; sparse sampling, a missed change in profile shape, or inaccurate measurements can affect the estimate. The observed AUC also cannot describe unobserved exposure after its endpoint.
Use the first dose time as zero and find the latest time with a quantifiable concentration.
Use a justified concentration at dose time; for a drug-free single-dose profile, this is zero. Do not substitute zero for a later BLQ result.
Apply the linear trapezoid to rises, ties, and zero endpoints; apply the log trapezoid to positive declines.
Add the interval areas, state \(t_{\mathrm{last}}\), and report concentration-time units without implying that a tail was included.
Check Your Understanding
Try each question, then compare your reasoning with the solutions.
- A drug-free participant has hypothetical concentrations \(0,1.00,0.40\ \mathrm{mg/L}\) at 0, 2, and 5 h. Calculate \(AUC_{0-5}\) using the stated interval rules.
- Concentrations at 0, 1, 3, and 4 h are \(0,0.50,0.50,0.20\ \mathrm{mg/L}\). Calculate the observed area through 4 h.
- Samples are collected at 0, 2, 4, and 6 h; the 4 h concentration is quantifiable and the 6 h result is BLQ. Which time is \(t_{\mathrm{last}}\), and should a 4-to-6 h trapezoid be included?
- A report gives an observed AUC through 12 h but no terminal extrapolation. Does that value include exposure after 12 h? Explain.
Solutions. (1) From 0 to 2 h, \(A=2(0+1.00)/2=1.00\). From 2 to 5 h, \(A=3(1.00-0.40)/\ln(1.00/0.40)=1.80/\ln(2.5)\approx1.964\). The total is approximately \(2.964\ \mathrm{mg\cdot h/L}\). (2) The 0-to-1 h linear area is \(1(0+0.50)/2=0.250\). The 1-to-3 h equal-concentration area is \(2(0.50+0.50)/2=1.00\). The 3-to-4 h log area is \(1(0.50-0.20)/\ln(2.5)\approx0.3274\). The total is \(1.577\ \mathrm{mg\cdot h/L}\). (3) \(t_{\mathrm{last}}=4\) h; no 4-to-6 h trapezoid is included because the later result is BLQ, not a quantifiable endpoint concentration. (4) No. The observed area ends at 12 h; exposure after 12 h is excluded unless a separate extrapolation is performed and reported.