Deriving the Log Trapezoidal Rule

Why the Shape Between Observations Matters

A concentration-time profile is observed at discrete sampling times, but area under the concentration-time curve (AUC) includes the area between those observations. The linear trapezoidal rule treats concentration as changing linearly within an interval. A different local model is useful when concentration is declining and is reasonably represented as an exponential function of time: integrate that curve instead of a straight line.

The log trapezoidal rule is therefore a numerical-integration rule tied to a specific interpolation assumption. It does not establish that the underlying pharmacokinetics follow a single exponential model over the entire study. It estimates the area within one interval from its two endpoint concentrations and the time separating them.

Definition: For a declining segment from time $t_i$ to $t_{i+1}$, let $C_i$ and $C_{i+1}$ be positive endpoint concentrations, with $C_i>C_{i+1}$, and let $\Delta t=t_{i+1}-t_i>0$. The log trapezoidal area for that segment is $$ A_i=\Delta t\frac{C_i-C_{i+1}}{\ln(C_i/C_{i+1})}. $$ The logarithmic mean of the endpoint concentrations is $(C_i-C_{i+1})/\ln(C_i/C_{i+1})$. Thus, the area is the elapsed time multiplied by that mean concentration.

The formula requires positive, unequal endpoint concentrations and a declining segment. Its units are concentration multiplied by time; for concentrations in mg/L and time in hours, the area is in mg·h/L. The method uses the actual $\Delta t$, not a standard interval or an average sampling interval.

Deriving the Rule

Suppose the concentration between two observations follows an exponential curve that passes through both endpoints. Writing elapsed time from the first observation as $\tau=t-t_i$, a convenient form is $C(\tau)=C_i e^{-k\tau}$. The constant $k$ is chosen so that the curve reaches $C_{i+1}$ after $\Delta t$. This interpolation is the assumption used for the calculation; it is not proof that the concentration has followed this curve between samples.

Fit the exponential through the endpoints.
At $\tau=\Delta t$, require $C_{i+1}=C_i e^{-k\Delta t}$. Taking natural logarithms gives $k=\ln(C_i/C_{i+1})/\Delta t$. Since $C_i>C_{i+1}$, $k>0$ and the interpolated concentration declines.

Integrate over the interval.
The area is the integral of concentration over elapsed time from zero to $\Delta t$. For $k>0$, this integral is $C_i(1-e^{-k\Delta t})/k$.

Substitute the endpoint relation.
Because $e^{-k\Delta t}=C_{i+1}/C_i$, the integral becomes $(C_i-C_{i+1})/k$. Substituting the expression for $k$ yields the log trapezoidal formula.

\begin{aligned} A_i &=\int_0^{\Delta t} C_i e^{-k\tau}\,d\tau\\ &=\frac{C_i(1-e^{-k\Delta t})}{k}\\ &=\frac{C_i-C_{i+1}}{k}\\ &=\Delta t\frac{C_i-C_{i+1}}{\ln(C_i/C_{i+1})}. \end{aligned}

This derivation also explains the logarithmic mean interpretation: dividing area by interval length gives the average concentration over the interval under the exponential interpolation. The value is between the two endpoint concentrations, as an average along a curve that stays between them should be. For example, if the formula produced an average concentration below the lower endpoint or above the higher endpoint, that would signal a calculation or implementation error.

Worked Calculations

Worked Example: One Declining Segment

Hypothetical data. At 2 h the concentration is 8.0 mg/L; at 5 h it is 3.2 mg/L. Assume an exponential decline between these observations. Calculate the segment area.

The interval is $\Delta t=5-2=3$ h. The endpoints are positive and unequal, and the later concentration is lower, so the segment meets the basic conditions for the declining-segment formula. The ratio is $8.0/3.2=2.5$, and $\ln(2.5)=0.91629$.

A_i=3\ \text{h}\times \frac{8.0-3.2\ \text{mg/L}}{\ln(8.0/3.2)} =\frac{14.4}{0.91629} =15.71\ \text{mg·h/L}.

Under the assumed exponential interpolation, the area from 2 to 5 h is 15.71 mg·h/L. The concentration units appear in the numerator; the logarithm of a concentration ratio is dimensionless, so multiplying by hours gives the required concentration-time units.

Worked Example: A Longer, Unequally Spaced Interval

Hypothetical data. A concentration falls from 4.5 mg/L at 6 h to 1.5 mg/L at 10 h. The sample gap is 4 h. Calculate the area, using the interval length rather than assuming a fixed sampling schedule.

The endpoints are positive, unequal, and declining. Here $\Delta t=10-6=4$ h, $C_i/C_{i+1}=4.5/1.5=3$, and $\ln(3)=1.09861$.

A_i=4\ \text{h}\times \frac{4.5-1.5\ \text{mg/L}}{\ln(4.5/1.5)} =\frac{12}{1.09861} =10.92\ \text{mg·h/L}.

The computed segment contributes 10.92 mg·h/L. A different interval length would change the area even if the endpoint concentrations were unchanged; using the actual times is essential when observations are unequally spaced.

Worked Example: Summing Areas Across Unequal Intervals

Hypothetical data. A declining profile has concentrations of 10 mg/L at 0 h, 8 mg/L at 1 h, 5 mg/L at 3.5 h, and 2 mg/L at 8 h. Use the log trapezoidal rule separately for each interval, then sum the segment areas from 0 to 8 h.

Each interval is declining and has positive, unequal endpoint concentrations, so the formula applies to all three under the stated exponential-interpolation assumption. The interval lengths are 1 h, 2.5 h, and 4.5 h.

\begin{aligned} A_{0,1} &=1\frac{10-8}{\ln(10/8)} =\frac{2}{0.223144} =8.963\ \text{mg·h/L},\\ A_{1,3.5} &=2.5\frac{8-5}{\ln(8/5)} =\frac{7.5}{0.470004} =15.96\ \text{mg·h/L},\\ A_{3.5,8} &=4.5\frac{5-2}{\ln(5/2)} =\frac{13.5}{0.916291} =14.73\ \text{mg·h/L}. \end{aligned}

The interval areas add because the intervals are adjacent and do not overlap:

A_{0,8}=8.963+15.96+14.73=39.65\ \text{mg·h/L}.

The unequal gaps have been handled individually rather than averaged or replaced with one common interval. This sum is the area for the observed 0-to-8 h span under the chosen interval-by-interval interpolation; it is not an extrapolation beyond the last observation.

Assumptions, Diagnostics, and Interpretation

The mathematical requirement is clear: the log trapezoidal formula is derived for a positive exponential decline between unequal endpoint concentrations. Its practical use also relies on the judgment that exponential interpolation is a reasonable representation of the unobserved concentrations in that interval. A declining pair of observations satisfies the endpoint pattern, but by itself cannot demonstrate the actual shape between samples. Sparse sampling, measurement error, or a change in the underlying concentration-time behavior may make any interpolation uncertain.

Useful checks include confirming that time increases, concentrations use consistent units, and each segment used in this calculation has positive, unequal endpoints with the earlier concentration larger. Check the concentration ratio and its natural logarithm: with a decline, $C_i/C_{i+1}>1$, so $\ln(C_i/C_{i+1})>0$. The resulting area must be positive. It should also be no smaller than $\Delta t C_{i+1}$ and no larger than $\Delta t C_i$, because the assumed curve stays between its endpoint concentrations. These checks can identify sign errors, a reversed ratio, or a misplaced time interval.

A useful comparison is the linear trapezoidal rule. That rule assumes a straight line between the endpoints; the log trapezoidal rule assumes an exponential decline. For the first worked example, the linear estimate would be $3(8.0+3.2)/2=16.8$ mg·h/L, compared with the log estimate of 15.71 mg·h/L. The difference reflects different assumed paths, not an arithmetic error. Neither interpolation can be selected solely because its result is larger or smaller. The scientific context and the intended numerical-integration convention matter.

Common pitfall: Treating “log trapezoidal” as an instruction to take the logarithm of the concentrations and then average them. That is not the area formula: the logarithm is applied to the dimensionless ratio $C_i/C_{i+1}$, and the formula uses the concentration difference in the numerator. Another error is substituting a nominal sampling interval when actual times differ. Use $\Delta t=t_{i+1}-t_i$ for each segment.

The method has limits. Two endpoint measurements do not reveal whether the true curve was exponential throughout the gap, and the formula cannot repair poor sampling or concentration measurements. In addition, this tutorial’s formula is specifically for positive, unequal, declining endpoints. Do not apply it mechanically to a segment that fails those conditions. A numerical integration method is part of the definition of the estimated exposure, so changing interpolation assumptions can change the resulting AUC even when the observed data are identical.

Check Your Understanding

Answer each question, then compare with the solutions below.

For a declining segment from 1 h at 6 mg/L to 4 h at 2 mg/L, calculate the log trapezoidal area in mg·h/L.
For an exponential decline, explain why the denominator $\ln(C_i/C_{i+1})$ is positive when the segment meets the declining condition.
A segment has an earlier concentration of 3 mg/L and a later concentration of 5 mg/L. Can you apply the declining-segment formula as defined here? Give the relevant reason.
Why must separate areas be calculated using each actual time interval when sampling times are unequally spaced?

Solutions. (1) The interval is $4-1=3$ h, and the endpoints are positive and declining. Thus $A_i=3(6-2)/\ln(6/2)=12/\ln(3)=10.92$ mg·h/L. (2) Since $C_i>C_{i+1}>0$, the ratio $C_i/C_{i+1}>1$, and the natural logarithm of a number greater than 1 is positive. (3) No. Although both concentrations are positive and unequal, the later concentration is higher, so this is not a declining segment of the type defined for the formula here. (4) The segment area depends on its elapsed time, $\Delta t$. Each actual interval has its own duration, so replacing those durations with a common or average interval would generally calculate a different area.

Deriving the Log Trapezoidal Rule

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