Deriving the Linear Trapezoidal Rule

From Two Observations to an Interval Area

An AUC calculation begins with concentration measurements at specified times, but the area between measurements is not directly observed. A rule is needed to represent the concentration-time curve between each pair of samples. The linear trapezoidal rule makes a local approximation: between two adjacent observations, concentration changes along a straight line. Integrating that line gives the area of a trapezoid.

The earlier material on AUC as a numerical integral establishes the goal: approximate the area under the concentration-time curve from the available observations. This tutorial derives the particular approximation obtained by connecting each adjacent pair of observations with a straight line. The time intervals do not need to be equally spaced.

Definition: Let \(t_i\) and \(t_{i+1}\) be adjacent observation times, with measured concentrations \(C_i\) and \(C_{i+1}\). The linear trapezoidal estimate of the area over this interval is the area under the straight-line interpolation joining \((t_i,C_i)\) and \((t_{i+1},C_{i+1})\). Write the elapsed time as \(\Delta t_i=t_{i+1}-t_i\).

The estimate depends on the interval width and on both endpoint concentrations. In particular, \(\Delta t_i\) is the actual time difference for that pair of observations. Equal spacing is not required and must not be assumed merely because the data are listed in a table.

Deriving the Linear Trapezoidal Formula

Assume \(t_{i+1}>t_i\). Under linear interpolation, the estimated concentration at any time \(t\) in the interval is the value at the left endpoint plus a constant slope multiplied by the elapsed time from that endpoint:

\widehat{C}(t)=C_i+\frac{C_{i+1}-C_i}{\Delta t_i}(t-t_i), \qquad t_i\leq t\leq t_{i+1}.

The slope is \((C_{i+1}-C_i)/\Delta t_i\), so this expression reaches \(C_i\) at \(t_i\) and \(C_{i+1}\) at \(t_{i+1}\). Integrating the interpolated concentration over the interval gives:

\begin{aligned} AUC_{i,i+1} &=\int_{t_i}^{t_{i+1}}\widehat{C}(t)\,dt\\ &=C_i\Delta t_i+ \frac{C_{i+1}-C_i}{\Delta t_i} \int_{t_i}^{t_{i+1}}(t-t_i)\,dt\\ &=C_i\Delta t_i+ \frac{C_{i+1}-C_i}{\Delta t_i}\frac{(\Delta t_i)^2}{2}\\ &=\frac{C_i+C_{i+1}}{2}\Delta t_i. \end{aligned}

The same result follows geometrically: the trapezoid has parallel sides of lengths \(C_i\) and \(C_{i+1}\), and width \(\Delta t_i\). Its area is the average of the two concentrations multiplied by the time width.

Result: For two adjacent observations, the linear trapezoidal estimate is

AUC_{i,i+1}=\frac{C_i+C_{i+1}}{2}(t_{i+1}-t_i).

The formula assumes linear change between those observations. It is an estimate of the area over the interval, not proof that the actual concentration curve is linear.

If concentration is measured in mg/L and time in hours, the interval area is in mg·h/L. The result is an area with units of concentration multiplied by time, not a concentration. If the concentration and time units differ, the AUC unit must reflect both units.

Applying the Rule With Actual Time Differences

Worked Example: One Irregularly Spaced Interval

In a hypothetical concentration-time profile, concentration is 8 mg/L at 2.0 hours and 3 mg/L at 4.25 hours. Estimate the interval AUC using linear interpolation.

First calculate the actual interval width: \(\Delta t=4.25-2.0=2.25\) hours. The mean of the two endpoint concentrations is \((8+3)/2=5.5\) mg/L. Therefore:

AUC_{2.0,4.25} =\frac{8+3}{2}(4.25-2.0) =5.5\times2.25 =12.375\ \text{mg·h/L}.

The estimated area is 12.375 mg·h/L, or 12.38 mg·h/L to four significant figures. The time width is 2.25 hours, not a presumed one-hour or two-hour interval.

For a series of ordered observation times \(t_0<t_1<\cdots<t_m\), calculate each adjacent interval separately and add the interval areas. This produces a piecewise-linear approximation: the profile is represented by a succession of straight segments, one for each observed pair.

AUC_{t_0,t_m}\approx \sum_{i=0}^{m-1} \frac{C_i+C_{i+1}}{2}(t_{i+1}-t_i).

Worked Example: Cumulative AUC With Unequal Sampling Gaps

Consider hypothetical measurements of 0 mg/L at 0 hours, 6 mg/L at 1.5 hours, 4 mg/L at 4 hours, and 1 mg/L at 9 hours. Calculate the linear trapezoidal AUC from 0 to 9 hours.

There are three adjacent intervals. Use the time difference for each interval, rather than a single spacing:

\begin{aligned} AUC_{0,1.5}&=\frac{0+6}{2}(1.5-0)=3(1.5)=4.5\ \text{mg·h/L},\\ AUC_{1.5,4}&=\frac{6+4}{2}(4-1.5)=5(2.5)=12.5\ \text{mg·h/L},\\ AUC_{4,9}&=\frac{4+1}{2}(9-4)=2.5(5)=12.5\ \text{mg·h/L}. \end{aligned}

Add the interval estimates:

AUC_{0,9}\approx4.5+12.5+12.5=29.5\ \text{mg·h/L}.

The sampling gaps are 1.5, 2.5, and 5 hours. Each gap is multiplied by its own average endpoint concentration. The result is the area under the piecewise-linear curve through the four observations.

Why Irregular Spacing Must Be Handled Explicitly

The interval-width factor is essential. Omitting it, or replacing it with an assumed common spacing, changes the estimate. For example, in the 1.5-to-4-hour interval above, the endpoint average is 5 mg/L. Multiplying by 1 hour would give 5 mg·h/L, but the observed times are 2.5 hours apart, so the correct linear-trapezoid contribution is \(5\times2.5=12.5\) mg·h/L.

Worked Example: Detecting an Equal-Spacing Error

Suppose a hypothetical profile has concentrations of 2 mg/L at 0 hours, 5 mg/L at 1 hour, and 1 mg/L at 3 hours. The samples are not equally spaced: the first gap is 1 hour and the second is 2 hours. Calculate the interval areas correctly, then compare with the error of treating both gaps as 1 hour.

For the first interval:

AUC_{0,1}=\frac{2+5}{2}(1-0)=3.5\ \text{mg·h/L}.

For the second interval, the endpoint average is 3 mg/L and the elapsed time is 2 hours:

AUC_{1,3}=\frac{5+1}{2}(3-1)=3(2)=6\ \text{mg·h/L}.

Thus the correct total is \(3.5+6=9.5\) mg·h/L. If the second interval were mistakenly treated as 1 hour, its area would be recorded as \(3\times1=3\) mg·h/L and the incorrect total would be 6.5 mg·h/L. The discrepancy arises from using the wrong time width, not from rounding.

Assumptions, Diagnostics, and Limits

The arithmetic formula is simple, but its interpretation depends on the data and approximation. Check the following conditions before treating the result as a useful estimate:

Adjacent times are correctly ordered. The intervals should have positive width. A duplicated time gives a zero-width interval; reversed times indicate an ordering or data-entry problem that should be resolved rather than silently used.
Concentrations and times have consistent units. For example, do not multiply an interval in minutes by concentrations while labeling the result as mg·h/L without converting minutes to hours.
Each pair is adjacent in the sampled profile. The calculation connects the observations supplied for that interval; it does not account for an unrecorded feature between them.
The straight-line approximation is acceptable for the purpose. A long interval, sparse sampling, or rapid curvature can make the linear segment a poor representation of the underlying concentration-time path.

The measured endpoint values alone do not reveal the exact curve between samples. To see why, take hypothetical endpoints of 2 mg/L at 0 hours and 6 mg/L at 2 hours. The straight line has area \(2(2)+(6-2)(2)/2=8\) mg·h/L. A different smooth curve could have the same endpoints and a different area. For example, with time \(t\) expressed in hours, \(C(t)=2+2t+t(2-t)\) mg/L also equals 2 and 6 mg/L at the endpoints, but its area is:

\int_0^2\left[2+2t+t(2-t)\right]dt =8+\frac{4}{3} =9.333\ \text{mg·h/L}.

Thus, knowing the endpoints does not determine the true interval area without a model for the shape between them. The linear trapezoidal result is specifically the area implied by linear interpolation. It has no universal direction of error: depending on how the actual curve bends, the straight-line estimate can be above or below the true area.

A useful diagnostic is to inspect the sampling grid alongside the concentration profile. Widely separated observations around a region of rapid change deserve particular caution. Rechecking the time differences, plotting the measurements, and comparing the interval widths can reveal calculation or sampling problems. A numerical answer with several decimal places does not remove uncertainty caused by an inadequate representation between observations.

Alternatives and Practical Significance

The principal alternatives address different sources of uncertainty. Collecting more observations can reduce the widths of intervals and better resolve changes in the profile, although it cannot guarantee that every feature is captured. A pharmacokinetic model can instead specify a functional concentration-time relationship and integrate that curve; this requires an adequately chosen model and defensible parameter estimates. The linear trapezoidal rule is more direct and makes fewer structural commitments, but its local straight-line assumption may be inadequate when the curve bends substantially between widely spaced samples.

The rule is valuable because it is transparent and reproducible: each interval contribution can be checked from two concentrations and one time difference. It also makes clear why AUC depends on both the observed concentrations and when they were measured. When comparing calculated exposure values, consistent handling of time units and the same stated integration convention are important. The calculation alone does not establish that the sampling schedule adequately represents the underlying profile.

Common pitfall: “The samples are listed in order, so I can use one common interval width.” That reasoning is wrong unless the observation times are in fact equally spaced. Use \((t_{i+1}-t_i)\) separately for every adjacent pair, then sum those interval areas.

Check Your Understanding

Use the linear trapezoidal rule and show the time width used in each calculation.

Concentrations are 4 mg/L at 1 hour and 10 mg/L at 3 hours. What is the interval AUC?
A hypothetical profile has concentrations 1, 3, and 2 mg/L at 0, 2, and 5 hours, respectively. Calculate the cumulative AUC from 0 to 5 hours.
Why can two curves with the same concentrations at the sampled times have different true AUC values between those times?
A colleague calculates an interval area using the average endpoint concentration but omits the time difference. What is missing, and what are the resulting units if concentration is in mg/L?

Solutions. (1) The width is \(3-1=2\) hours, so \(AUC=(4+10)(2)/2=14\) mg·h/L. (2) The first interval contributes \((1+3)(2-0)/2=4\) mg·h/L; the second contributes \((3+2)(5-2)/2=7.5\) mg·h/L. The total is 11.5 mg·h/L. (3) The samples specify only endpoint values. Different possible curves between those endpoints can have different areas; the linear trapezoid is one interpolation assumption, not an observation of the intervening curve. (4) The elapsed time is missing. An average concentration alone has units mg/L, not mg·h/L, and therefore is not an interval AUC.