Switching Between Linear and Log Trapezoids

Why the Choice Can Change From One Interval to the Next

A concentration–time profile can rise during absorption, fall during elimination, and include a flat interval or an observation at zero. A single trapezoid formula need not represent all these segments equally well. The linear-up, log-down rule makes the choice interval by interval: use a linear trapezoid for a rising segment, and, when both endpoint concentrations are positive and unequal, a logarithmic trapezoid for a declining segment. Equal concentrations and intervals with a zero endpoint use the linear formula.

The rule estimates the area under the concentration–time curve between observed times. It is a computational convention based on the shape assigned to each interval, not a claim that the true concentration curve is known between samples. The calculated interval areas have units of concentration multiplied by time—for example, mg·h/L when concentration is in mg/L and time is in hours.

Definition: Let $t_i<t_{i+1}$ be consecutive observation times, let $C_i$ and $C_{i+1}$ be the corresponding nonnegative concentrations, and let $\Delta t_i=t_{i+1}-t_i$. Under the linear-up, log-down convention, the interval area is $$ AUC_i = \begin{cases} \dfrac{\Delta t_i(C_i+C_{i+1})}{2}, & C_{i+1}\geq C_i,\\[6pt] \dfrac{\Delta t_i(C_i-C_{i+1})}{\ln(C_i/C_{i+1})}, & 0<C_{i+1}<C_i,\\[6pt] \dfrac{\Delta t_i(C_i+C_{i+1})}{2}, & C_i>C_{i+1}=0. \end{cases} $$ The first and third cases are linear trapezoids. The second is a logarithmic trapezoid. The total over the observed intervals is the sum of the interval areas.

The cases can be summarized in words. If concentration rises, use the linear formula. If it declines and both concentrations are positive, use the logarithmic formula. If the concentrations are equal, use the linear formula; this gives the area of a rectangle. If a declining interval ends at zero, use the linear formula because the logarithm would be undefined. A rising interval from zero is also handled by the linear formula.

What the Two Trapezoids Represent

For the linear trapezoid, concentration is approximated as changing along a straight line between the two observations. The area is the width of the interval multiplied by the average of its endpoint concentrations:

AUC_{\mathrm{linear}}=\Delta t\frac{C_1+C_2}{2}.

For a logarithmic trapezoid, a declining positive segment is approximated by an exponential curve. This shape corresponds to a constant proportional rate of decline over the interval. If $C_1$ is the concentration at the earlier time and $C_2$ is the concentration at the later time, with $C_1>C_2>0$, the exponential approximation has the form $C(t)=C_1e^{-k(t-t_1)}$. The decline parameter $k$ is chosen to match the endpoint:

C_2=C_1e^{-k\Delta t} \qquad\Longrightarrow\qquad k=\frac{\ln(C_1/C_2)}{\Delta t}.

Integrating that curve over the interval gives the logarithmic trapezoid. This short derivation also explains the conditions for using it: the endpoint concentrations must be positive, unequal, and declining in time.

\begin{aligned} AUC_{\mathrm{log}} &=\int_0^{\Delta t} C_1e^{-ku}\,du\\ &=\frac{C_1(1-e^{-k\Delta t})}{k}\\ &=\frac{C_1-C_2}{k}\\ &=\Delta t\frac{C_1-C_2}{\ln(C_1/C_2)}. \end{aligned}

The logarithmic formula is not an instruction to take the logarithm of a zero or negative value. It is defined for the specified declining, positive endpoints. Concentrations recorded in ordinary concentration units can be used directly in the ratio $C_1/C_2$, which is dimensionless; both endpoint concentrations must use the same units.

Worked Application of the Switching Rule

Worked Example: Choosing a Formula Across a Mixed Profile

Consider hypothetical observations at equally spaced times. Concentration is measured in mg/L and time in hours.

Time (h)	Concentration (mg/L)	Change to next observation
0	0	Rising
1	4	Rising
2	10	Declining
3	6	Equal
4	6	Declining to zero
5	0

For $0$ to $1$ h, concentration rises from 0 to 4 mg/L, so use a linear trapezoid: $1(0+4)/2=2.000$ mg·h/L.

For $1$ to $2$ h, concentration rises from 4 to 10 mg/L, again use the linear formula: $1(4+10)/2=7.000$ mg·h/L.

For $2$ to $3$ h, concentration declines from 10 to 6 mg/L, and both endpoints are positive. Use the logarithmic formula: $1(10-6)/\ln(10/6)=4/0.510826=7.830$ mg·h/L.

For $3$ to $4$ h, the concentrations are equal at 6 mg/L. Use the linear formula: $1(6+6)/2=6.000$ mg·h/L.

For $4$ to $5$ h, concentration declines from 6 mg/L to zero. Use the linear formula, since the logarithmic formula cannot be evaluated at zero: $1(6+0)/2=3.000$ mg·h/L.

Adding the five interval areas gives $2.000+7.000+7.830+6.000+3.000=25.83$ mg·h/L. Each interval was classified using its own endpoint concentrations; the formula was not chosen once for the entire profile.

Equal Concentrations and the Boundary Between Rules

When both concentrations are equal to $C$, the linear trapezoid is $\Delta t(C+C)/2=\Delta tC$: the rectangle of height $C$ and width $\Delta t$. The logarithmic expression, in contrast, would have zero in both its numerator and denominator. It is therefore not evaluated by direct substitution. The linear result is also the limiting value of the logarithmic expression as the two positive concentrations approach one another.

\lim_{C_2\to C_1} \frac{C_1-C_2}{\ln(C_1/C_2)} =C_1, \qquad \text{so}\qquad \lim_{C_2\to C_1} AUC_{\mathrm{log}}=\Delta t C_1.

This limit gives a mathematical reason the formulas meet at equal concentrations. It does not mean that software should evaluate the logarithmic expression at equality: direct computation there is undefined. The rule explicitly assigns equal endpoints to the linear case.

Worked Example: A Positive Decline Versus a Linear Trapezoid

Suppose a hypothetical concentration declines from 12 to 8 mg/L over 2 h. Both concentrations are positive and unequal, so the log-down rule selects:

AUC_{\mathrm{log}} =2\frac{12-8}{\ln(12/8)} =\frac{8}{\ln(1.5)} =\frac{8}{0.405465} =19.73\ \text{mg·h/L}.

For comparison, the linear trapezoid would be $2(12+8)/2=20.00$ mg·h/L. The difference is $20.00-19.73=0.27$ mg·h/L (approximately). The log-down convention selects the smaller area here because the exponential curve joining these declining endpoints lies below the straight chord. The alternative calculation is useful for comparison, but it is not the selected rule for this positive declining interval.

Worked Example: Nearly Equal Endpoints and a Zero Endpoint

Consider a hypothetical positive decline from 10.00 to 9.99 mg/L over 0.5 h. The log-down formula gives:

AUC_{\mathrm{log}} =0.5\frac{10.00-9.99}{\ln(10.00/9.99)} =0.5\frac{0.01}{0.00100050} \approx 4.9975\ \text{mg·h/L}.

The linear area is $0.5(10.00+9.99)/2=4.9975$ mg·h/L to the shown precision, consistent with the limit as the endpoints approach equality. In computation, however, subtracting nearly equal logarithms can lose numerical precision. A robust implementation should handle exact equality with the linear formula and use numerically stable functions or adequate precision for very small relative changes.

Now suppose a separate hypothetical interval declines from 10 to 0 mg/L over 0.5 h. The log formula would require $\ln(10/0)$, which is undefined. The rule uses the linear trapezoid: $0.5(10+0)/2=2.500$ mg·h/L. A zero endpoint is a formula-selection issue; it does not justify replacing zero by an arbitrary small positive number.

Assumptions, Checks, and Common Failure Modes

The switching rule is straightforward only when the interval data and their interpretation are clear. Before calculating, confirm that the observation times are ordered, each interval has positive duration, and concentrations are nonnegative and expressed in the same units. The rule applies to adjacent observed points; it does not supply missing observations or decide how an unquantifiable measurement should be handled.

Its shape assumptions differ by interval. Linear interpolation treats concentration as changing at a constant absolute rate between endpoints. Logarithmic interpolation treats a positive decline as exponential, with a constant proportional rate of decline over that interval. These are local approximations. Sparse sampling, rapid changes in absorption, a changing elimination rate, or a segment that actually changes direction between observations can make either approximation a poor representation of the unobserved curve. Endpoint data alone cannot diagnose every such mismatch.

Several practical checks help identify mistakes:

Check direction using time order. A rising segment has $C_{i+1}>C_i$; a declining segment has $C_{i+1}<C_i$. Reversing the endpoints in a logarithmic calculation can produce an erroneous sign.
Check the logarithmic domain. Use the log formula only when $C_i>C_{i+1}>0$. A zero concentration, an equal pair, or a negative value does not meet its conditions.
Check units. Multiplying concentration by elapsed time must produce concentration-time units. The logarithm's argument must be a ratio of concentrations in the same units.
Check plausibility. For a positive decline, the logarithmic interval area should lie between $\Delta t C_{i+1}$ and $\Delta t C_i$. A negative area or one outside those bounds signals an arithmetic, ordering, or formula error.
Check the sampling interpretation. The calculation uses the concentrations supplied at the observation times. If a reported zero represents a below-quantification result rather than a measured zero, its handling requires a stated data convention; the log formula cannot be made valid simply by treating the reported value as positive.

A frequent error is to use the logarithmic formula for every interval because it was derived for exponential decline. That fails on rising intervals and at equal or zero endpoints. The correction is to classify each interval first, then apply the assigned formula. Another error is to choose a formula based on whether the whole profile is “mostly rising” or “mostly falling.” The switching rule is local: each pair of consecutive observations has its own direction and endpoint conditions.

How the Hybrid Rule Compares With Alternatives

Using linear trapezoids for every interval is simple, applies at zero and equal endpoints, and does not require a logarithm. It can be less representative of a positive exponential decline, however, because a straight chord can overstate the area under a declining exponential curve. Using logarithmic trapezoids for every interval is not a viable alternative: the declining exponential derivation does not apply to a rise, and the formula is undefined for equal or zero endpoints.

The hybrid rule combines the linear treatment of rising segments with the logarithmic treatment of eligible declines, while retaining a defined calculation for equal or zero cases. It is useful as a consistent numerical convention, but it does not remove uncertainty caused by sparse sampling or establish that either assumed within-interval shape is the true curve. When comparing results, the chosen integration convention should be applied consistently and reported clearly.

The central practical distinction is between a formula-selection rule and a model of the full concentration profile. This convention selects an area formula between adjacent observations. It does not, by itself, estimate the concentration beyond the observed time range, reconstruct an unobserved peak, or resolve how below-quantification data should be treated.

Check Your Understanding

For each problem, identify the interval rule and calculate or explain the result. Solutions are provided below the questions.

A concentration rises from 2 to 7 mg/L over 1.5 h. Which formula applies, and what is the interval area?
A concentration declines from 9 to 3 mg/L over 2 h. Calculate the log-down area to 3 significant figures.
Concentration is 5 mg/L at both ends of a 3 h interval. Which formula applies, and why is direct use of the logarithmic formula inappropriate?
A concentration declines from 4 mg/L to zero over 0.75 h. Calculate the interval area and state why the log formula is not used.

Solutions. (1) The segment rises, so use the linear formula: $1.5(2+7)/2=6.75$ mg·h/L. (2) The endpoints are positive and declining, so use the log formula: $2(9-3)/\ln(9/3)=12/\ln(3)=10.9$ mg·h/L. (3) Use the linear formula: $3(5+5)/2=15.0$ mg·h/L. Direct substitution into the logarithmic expression gives zero divided by zero, so it is undefined. (4) Use the linear formula: $0.75(4+0)/2=1.50$ mg·h/L. The log formula is undefined because its endpoint ratio requires division by zero.