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Exposure Metrics: AUC, Cmax and Partial AUC · Tutorial 31 of 399

Combining Nonoverlapping Partial-AUC Windows

You will learn when partial AUCs can be added directly, how gaps and overlaps change the result, and how to check a combined calculation.

Intermediate 10 min read

What You'll Learn

  • Define the AUC over a collection of adjacent, separated, or overlapping time windows.
  • Explain why shared boundary points do not add exposure twice.
  • Distinguish the sum over selected windows from AUC across the full time span.
  • Correct a sum of two overlapping-window AUCs using the AUC over their intersection.
  • Apply a consistent trapezoidal calculation to boundaries and component intervals.
  • Diagnose gaps, duplicated coverage, and inconsistent boundary calculations.

Why Combining Windows Requires a Coverage Check

A partial AUC describes exposure within a specified time window. When several windows are reported, adding their partial AUCs is meaningful only if we are clear about which times are covered, which are omitted, and whether any time is covered more than once. Adjacent windows can form one continuous interval; separated windows leave a gap; overlapping windows count some exposure twice if their AUCs are simply added.

These distinctions matter when a combined quantity is intended to represent exposure over selected periods rather than over the entire elapsed time. A sum across separated windows can be correct for the selected periods while being smaller than AUC across the full span. Conversely, a sum across overlapping windows can exceed AUC over their union. The comparison concerns time coverage, not a change in the concentration profile.

Definition: For a concentration-time profile \(C(t)\), the AUC over a window \(W=[a,b]\), where \(a\leq b\), is \(\int_a^b C(t)\,dt\). For several windows \(W_1,\ldots,W_k\), the sum of window-specific AUCs is \(\sum_{j=1}^k\int_{W_j}C(t)\,dt\). The AUC over their union is \(\int_{\cup_j W_j}C(t)\,dt\), which counts each time in the union once.

The integral definition assumes that \(C(t)\) is integrable over the finite windows. In pharmacokinetic applications the concentration profile is ordinarily nonnegative, and AUC has units of concentration multiplied by time, such as mg·h/L. An endpoint shared by two adjacent windows has zero duration, so that single point contributes no additional area to the integral. In a numerical calculation, however, one must still avoid accidentally treating a finite interval as if it were only a shared endpoint.

Adjacent Windows and Separated Windows

Suppose \(W_1=[a,b]\) and \(W_2=[b,c]\), with \(a\leq b\leq c\). They meet at \(b\) but have no overlap of positive duration. Additivity of the integral gives

$$ \int_a^b C(t)\,dt+\int_b^c C(t)\,dt=\int_a^c C(t)\,dt. $$

Thus, adjacent windows can be added to obtain AUC across the combined continuous interval. The same principle applies to a collection of windows whose interiors do not overlap. If those windows are separated, their sum gives AUC over the union of the selected windows, not automatically AUC across the span from the earliest start to the latest end.

The gap distinction can be written directly. For \(a<b<c<d\), the selected windows \([a,b]\) and \([c,d]\) omit \((b,c)\). Therefore

$$ \int_a^b C(t)\,dt+\int_c^d C(t)\,dt = \int_a^d C(t)\,dt-\int_b^c C(t)\,dt. $$

The right-hand side shows that the selected-window sum equals the AUC across the full span only when the omitted gap has zero AUC. A gap with positive duration does not, by itself, prove that its AUC is positive; the concentration could be zero throughout that gap. In ordinary profiles with positive concentrations over the gap, the selected-window sum is less than the full-span AUC.

Worked Example: Adding Adjacent Windows

Suppose a hypothetical concentration profile has measured concentrations of 0 mg/L at 0 h, 8 mg/L at 2 h, and 2 mg/L at 5 h. Use the linear trapezoidal rule to calculate AUC over the adjacent windows [0, 2] h and [2, 5] h, then check the result against the combined interval. Assume the same linear interpolation is used throughout.

For a segment from \(t_i\) to \(t_{i+1}\), the linear trapezoidal area is \((t_{i+1}-t_i)(C_i+C_{i+1})/2\). For [0, 2] h:

$$ \frac{2-0}{2}(0+8)=8\ \text{mg·h/L}. $$

For [2, 5] h:

$$ \frac{5-2}{2}(8+2)=15\ \text{mg·h/L}. $$

The sum is \(8+15=23\) mg·h/L. Calculating directly across the combined interval gives the same two trapezoids, \(8+15=23\) mg·h/L. The windows share the concentration at 2 h, but no area is added twice: both calculations include the segment on each side of that time exactly once.

Worked Example: Separated Windows and an Omitted Gap

Consider a hypothetical profile measured at 0, 1, 3, and 4 h, with concentrations 2, 6, 4, and 2 mg/L, respectively. Using linear trapezoids, calculate the sum for windows [0, 1] h and [3, 4] h, and compare it with the AUC from 0 to 4 h. For this calculation, use linear interpolation between the listed observations, including across the gap.

The first selected window has area

$$ \frac{1-0}{2}(2+6)=4\ \text{mg·h/L}. $$

The second selected window has area

$$ \frac{4-3}{2}(4+2)=3\ \text{mg·h/L}. $$

The sum for the selected windows is \(4+3=7\) mg·h/L. The gap [1, 3] h has estimated area

$$ \frac{3-1}{2}(6+4)=10\ \text{mg·h/L}. $$

Therefore AUC from 0 to 4 h is \(4+10+3=17\) mg·h/L. The selected-window sum, 7 mg·h/L, is not the full-span AUC: it excludes the 10 mg·h/L assigned to the gap. This conclusion depends on the specified trapezoidal interpolation for that gap; the observed endpoint concentrations alone do not determine the exact continuous profile between them.

Overlapping Windows: Count the Intersection Once

For two windows \(A\) and \(B\) that overlap, adding their individual AUCs counts the overlap in both windows. The general identity is

$$ \int_A C(t)\,dt+\int_B C(t)\,dt = \int_{A\cup B}C(t)\,dt+\int_{A\cap B}C(t)\,dt. $$

One justification is to consider each time \(t\). If it belongs to neither window, both sides contribute zero. If it belongs to exactly one window, both sides count \(C(t)\) once. If it belongs to both, the left side counts \(C(t)\) twice, while the right side counts it once in the union and once in the intersection. Integrating these equal contributions over time proves the identity. Rearranging gives

$$ \int_{A\cup B}C(t)\,dt = \int_A C(t)\,dt+\int_B C(t)\,dt-\int_{A\cap B}C(t)\,dt. $$

For a nonnegative concentration profile, the intersection AUC is nonnegative. It is generally necessary to subtract the intersection AUC once to obtain the union AUC. However, positive overlap duration alone does not guarantee a positive intersection AUC: if concentration is zero almost everywhere in the overlap, that integral is zero, and the sum already equals the union AUC. More generally, the sum equals the union AUC exactly when the intersection AUC is zero.

Worked Example: Correcting a Sum for an Overlap

Suppose a hypothetical profile has concentrations 2, 4, 2, and 0 mg/L at 0, 1, 2, and 3 h. Use linear interpolation between these points. Calculate AUC for windows [0, 2] h and [1, 3] h, and then calculate AUC over their union.

For [0, 2] h, the two trapezoids have areas

$$ \frac{1}{2}(2+4)=3,\qquad \frac{1}{2}(4+2)=3, $$

so the first window AUC is \(3+3=6\) mg·h/L. For [1, 3] h, the trapezoids have areas

$$ \frac{1}{2}(4+2)=3,\qquad \frac{1}{2}(2+0)=1, $$

so the second window AUC is \(3+1=4\) mg·h/L. Their sum is 10 mg·h/L. The intersection is [1, 2] h, with AUC \(\frac{1}{2}(4+2)=3\) mg·h/L. Subtracting that intersection once gives

$$ \text{AUC over }[0,3]=6+4-3=7\ \text{mg·h/L}. $$

A direct check over [0, 3] h gives \(3+3+1=7\) mg·h/L. The values agree, and the overlap has been counted exactly once in the union result.

Practical Calculation and Checks

When AUCs are calculated from sampled concentrations, the accounting principle applies to the curve represented by the numerical method. If a window boundary falls between sampling times, use the specified boundary-interpolation method before calculating the window AUC. For a combined calculation, use consistent boundary concentrations and a consistent integration rule on the shared profile. If a sampled interval must be split at a window boundary, calculate the pieces so that each part of the interval is represented once in the selected union.

For more than two windows, define \(m(t)\) as the number of windows containing time \(t\). The sum of the individual AUCs weights concentration by this coverage count:

$$ \sum_{j=1}^k\int_{W_j}C(t)\,dt = \int_{\cup_j W_j}m(t)C(t)\,dt. $$

The union AUC instead counts each time in the union once. This makes a useful audit possible: list the windows in chronological order, mark every start and end, and partition the time axis into pieces on which coverage is constant. A piece covered twice contributes twice to the uncorrected sum; a piece covered zero times inside the overall span is a gap.

1
Write every window explicitly.
Record its start and end on the same time scale, with units, before adding any AUC values.
2
Check coverage.
Determine whether the windows have disjoint interiors, are separated by gaps, or overlap over intervals of positive duration.
3
Choose the target quantity.
Decide whether the requested result is the sum over selected windows, AUC over their union, or AUC over the full span including gaps.
4
Integrate each needed piece once.
Use consistent concentrations at boundaries and a consistent numerical rule; subtract the intersection AUC when correcting a two-window overlap.

A common mistake is to assume that adding two partial AUCs always gives the AUC from the earliest start to the latest end. That is wrong if the windows are separated, because the gap is omitted, and it is wrong if they overlap, because the overlap is duplicated. The correction is to state the target first: for separated windows, add the gap only if the target includes it; for overlapping windows, subtract the intersection once if the target is the union.

An alternative to summing precomputed window AUCs is to partition the requested union into nonoverlapping pieces and integrate each piece once. This is often the clearest method when there are several overlaps or when boundaries cut through sampled intervals. Inclusion-exclusion is convenient for two windows; direct partitioning is usually easier to audit for more complicated collections. Neither approach removes numerical integration error: both depend on the available observations, boundary interpolation, and chosen trapezoidal rule. The arithmetic can be correct even when the profile is sparsely sampled or the numerical approximation is imperfect.

Keep the interpretation narrow. A sum over disjoint windows is the exposure accumulated in those windows under the chosen integration method. It does not imply that the intervening exposure is zero, and it is not the full-span AUC unless the gaps have zero AUC. Likewise, a union AUC from overlapping windows is not the uncorrected sum unless the intersection AUC is zero. Reporting the windows, units, and whether gaps or overlaps were handled makes the quantity interpretable and reproducible.

Check Your Understanding

For each question, identify the coverage issue and calculate or state the appropriate AUC relationship. Solutions are provided below.

  1. Two adjacent windows are [0, 2] h and [2, 5] h. Their AUCs are 8 and 15 mg·h/L. What is the AUC over the combined interval?
  2. Windows [0, 1] h and [3, 4] h have AUCs of 4 and 3 mg·h/L. The gap AUC is 10 mg·h/L. What is the sum over selected windows, and what is AUC across [0, 4] h?
  3. Two windows overlap. Their AUCs are 6 and 4 mg·h/L, and the intersection AUC is 3 mg·h/L. What is the AUC over their union?
  4. Two windows overlap for one hour, but concentration is zero throughout that overlap. Must the sum of their AUCs exceed their union AUC? Explain.

Solutions. (1) The windows are adjacent, so the combined AUC is \(8+15=23\) mg·h/L; the shared endpoint adds no area twice. (2) The selected-window sum is \(4+3=7\) mg·h/L. Including the gap gives \(7+10=17\) mg·h/L across [0, 4] h. (3) Subtract the overlap once: \(6+4-3=7\) mg·h/L over the union. (4) No. The sum equals the union AUC plus the intersection AUC, and the intersection AUC is zero, so the sum and union AUC are equal despite the positive overlap duration.