Average Values for Exponential and Logarithmic Models
Exponential and logarithmic functions describe different kinds of change. An exponential model can represent a quantity that grows or decays by a constant proportion over equal time intervals. A logarithmic model can describe a quantity that increases while its rate of increase becomes smaller.
For either model, the average over an interval is determined by the function’s accumulated value across the whole interval, not just its endpoint values. Definite integrals provide a precise way to calculate that average.
The factor \(1/(b-a)\) divides the total accumulated value by the interval length. Thus the average has the same units as \(f\). The average is also the constant function value that would produce the same integral over \([a,b]\).
Average Value of an Exponential Model
Consider \(f(t)=A e^{kt}\) on \([a,b]\), where \(A\) and \(k\) are constants. If \(k\ne 0\), an antiderivative is \(A e^{kt}/k\), so the average is
When \(k>0\), the model grows as \(t\) increases; when \(k<0\), it decays. The same formula applies in either case. If \(k=0\), the function is constant, \(f(t)=A\), and its average is \(A\).
Often an exponential model is given by its value \(C\) at the start of the interval. Writing \(f(t)=C e^{k(t-a)}\), and letting \(L=b-a\), gives the equivalent form
This formula applies to decay as well as growth: for decay, \(k<0\), so both \(e^{kL}-1\) and \(kL\) are negative, and their ratio is positive.
Worked Example: Average Value for Exponential Growth
A quantity is modeled by \(Q(t)=3e^{0.4t}\) on \(0\le t\le 5\), where \(t\) is measured in hours and \(Q\) in kilograms. Find its average quantity on this interval.
Here \(A=3\), \(k=0.4=2/5\), and the interval length is \(5\) hours. The exponential average formula gives
Thus the exact average is \(\frac{3}{2}(e^2-1)\) kilograms, approximately \(9.58\) kilograms (rounded to two decimal places).
To check the calculation, an antiderivative is
The integral is \(F(5)-F(0)=7.5(e^2-1)\) kilogram-hours. Dividing by \(5\) hours gives \(1.5(e^2-1)\) kilograms, as above. Multiplying the average by the interval length recovers the integral.
Worked Example: Average Value for Exponential Decay
Suppose a substance has amount \(M(t)=80e^{-0.3t}\) grams after \(t\) hours. Find its average amount during the first \(10\) hours.
The growth constant is \(k=-0.3\), and the interval is \([0,10]\). Therefore
Numerically, this is approximately \(25.34\) grams, rounded to two decimal places.
For a direct check, \(G(t)=-(80/0.3)e^{-0.3t}\) is an antiderivative because \(G'(t)=80e^{-0.3t}\). Hence
Dividing by \(10\) hours gives \(\frac{80}{3}(1-e^{-3})\) grams, confirming the result. The negative signs in the antiderivative and in the exponential formula are essential; the average amount is positive.
Why the Exponential Formula Works
The formula is an application of the average-value definition and the Fundamental Theorem of Calculus. Its derivation also shows exactly where the interval length enters.
For \(f(t)=Ae^{kt}\), its average on \([a,b]\) is the integral divided by \(b-a\).
When \(k\ne 0\), \(F(t)=Ae^{kt}/k\) satisfies \(F'(t)=Ae^{kt}\).
The Fundamental Theorem gives \(\int_a^b Ae^{kt}\,dt=F(b)-F(a)=A(e^{kb}-e^{ka})/k\). Dividing by \(b-a\) yields the stated average formula.
This derivation requires a finite interval with \(a<b\), an integrable function there, and \(k\ne 0\) for the displayed antiderivative formula. The constant case \(k=0\) is handled separately. No assumption that the model is increasing is needed.
For a positive exponential model, the average can also be expressed using the two endpoint values. If \(y_a=Ae^{ka}\) and \(y_b=Ae^{kb}\), with \(A>0\), then \(\ln y_b-\ln y_a=k(b-a)\). Consequently,
This expression is sometimes called the logarithmic mean of the endpoint values. It is another description of the same integral average, not the arithmetic mean of the endpoints.
Average Value of a Logarithmic Function
The natural logarithm \(\ln x\) is defined for \(x>0\). Its antiderivative is \(x\ln x-x\), since
Thus, on an interval \([a,b]\) with \(0<a<b\), the average of \(\ln x\) is
For a logarithm with base \(c>0\), \(c\ne1\), use \(\log_c x=\ln x/\ln c\). The average is the natural-logarithm average divided by \(\ln c\). Any logarithmic model must be defined throughout the interval; in particular, its input must remain positive.
Worked Example: Average Value of a Logarithmic Model
Find the average value of \(f(x)=\ln x\) on \([1,e^2]\). Here \(x\) is treated as a dimensionless input, so the logarithm is defined on the entire interval.
Using the antiderivative \(H(x)=x\ln x-x\), evaluate the integral at both endpoints:
Therefore
The interval length is \(e^2-1\), so
The approximation is rounded to two decimal places. Differentiating \(H\) gives \(H'(x)=\ln x\), confirming the antiderivative. Also, the average multiplied by the interval length is \(\frac{e^2+1}{e^2-1}(e^2-1)=e^2+1\), the integral just calculated.
Interpretation and Common Pitfalls
An average value is the constant level that has the same accumulated integral as the original model over the interval. If \(f\) represents a positive quantity over time, then \(f_{\mathrm{avg}}(b-a)\) is the total accumulated quantity in the corresponding quantity-time units. This interpretation is useful when a varying model needs to be summarized by one representative level.
A common error is to average the two endpoint values instead of integrating. For the growth model \(Q(t)=3e^{0.4t}\) on \([0,5]\), the endpoint values are \(3\) and \(3e^2\). Their arithmetic mean is
which is not the integral average \(\frac{3}{2}(e^2-1)\). The endpoint calculation gives equal weight to the two endpoints; it does not account for the function’s values throughout the interval.
Another frequent mistake is to find the definite integral but forget to divide by the interval length. The integral is the accumulated total, not the average. A reliable calculation names the interval length, evaluates the integral, and then divides. For a logarithm, also check the domain before integrating: an interval containing zero or negative inputs is not in the domain of \(\ln x\).
Check Your Understanding
Use the average-value definition and the model’s domain to answer these questions.
- What is the average of \(4e^{0.2t}\) on \([0,5]\)?
- For a decaying exponential with exponent \(-\lambda t\), where \(\lambda>0\), why does the average formula still give a positive value when the model’s coefficient is positive?
- What is the average of \(\ln x\) on \([1,e]\)?
- Why is the arithmetic mean of the endpoint values generally not the integral average?
- What condition must an interval \([a,b]\) satisfy for \(\ln x\) to be defined throughout it?
