From a Concentration–Time Profile to a Window-Specific Area
A partial area under the concentration–time curve summarizes exposure over a defined interval, rather than over the entire observed profile. The interval might be chosen to focus on an early or late portion of a profile. Calculating its area requires using only the portions of the concentration–time curve inside the specified window, including the parts between the window boundaries and the nearest sampled times.
The calculation is a numerical approximation. It depends on the recorded concentrations, the sampling times, and the interpolation rule used between observations. A reliable calculation first places the window boundaries on the profile, then calculates an area for every segment inside the window, and finally sums those areas.
The window must be specified before the calculation: its lower boundary must be earlier than its upper boundary, and both boundaries must fall within the time range being used. A boundary can coincide with a sampled time, or lie between two sampled times. In the latter case, the concentration at that boundary must be interpolated before calculating the segment areas. The earlier material on interpolating concentrations at partial-AUC boundaries gives the boundary-value formulas; here those values are used as endpoints in the area calculation.
Segment Areas and Their Assumptions
For a linear segment between concentrations \(C_1\) and \(C_2\), measured over a time interval of length \(\Delta t\), the area is the area of a trapezoid. This is the natural choice for a rising segment under the linear-up/log-down convention. Equal positive concentrations also give a rectangle, which is the equal-endpoint case of the linear trapezoid.
For a declining segment with positive endpoints, a log trapezoid represents concentration as changing exponentially between the endpoints. This is the log-down part of the linear-up/log-down convention. It requires both endpoint concentrations to be positive.
These expressions rely on different shapes between measured points: a straight line for the linear trapezoid and a straight line in log concentration for the log trapezoid. They are numerical interpolation assumptions, not claims that the true concentration curve is known exactly between samples. If an endpoint is zero, the log expression is undefined; use a linear trapezoid for that segment under this convention. Use the actual time difference for each segment, even when sampling intervals are unequal.
To assemble the estimate, list the window boundary times and every sampled time strictly inside the window in chronological order. Associate each time with its concentration, interpolating a boundary value if needed. Calculate the area between each adjacent pair using the applicable segment rule, then add those areas. Every interval in the sum must lie wholly inside the specified window.
Record \(t_L\), \(t_U\), and all sampled times between them.
Use observed values at sampled boundaries and interpolate values at unsampled boundaries.
Use linear trapezoids for rising or equal-concentration segments, and log trapezoids for positive declining segments under the stated convention.
Add the segment areas and confirm that time, concentration, and area units are consistent.
Worked Calculations
Worked Example: A Window With Two Interpolated Boundaries
Suppose a hypothetical profile has concentrations of 0, 3, 6, 3, and 1.5 mg/L at 0, 2, 4, 6, and 9 hours, respectively. Estimate \(AUC_{[1,8]}\) using linear-up/log-down trapezoids. The window boundaries are both between sampled times, and the observations at 2, 4, and 6 hours fall inside the window.
At 1 hour, the profile is rising linearly from 0 mg/L at 0 hours to 3 mg/L at 2 hours. Linear interpolation gives \(C(1)=0+(1/2)(3-0)=1.5\) mg/L. At 8 hours, the profile is declining from 3 mg/L at 6 hours to 1.5 mg/L at 9 hours. Log-linear interpolation gives \(C(8)=3(1.5/3)^{(8-6)/(9-6)}=3(0.5)^{2/3}=1.8899\) mg/L.
The ordered points within the window are therefore \((1,1.5)\), \((2,3)\), \((4,6)\), \((6,3)\), and \((8,1.8899)\), with time in hours and concentration in mg/L. The first segment is rising, so its linear area is \(A_1=(2-1)(1.5+3)/2=2.250\) mg·h/L.
The second segment is also rising: \(A_2=(4-2)(3+6)/2=9.000\) mg·h/L. The third and fourth segments are positive declines, so both use log trapezoids: \(A_3=(6-4)(6-3)/\ln(6/3)=6/\ln 2=8.656\) mg·h/L, and \(A_4=(8-6)(3-1.8899)/\ln(3/1.8899)\approx4.805\) mg·h/L.
The estimate is the sum \(2.250+9.000+8.656+4.805=24.71\) mg·h/L. The interpolated endpoints define the edges of the window; the areas outside 1 to 8 hours are not included.
Worked Example: A Decline to Zero
Consider a hypothetical profile with concentrations of 0, 2, 2, and 0 mg/L at 0, 1, 3, and 5 hours. Estimate the partial AUC from 0.5 to 4 hours using linear-up/log-down where applicable.
The first segment, from 0 to 1 hour, rises linearly from 0 to 2 mg/L. Interpolation gives \(C(0.5)=1\) mg/L. The segment from 3 to 5 hours declines to zero, so a log trapezoid cannot be used for that interval. Under the stated convention, use linear interpolation, giving \(C(4)=1\) mg/L.
The window points are \((0.5,1)\), \((1,2)\), \((3,2)\), and \((4,1)\). The first segment rises, the second has equal concentrations, and the final segment is a decline to zero; all three use linear trapezoids. Their areas are \(A_1=(1-0.5)(1+2)/2=0.750\) mg·h/L, \(A_2=(3-1)(2+2)/2=4.000\) mg·h/L, and \(A_3=(4-3)(2+1)/2=1.500\) mg·h/L.
Thus \(AUC_{[0.5,4]}=0.750+4.000+1.500=6.250\) mg·h/L. The zero endpoint does not invalidate the whole partial AUC; it only rules out the logarithmic formula for the segment that includes that endpoint.
Worked Example: Interpolating Both Ends of a Declining Window
Suppose a hypothetical profile declines from 4 mg/L at 2 hours to 1 mg/L at 5 hours. Estimate the area from 2.5 to 4.5 hours, assuming log-linear interpolation across this positive declining interval.
At 2.5 hours, the elapsed fraction of the 3-hour interval is \(0.5/3=1/6\). At 4.5 hours, it is \(2.5/3=5/6\). Interpolation on the log scale gives \(C(2.5)=4(1/4)^{1/6}=3.1748\) mg/L and \(C(4.5)=4(1/4)^{5/6}=1.2599\) mg/L.
The window duration is 2 hours. Its log-trapezoid area is \(AUC_{[2.5,4.5]}=2(3.1748-1.2599)/\ln(3.1748/1.2599)\approx4.144\) mg·h/L.
Both interpolated concentrations are positive, and the endpoint ordering confirms a decline, so the log trapezoid is applicable. The calculation estimates only the area between 2.5 and 4.5 hours, not the area over the full 2-to-5-hour interval.
Checks, Alternatives, and Common Errors
Before accepting a result, check the time ordering and the segment classifications. Each segment should have a positive duration, the boundary concentrations should fall between the adjacent observed concentrations under the chosen interpolation rule, and the area should be nonnegative for nonnegative concentrations. A segment’s duration must be the difference between its own endpoint times, not an assumed common sampling interval. A final units check is also useful: multiplying mg/L by hours gives mg·h/L.
The linear trapezoid and log trapezoid are alternatives for representing the profile between observations, not interchangeable arithmetic shortcuts. For a positive declining segment, the log trapezoid generally gives a different area from the linear trapezoid because it assumes a different shape between endpoints. When concentrations are equal, the linear trapezoid gives the rectangle area; the log expression has a zero-over-zero form and should not be evaluated directly. When a segment includes zero, the log expression is unavailable. Neither rule can recover unobserved changes in the true profile, so sparse sampling or a sharply changing profile can make the estimate sensitive to interpolation.
| Segment situation | Area calculation under linear-up/log-down | Check |
|---|---|---|
| Rising concentrations | Linear trapezoid | Use the time difference for that segment |
| Equal concentrations | Linear trapezoid, equivalent to a rectangle | Area is concentration multiplied by duration |
| Declining, positive concentrations | Log trapezoid | Both endpoints must be greater than zero |
| Decline with an endpoint at zero | Linear trapezoid for that segment | The logarithmic expression is undefined |
Common mistake: treating every positive decline as a linear trapezoid after specifying linear-up/log-down. For example, a decline from 4 to 2 mg/L over 3 hours has a linear area of \(3(4+2)/2=9.000\) mg·h/L, but its log-trapezoid area is \(3(4-2)/\ln(4/2)=8.656\) mg·h/L. The correction is to apply the stated rule segment by segment and use the logarithmic formula for positive declines. Mixing rules without stating or justifying that choice changes the estimate.
Check Your Understanding
Try each question, then compare your reasoning with the solutions.
- A positive concentration declines from 6 to 3 mg/L over 2 hours. What is its log-trapezoid area?
- Concentrations are 1 mg/L at 0 hours and 5 mg/L at 4 hours. Under linear interpolation, what concentration is assigned at 1 hour, and what is the area from 0 to 1 hour?
- Why should a log trapezoid not be used for a segment with one endpoint equal to zero?
- A profile has equal concentrations of 2 mg/L at 3 and 5 hours. What is the segment area, and which rule gives it?
Solutions. (1) \(2(6-3)/\ln(6/3)=6/\ln 2=8.656\) mg·h/L. (2) Linear interpolation gives \(C(1)=1+(1/4)(5-1)=2\) mg/L. The area is \(1(1+2)/2=1.500\) mg·h/L. (3) The logarithmic expression requires a ratio of positive endpoint concentrations; with a zero endpoint, the logarithm is undefined. Use a linear trapezoid for that segment under the stated convention. (4) The equal concentrations form a rectangle, which is the linear trapezoid: \((5-3)(2+2)/2=4.000\) mg·h/L.