Integration in Economics, Ecology, and Physiology, calculus lesson

A doctor wants to know how much blood a heart pumps per minute without opening the chest. A forester wants to know whether a cutting plan will leave the forest bigger or smaller in twenty years. Neither can measure the total directly, but both can measure a rate. Integrating the rate turns it into the total, as long as the model behind it fits the situation.

Key terms

Accumulation from measured rates

Definition If $F$ is differentiable on $[a, b]$ and its rate of change $F' = f$ is continuous there, the net change theorem gives

F(b) - F(a) = \int_a^b f(t) \, dt.

When $f$ is known only at measured times $t_0 < t_1 < \dots < t_n$ , the integral is estimated by a Riemann sum or by the trapezoid estimate, the average of the left and right sums.

In plain words A rate tells you how fast something builds up. Add the rate over time and you get how much built up.

Example Water flows into a reservoir at these measured rates.

Time (h)	$0$	$2$	$4$	$6$	$8$
Rate $(\text{m}^3/\text{h})$	$40$	$52$	$60$	$56$	$48$

The left sum is $2(40 + 52 + 60 + 56) = 416$ and the right sum is $2(52 + 60 + 56 + 48) = 432$ . Their average estimates the inflow over the $8$ hours as $424 \ \text{m}^3$ .

Consumer surplus

Definition Let the demand function $p = D(x)$ give the price at which consumers would buy $x$ units, with $D$ continuous and decreasing on $[0, X]$ . If the market sells $X$ units at price $P = D(X)$ , the consumer surplus is

\text{CS} = \int_0^X \big( D(x) - P \big) \, dx.

In plain words Some buyers would have paid more than the market price. Consumer surplus adds up what they saved.

Example Take $D(x) = 50 - 0.5x$ and $X = 40$ , so $P = 30$ . Then

\text{CS} = \int_0^{40} (20 - 0.5x) \, dx = 800 - 400 = 400.

Producer surplus

Definition Let the supply function $p = S(x)$ give the price at which producers would supply $x$ units, with $S$ continuous and increasing on $[0, X]$ . At sales of $X$ units and price $P = S(X)$ , the producer surplus is

\text{PS} = \int_0^X \big( P - S(x) \big) \, dx.

At the market equilibrium, $X$ is the quantity where $D(X) = S(X)$ .

In plain words Some producers would have sold for less than the market price. Producer surplus adds up what they gained.

Example With supply $S(x) = 10 + 0.5x$ and the demand above, $50 - 0.5x = 10 + 0.5x$ gives $X = 40$ and $P = 30$ . Then

\text{PS} = \int_0^{40} \big( 30 - (10 + 0.5x) \big) \, dx = \int_0^{40} (20 - 0.5x) \, dx = 400.

Present value

Definition Suppose money arrives as a continuous stream at a rate of $f(t)$ dollars per year for $0 \le t \le T$ , with $f$ continuous, and interest is compounded continuously at a constant annual rate $r > 0$ . The present value of the stream is

\text{PV} = \int_0^T f(t) \, e^{-rt} \, dt.

In plain words A dollar that arrives in $t$ years is worth $e^{-rt}$ dollars today, because that smaller amount would grow to one dollar by then. The integral discounts every bit of the stream and adds the results.

Example An income of $1000$ dollars per year for $5$ years at $r = 0.05$ has present value

\int_0^5 1000 e^{-0.05t} \, dt = 20000 \left(1 - e^{-0.25}\right) \approx 4424 \ \text{dollars},

less than the $5000$ dollars paid in total.

Renewable resource models

Definition Let $S(t)$ be the size of a renewable stock, such as timber or a fish population, that regrows at rate $g(t)$ and is harvested at rate $h(t)$ , both continuous. Then $S'(t) = g(t) - h(t)$ and

S(T) = S(0) + \int_0^T \big( g(t) - h(t) \big) \, dt.

The harvest is sustainable over $[0, T]$ when $S(T) \ge S(0)$ .

In plain words The stock goes up by what grows and down by what is taken. The integral of the difference gives the net change.

Example A forest holds $5000 \ \text{m}^3$ of timber. It grows at $g(t) = 200 + 10t$ cubic meters per year and is cut at $250$ cubic meters per year. After $20$ years it holds

5000 + \int_0^{20} (10t - 50) \, dt = 5000 + 2000 - 1000 = 6000 \ \text{m}^3.

Blood flow

Definition Poiseuille’s law models steady laminar flow of blood through a straight cylindrical vessel of radius $R$ and length $l$ , with pressure difference $P$ between the ends and viscosity $\eta$ . The velocity at distance $r$ from the axis is

v(r) = \frac{P}{4\eta l} \left( R^2 - r^2 \right), \qquad 0 \le r \le R.

A thin ring of radius $r$ and width $\Delta r$ has area about $2\pi r \, \Delta r$ , so the flux, the volume of blood per unit time, is

Q = \int_0^R 2\pi r \, v(r) \, dr = \frac{\pi P R^4}{8 \eta l}.

In plain words Blood moves fastest along the center and not at all against the wall. Adding the flow through thin rings gives the total, and that total grows with the fourth power of the radius.

Example With $\eta = 0.027$ , $R = 0.008$ cm, $l = 2$ cm and $P = 4000 \ \text{dyn}/\text{cm}^2$ ,

Q = \frac{\pi \cdot 4000 \cdot (0.008)^4}{8 \cdot 0.027 \cdot 2} \approx 1.19 \times 10^{-4} \ \text{cm}^3/\text{s}.

Cardiac output

Definition In the dye dilution method, an amount $A$ of dye is injected into a vein near the heart, and the dye concentration $c(t)$ is measured in an artery for $0 \le t \le T$ , until the dye has passed. If $c$ is continuous, $\int_0^T c(t) \, dt > 0$ , and the heart pumps at a constant rate $F$ , then all the dye passes the sensor, so

A = F \int_0^T c(t) \, dt

and therefore

F = \frac{A}{\int_0^T c(t) \, dt}.

In plain words All the dye that went in comes out past the sensor. A strong heart washes it through quickly, so the area under the concentration curve is small.

Example Inject $A = 6$ mg and suppose $c(t) = 0.5\,t(10 - t)$ mg per liter for $0 \le t \le 10$ seconds. Then

\int_0^{10} 0.5\,t(10 - t) \, dt = 0.5 \left( 500 - \frac{1000}{3} \right) = \frac{250}{3},

so $F = 18/250 = 0.072$ liters per second, which is $4.32$ liters per minute.

Model assumptions

Definition Each formula above is exact only for its model. Surplus assumes continuous demand and supply curves and a single market price. Present value assumes a continuous stream and a constant rate $r$ . Poiseuille’s law assumes steady laminar flow in a rigid straight tube. The dye method assumes the dye mixes fully, the output $F$ is constant, and no dye comes around a second time before $T$ .

In plain words The integral is only as good as the model behind it. Before trusting a number, ask whether the assumptions hold well enough.

Example By Poiseuille’s law $Q$ is proportional to $R^4$ . If a vessel narrows by $10\%$ , the flux becomes $0.9^4 \approx 0.656$ of its former value, about $34\%$ less. That large drop is what the model predicts, but a real artery is elastic and the flow pulses, so the true change may differ.

Check yourself

Find the consumer surplus for demand $D(x) = 100 - 2x$ when $20$ units are sold.

Show the answer

The price is $P = D(20) = 60$ , so

\text{CS} = \int_0^{20} (40 - 2x) \, dx = 800 - 400 = 400.

What is the present value of $2000$ dollars per year for $3$ years at $r = 0.04$ ?

Show the answer

The present value is

\int_0^3 2000 e^{-0.04t} \, dt = 50000\left(1 - e^{-0.12}\right) \approx 5654 \ \text{dollars}.

In a dye test $8$ mg of dye is injected and $\int_0^T c(t) \, dt = 160 \ \text{mg} \cdot \text{s}/\text{L}$ . Find the cardiac output in liters per minute.

Show the answer

The output is

F = \frac{8}{160} = 0.05

liters per second, which is $3$ liters per minute.

Integration in Economics, Ecology, and Physiology

Key terms

Accumulation from measured rates

Consumer surplus

Producer surplus

Present value

Renewable resource models

Blood flow

Cardiac output

Model assumptions

Check yourself

How this lesson connects

Builds on