Cumulative AUC Versus Window-Specific AUC

Two AUC Questions That Sound Similar

An AUC reported from time zero answers how much concentration-time exposure has accumulated up to a specified time. A window-specific AUC answers how much exposure occurred between two specified times. Both have concentration-times-time units, but they describe different intervals. Confusing them can make an exposure result appear to refer to an earlier or later part of a profile than it actually does.

For example, a cumulative AUC from 0 to 8 hours includes exposure during the first 2 hours. A window-specific AUC from 2 to 8 hours does not. The distinction matters whenever an exposure measure is tied to a time window: the time boundaries determine which part of the concentration-time profile contributes to the result.

Definition: For a concentration-time profile \(C(t)\), the cumulative AUC from time zero to \(t\) is the area \(AUC_{0,t}=\int_0^t C(u)\,du\). The window-specific AUC from \(a\) to \(b\), where \(0\leq a<b\), is \(AUC_{a,b}=\int_a^b C(u)\,du\). AUC has units of concentration multiplied by time, such as \(\mathrm{mg\cdot h/L}\).

In concentration-time data, the integral is generally estimated numerically. The earlier material on Calculating Partial AUC With Trapezoids covers direct area calculation between window boundaries. Here the focus is how that window area relates to cumulative AUC values, and why the two calculations should agree when they use the same profile representation.

The Relationship Between Cumulative and Window-Specific AUC

The relationship follows by splitting the cumulative integral at the start of the window. For \(0\leq a<b\), the area from zero to \(b\) consists of the area from zero to \(a\), plus the area from \(a\) to \(b\):

AUC_{0,b} = \int_0^a C(u)\,du+\int_a^b C(u)\,du = AUC_{0,a}+AUC_{a,b}.

Rearranging gives the conversion:

AUC_{a,b}=AUC_{0,b}-AUC_{0,a}.

Key result: A window-specific AUC is the cumulative AUC at the window end minus the cumulative AUC at the window start. Both cumulative values must refer to the same concentration-time profile and be calculated consistently. In particular, \(AUC_{0,b}\) alone is not the AUC from \(a\) to \(b\).

This identity is exact for integrals of the same profile. In practice, the profile between observed concentrations is represented by an interpolation rule, and its area is estimated. The subtraction identity remains valid for the estimates when both cumulative AUC values are built from the same piecewise representation and the boundaries are handled consistently.

For instance, if the profile between observations is treated as linear, an interior boundary concentration should be obtained from that same linear segment. If the declining segment is represented log-linearly, boundary interpolation and area calculations should use that same log-linear representation. Splitting one estimated segment at a boundary does not change its total area: the two subareas add to the area of the original segment. If different rules or boundary values are used in the two calculations, however, subtraction may not match direct integration.

Worked Calculations

Worked Example: Subtracting Cumulative AUC Values

Suppose a hypothetical concentration-time profile, calculated using linear trapezoids, has concentrations of 0, 6, 4, and 1 mg/L at 0, 2, 5, and 8 hours, respectively. Find the window-specific AUC from 2 to 8 hours, both directly and by subtracting cumulative AUC values.

First calculate the trapezoid areas. For a segment with endpoint concentrations \(C_1\) and \(C_2\), and duration \(\Delta t\), the linear trapezoid area is \(\Delta t(C_1+C_2)/2\).

From 0 to 2 hours, the area is \(2(0+6)/2=6\ \mathrm{mg\cdot h/L}\). From 2 to 5 hours, it is \(3(6+4)/2=15\ \mathrm{mg\cdot h/L}\). From 5 to 8 hours, it is \(3(4+1)/2=7.5\ \mathrm{mg\cdot h/L}\).

Thus \(AUC_{0,2}=6\ \mathrm{mg\cdot h/L}\), and \(AUC_{0,8}=6+15+7.5=28.5\ \mathrm{mg\cdot h/L}\). Subtraction gives \(AUC_{2,8}=28.5-6=22.5\ \mathrm{mg\cdot h/L}\). Direct calculation over the window gives \(15+7.5=22.5\ \mathrm{mg\cdot h/L}\), as required.

The cumulative result through 8 hours includes the 0-to-2-hour area; subtracting the cumulative value at 2 hours removes that early exposure.

Worked Example: A Window Boundary Inside an Observed Segment

Consider hypothetical concentrations of 0, 8, and 4 mg/L at 0, 4, and 8 hours. Use linear interpolation to calculate the AUC from 1 to 6 hours. The requested boundaries are inside observed segments, so their concentrations must be estimated from the piecewise-linear profile.

Between 0 and 4 hours, concentration rises by 8 mg/L over 4 hours, or 2 mg/L per hour. Therefore \(C(1)=0+2(1)=2\ \mathrm{mg/L}\). Between 4 and 8 hours, concentration falls by 4 mg/L over 4 hours, or 1 mg/L per hour. Therefore \(C(6)=8-1(2)=6\ \mathrm{mg/L}\).

Direct calculation from 1 to 4 hours gives \(3(2+8)/2=15\ \mathrm{mg\cdot h/L}\). From 4 to 6 hours it gives \(2(8+6)/2=14\ \mathrm{mg\cdot h/L}\). The window area is \(15+14=29\ \mathrm{mg\cdot h/L}\).

For a cumulative check, \(AUC_{0,1}=1(0+2)/2=1\ \mathrm{mg\cdot h/L}\). The cumulative area through 6 hours is \(AUC_{0,6}=4(0+8)/2+2(8+6)/2=16+14=30\ \mathrm{mg\cdot h/L}\). Thus \(AUC_{1,6}=30-1=29\ \mathrm{mg\cdot h/L}\), matching the direct calculation.

Worked Example: Applying the Same Check to a Log-Linear Segment

Suppose hypothetical concentrations decline from 16 mg/L at 0 hours to 12 mg/L at 2 hours and 6 mg/L at 6 hours. Use log-linear trapezoids on these declining segments. Calculate the AUC from 2 to 6 hours and verify it using cumulative AUC values.

For positive endpoint concentrations on a log-linear segment, the area over duration \(\Delta t\) is \(\Delta t(C_1-C_2)/\ln(C_1/C_2)\). From 2 to 6 hours, the area is \(4(12-6)/\ln(12/6)=24/\ln(2)=34.62\ \mathrm{mg\cdot h/L}\).

From 0 to 2 hours, the area is \(2(16-12)/\ln(16/12)=8/\ln(4/3)=27.81\ \mathrm{mg\cdot h/L}\). Therefore \(AUC_{0,2}=27.81\ \mathrm{mg\cdot h/L}\), and \(AUC_{0,6}=27.81+34.62=62.43\ \mathrm{mg\cdot h/L}\). The difference is \(62.43-27.81=34.62\ \mathrm{mg\cdot h/L}\), agreeing with the direct window calculation.

The check works because both cumulative values and the direct window value use the same log-linear representation on the 2-to-6-hour segment. A zero or negative concentration cannot be inserted into this log-trapezoid formula; the interpolation or integration rule must be suitable for the data and chosen consistently.

Assumptions, Checks, and Failure Modes

The conversion is straightforward, but it depends on what the AUC values represent. Before applying it, verify that the cumulative values use the same concentration-time profile, time origin, concentration units, and area-estimation conventions. If either window boundary falls between observations, both approaches must use compatible interpolated concentrations there. Otherwise, the subtraction and direct calculation may refer to different estimated profiles.

A practical check is to calculate the window area in both ways when the cumulative values are available. The result should agree apart from small differences due to rounding. A substantial discrepancy is a signal to inspect the calculation, not a reason to choose whichever result seems more plausible. Common causes include:

Subtracting the wrong cumulative value. For a window from \(a\) to \(b\), subtract \(AUC_{0,a}\) from \(AUC_{0,b}\), not from \(AUC_{0,\text{end}}\) at some other time.
Mixing units or time scales. If one area uses hours and another uses minutes, subtraction is invalid until the units are made consistent.
Inconsistent boundary handling. A directly calculated window may use an interpolated boundary concentration that differs from the one underlying the cumulative values.
Using different integration rules. For example, a direct window calculation and cumulative subtraction can disagree if one calculation uses linear interpolation and the other uses log-linear interpolation on the same declining interval.
Rounding intermediate areas too early. Subtraction can magnify the visible effect of rounding, particularly when the window area is small relative to both cumulative values. Retain precision in intermediate calculations and round the reported result at the end.

For a nonnegative concentration profile and \(a<b\), a correctly calculated window area cannot be negative. A negative difference between cumulative AUC values is therefore a warning of reversed boundaries, inconsistent calculations, or a data-processing error. Small discrepancies can also arise when reported cumulative values have already been rounded.

Which Calculation Should Be Used?

Direct integration over the window makes its boundaries explicit. It is useful when only the target interval is needed or when the window boundaries require special interpolation. Cumulative subtraction is convenient when reliable cumulative AUC values at both boundaries are already available. With consistent methods, neither approach changes the underlying area; they are two calculations of the same quantity.

The methods are not interchangeable if their inputs or assumptions differ. In particular, subtracting AUC values calculated with different integration rules does not necessarily produce the area that would be obtained by directly integrating the window under either rule. A result should therefore be described with its actual time bounds and method, for example, “linear-trapezoidal \(AUC_{2,8}\) in \(\mathrm{mg\cdot h/L}\),” rather than simply “AUC.”

AUC has an additive structure, but a cumulative value and a window-specific value have distinct interpretations. Reporting the correct quantity helps keep the exposure interval clear and makes numerical checks possible. The key question is not only how large the AUC is, but also which times contributed to it.

Check Your Understanding

Answer each question, then compare with the solutions below.

If \(AUC_{0,10}=52\ \mathrm{mg\cdot h/L}\) and \(AUC_{0,3}=17\ \mathrm{mg\cdot h/L}\), what is \(AUC_{3,10}\)?
For a nonnegative profile, a calculation gives \(AUC_{2,6}=-1.4\ \mathrm{mg\cdot h/L}\). Name two features of the calculation you would check first.
Concentrations are 4 and 10 mg/L at 0 and 3 hours, respectively, with linear interpolation. Calculate the AUC from 1 to 3 hours.
Why might direct window integration fail to match cumulative-AUC subtraction even when both calculations use the same observed concentrations?

Solutions. (1) \(AUC_{3,10}=52-17=35\ \mathrm{mg\cdot h/L}\). (2) Check that the boundaries were subtracted in the correct order and that both cumulative values use compatible time units, profiles, and integration rules. A negative window area is not possible for a nonnegative profile. (3) The linear slope is \((10-4)/3=2\ \mathrm{mg/L}\) per hour, so \(C(1)=4+2(1)=6\ \mathrm{mg/L}\). The area from 1 to 3 hours is \(2(6+10)/2=16\ \mathrm{mg\cdot h/L}\). (4) The calculations may use different interpolation rules, boundary concentrations, or rounding. They agree when both represent the same profile and handle the boundaries consistently.