AUC With Rising Segments, Declines, and Equal Concentrations

Why Segment Direction Matters

AUC is assembled from areas between observed concentration-time points. The observations tell us the segment’s endpoints, but not the exact concentration path between them. A trapezoidal rule supplies an interpolation assumption for that unobserved interval. Choosing a rule without checking the endpoints can produce an inappropriate area—or a formula that cannot be evaluated at all.

For the conventional hybrid approach used here, use the linear trapezoid for rising segments, the log trapezoid for positive, unequal declining segments, and the linear trapezoid for equal concentrations. This is a segment-by-segment choice: a single profile may require more than one rule. The earlier material on Switching Between Linear and Log Trapezoids introduces the two rules; the focus here is how direction and endpoint values govern their use.

Definition: For observations \((t_1,C_1)\) and \((t_2,C_2)\), with \(t_2>t_1\), the segment duration is \(\Delta t=t_2-t_1\). The segment rises if \(C_2>C_1\), declines if \(C_2<C_1\), and is equal or flat if \(C_2=C_1\). A segment area has units of concentration multiplied by time, such as \(\mathrm{mg\cdot h/L}\).

The Rules and Their Conditions

The linear trapezoid treats concentration as changing at a constant rate between the endpoints. It is defined for any finite endpoint concentrations, including zero, and gives the straight-line area:

A_{\mathrm{lin}}=\Delta t\frac{C_1+C_2}{2}.

For a declining segment with \(C_1>C_2>0\), the log trapezoid assumes concentration changes exponentially between the observed values. Its area is:

A_{\log}=\Delta t\frac{C_1-C_2}{\ln(C_1/C_2)}.

This formula requires both concentrations to be positive and unequal. In the conventional hybrid rule, it is used for a decline, not automatically for a rise. A rising exponential curve can be integrated mathematically, but that does not by itself establish that exponential interpolation is suitable for the rising segment. The hybrid rule uses linear interpolation on rises and log interpolation on eligible declines.

Rule: For each segment, compare its endpoints before calculating area. If \(C_2>C_1\), use the linear trapezoid. If \(C_1>C_2>0\), the conventional hybrid rule uses the log trapezoid. If \(C_1=C_2\), use the linear trapezoid. If a declining segment reaches zero, use the linear trapezoid because the log formula is undefined.

When \(C_1=C_2=C>0\), the segment is a rectangle, so the area is \(C\Delta t\). Substituting equal values into the log formula would give \(0/0\), which is undefined. The linear formula, however, gives \(C\Delta t\) directly.

Why the Log Rule Works for a Decline

The log trapezoid follows by assuming an exponential path between two positive, unequal endpoints. Write that path as \(C(t)=C_1\exp(k(t-t_1))\). Matching its value at \(t_2\) requires \(C_2=C_1\exp(k\Delta t)\), so \(k=\ln(C_2/C_1)/\Delta t\). Integrating this path over the segment gives:

\begin{aligned} A &=\int_{t_1}^{t_2} C_1\exp(k(t-t_1))\,dt\\ &=\frac{C_2-C_1}{k}\\ &=\Delta t\frac{C_2-C_1}{\ln(C_2/C_1)}\\ &=\Delta t\frac{C_1-C_2}{\ln(C_1/C_2)}. \end{aligned}

For a decline, the numerator \(C_2-C_1\) and the logarithm \(\ln(C_2/C_1)\) are both negative, so their ratio is positive. The equivalent final form is also positive because \(C_1-C_2>0\) and \(\ln(C_1/C_2)>0\). This checks the sign of the calculated area.

As the endpoints approach the same positive value, the log-rule area approaches the rectangular area. To see this, write \(C_2=C_1+\varepsilon\), where \(\varepsilon\) approaches zero. Then \(\ln(C_2/C_1)=\ln(1+\varepsilon/C_1)\), and \(\ln(1+x)/x\) approaches 1 as \(x\) approaches zero. Therefore:

\lim_{\varepsilon\to 0} \Delta t\frac{\varepsilon}{\ln(1+\varepsilon/C_1)} =C_1\Delta t.

Thus the equal-concentration area is a well-defined limit, even though directly evaluating the log formula at equal endpoints is not possible. In practice, calculate the flat segment with the linear rule rather than attempting a numerical \(0/0\) calculation.

Applying the Rules to Observed Segments

Worked Example: A Rising Segment

Suppose hypothetical concentrations rise from \(1.0\ \mathrm{mg/L}\) at 1 hour to \(4.0\ \mathrm{mg/L}\) at 3 hours. The segment duration is \(3-1=2\) hours, and the endpoint comparison \(4.0>1.0\) identifies a rise.

Under the conventional hybrid rule, use the linear trapezoid:

A_{\mathrm{lin}} =(3-1)\frac{1.0+4.0}{2} =2\cdot 2.5 =5.00\ \mathrm{mg\cdot h/L}.

For comparison, mechanically inserting the same values into the log expression would give \(2(4.0-1.0)/\ln(4.0/1.0)=6/\ln(4)\approx4.328\ \mathrm{mg\cdot h/L}\). That calculation is possible, but it imposes exponential interpolation on this rise. It is not the selected rule under the conventional hybrid approach. The difference illustrates why the direction check is consequential.

Worked Example: A Positive Decline

For a hypothetical decline from \(12.0\ \mathrm{mg/L}\) at 4 hours to \(6.0\ \mathrm{mg/L}\) at 7 hours, the duration is 3 hours. Both concentrations are positive, they are unequal, and \(12.0>6.0\), so the log trapezoid is eligible under the hybrid rule.

A_{\log} =(7-4)\frac{12.0-6.0}{\ln(12.0/6.0)} =3\frac{6.0}{\ln(2)} =\frac{18.0}{0.693147} \approx25.97\ \mathrm{mg\cdot h/L}.

The linear trapezoid would give \(3(12.0+6.0)/2=27.00\ \mathrm{mg\cdot h/L}\). The two values differ because they represent different interpolated paths between the same observations. The log area is the one selected here because the segment is a positive decline and the assumed exponential interpolation is the basis of that rule.

Worked Example: Rising, Flat, and Declining Segments in One Profile

Consider the following hypothetical profile. Concentrations are in \(\mathrm{mg/L}\) and times are in hours.

Time (h)	Concentration (mg/L)	Segment to next observation
0	0	Rising
1	4	Rising
2	8	Equal
3	8	Declining
5	2	End

From 0 to 1 hour, the concentration rises from 0 to 4. Use the linear rule, which permits a zero endpoint:

A_{0\text{–}1} =(1-0)\frac{0+4}{2} =2.00\ \mathrm{mg\cdot h/L}.

From 1 to 2 hours, it rises from 4 to 8, so again use the linear rule:

A_{1\text{–}2} =(2-1)\frac{4+8}{2} =6.00\ \mathrm{mg\cdot h/L}.

From 2 to 3 hours, the concentration remains at 8. The flat segment has rectangular area:

A_{2\text{–}3} =(3-2)\frac{8+8}{2} =8.00\ \mathrm{mg\cdot h/L}.

From 3 to 5 hours, concentration declines from 8 to 2, and both endpoints are positive. Use the log rule:

A_{3\text{–}5} =(5-3)\frac{8-2}{\ln(8/2)} =\frac{12}{\ln(4)} \approx8.656\ \mathrm{mg\cdot h/L}.

Adding the segment areas gives the profile AUC over 0 to 5 hours:

AUC_{0\text{–}5} =2.00+6.00+8.00+8.656 =24.66\ \mathrm{mg\cdot h/L}.

The calculation uses the endpoint direction separately for each interval; no single interpolation rule is applied indiscriminately to the whole profile.

When a Log Trapezoid Is Unsuitable

The direction and values of the endpoints provide a practical diagnostic checklist. Before evaluating a logarithm, confirm that its arguments are valid and that the segment meets the rule’s conditions.

Rising endpoint concentrations: Do not automatically use the log trapezoid just because both values are positive. Under the conventional hybrid rule, use linear interpolation on the rise.
Equal positive concentrations: The direct log formula has a zero numerator and zero denominator. Use the linear area \(C\Delta t\).
A zero or negative concentration: A logarithm of zero or a negative number is not defined over the real numbers. The log rule cannot be evaluated; use the linear rule for that segment.
Incorrect order in the formula: For a decline, \(\ln(C_1/C_2)\) is positive when \(C_1>C_2\). Reversing only the numerator or only the ratio can produce a negative area, signaling an algebra or coding error.
Non-increasing time points: The segment duration must be positive. Check observation order and time units before calculating areas.

If a positive decline is very small, the log formula may be numerically unstable when computed directly: both the concentration difference and the logarithm are close to zero. The limiting area is approximately \(C\Delta t\), where \(C\) is near the common endpoint value. Numerical implementations should use a stable logarithmic-mean calculation or an appropriate small-difference limit rather than interpreting rounding noise as a meaningful difference.

Comparing the Approaches and Their Limits

The linear and log trapezoids are not rival formulas where one is universally correct. They encode different assumptions about the path between two observed values. The linear rule assumes a straight line in concentration over time; the log rule assumes a straight line in log concentration over time, equivalently an exponential concentration path. For a positive decline, exponential interpolation often produces a smaller area than the straight-line interpolation, as in the preceding example. Neither method recovers an unobserved concentration path exactly from two endpoints alone.

Segment condition	Conventional rule	Reason or caution
Rising concentrations	Linear trapezoid	Appropriate default in the hybrid rule; do not assume exponential growth from endpoints alone.
Positive, unequal decline	Log trapezoid	Uses exponential interpolation; both endpoint concentrations must be positive.
Equal concentrations	Linear trapezoid	Gives the rectangle directly; the log formula is indeterminate as written.
Decline to zero or a nonpositive value	Linear trapezoid	The logarithmic expression is undefined for a nonpositive endpoint.

The selection rules are conventions for numerical AUC calculation, not evidence that the true concentration curve is linear on rises or exponential on declines. Sparse sampling, abrupt changes, or complex profiles can leave substantial uncertainty about the curve between observations. The trapezoidal calculation estimates area from measured points; it does not replace careful sampling or establish the physiological mechanism behind a segment.

Common Mistakes

Applying the log rule to every pair of positive values. Positive endpoints alone are not enough: first check direction and the chosen interpolation convention.
Treating a flat segment as a special log calculation. Use \(C\Delta t\); the log formula’s \(0/0\) is a removable limit, not an instruction to divide by zero.
Using the log rule when one endpoint is zero. The logarithm is undefined. Use the linear trapezoid for that segment.
Summing areas with inconsistent units. If time is recorded in hours and concentration in \(\mathrm{mg/L}\), every segment and the total must be in \(\mathrm{mg\cdot h/L}\).
Assuming a larger calculated area is necessarily more accurate. The difference between linear and log areas reflects different interpolation assumptions, not a direct measure of which is closer to the unknown true area.

Key takeaway: Inspect each pair of adjacent observations before calculating area. Use a linear trapezoid for rises, equal values, and segments with nonpositive endpoints; use a log trapezoid only for a positive, unequal decline under the conventional hybrid approach. Confirm the area is positive and its units are concentration multiplied by time.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

A concentration rises from \(2.0\ \mathrm{mg/L}\) to \(5.0\ \mathrm{mg/L}\) over 2 hours. Which rule applies under the hybrid approach, and what is the area?
A concentration falls from \(9.0\ \mathrm{mg/L}\) to \(3.0\ \mathrm{mg/L}\) over 2 hours. Calculate the log trapezoidal area to 3 significant figures.
Why should equal positive concentrations not be inserted directly into the log trapezoid formula? What is the area for a 4-hour flat segment at \(3.0\ \mathrm{mg/L}\)?
A segment declines from \(6.0\ \mathrm{mg/L}\) to \(0\) over 1 hour. Which rule is usable, and what area does it give?

Solutions. (1) The segment rises, so use the linear rule: \(2(2.0+5.0)/2=7.00\ \mathrm{mg\cdot h/L}\). (2) It is a positive, unequal decline, so \(A_{\log}=2(9.0-3.0)/\ln(9.0/3.0)=12/\ln(3)\approx10.9\ \mathrm{mg\cdot h/L}\). (3) Equal endpoints make the log expression \(0/0\). Use the rectangle: \(3.0(4)=12.0\ \mathrm{mg\cdot h/L}\). (4) The log rule is undefined because it would require \(\ln(6.0/0)\). Use the linear rule: \(1(6.0+0)/2=3.00\ \mathrm{mg\cdot h/L}\).