Figure 5.1 The peregrine falcon (Falco peregrinus) feeds primarily on pigeons, doves, mice, and shorebirds.

Preview

“Nature laughs at the difficulties of integration.”

Pierre-Simon de Laplace (1749–1827)

Calculus has two parts—differential calculus, the topic of the previous chapters, and integral calculus. At the core of differential calculus is the concept of the instantaneous rate of change of a function. We have seen how this concept can be used to locally approximate functions and to identify maxima and minima. Integral calculus, on the other hand, deals with accumulated change and, thereby, recovering a function from a mathematical description of its instantaneous rate of change. This recovery process, interestingly enough, is related to the concept of finding the area under a curve.

Using integral calculus, we can calculate the velocity of a stooping peregrine falcon (see Figure 5.1) as it dives toward Earth at great speed to catch prey; calculate the blossom date of a tree as a function of anticipated temperature patterns, so that an orchard can be stocked with bees to facilitate pollination; or estimate the amount of a drug in the bloodstream of a patient connected to an IV.

A systematic method for estimating the area under the curve was devised by Riemann, one of the great mathematicians of the nineteenth century. This method is commonly known as the Riemann sum and yields in the limit an object called the definite integral. The fathers of calculus, Newton and Leibniz, proved a connection between the problem of finding antiderivatives and finding areas under a curve. This connection, the fundamental theorem of calculus, which is presented in Section 5.4, helps make calculus one of the most powerful mathematical tools for understanding biological and physical processes.

In Sections 5.5 through 5.7, we provide a short apprenticeship in various techniques used to compute and approximate integrals. Armed with these techniques, the chapter concludes with applications to cardiac output, survival and renewal equations, and the scientific notion of work.

5.1 Antiderivatives

Many mathematical operations have an inverse. For example, to undo the addition of b to a we subtract b: a + b − b = a. To undo division of a by b we multiply by b: b = a. To undo exponentiation, we take logarithms: ln e^a = a. The process of differentiation can be undone by a process called antidifferentiation.

To motivate antidifferentiation, consider how long it takes an organism to develop when the rate of development depends on environmental factors such as heat, light, and humidity. For example, the developmental rate of plants and insects, which lack internal thermal regulation mechanisms, depend critically on ambient temperature. For ambient temperatures within a range defined by developmental thresholds, a plant's or insect's organismal developmental rate can often be approximated by an increasing linear function of temperature. For example, Eileen Cullen, a doctoral student at the University of California, collected data shown in Table 5.1 on the developmental rate of a particular species of stinkbug (Figure 5.2) reared in the laboratory.

Figure 5.2 A green stinkbug

Table 5.1 Developmental rates of stinkbugs

Temperature (°F)	Developmental rate (1/days)
64.4	1/89
69.8	1/58
80.6	1/37
89.6	1/25

We see from this table that a stinkbug kept at 64.4°F completes of its development in one day and all of its development in eighty-nine days. Performing linear regression on these data (i.e., to find the “best-fitting” line as discussed in Section 4.3) yields

where T is temperature in degrees Fahrenheit. This relationship is illustrated in Figure 5.3. If T(x) is the temperature at time x and F(x) denotes the fraction of development completed by the stinkbug at time x, then the preceding equation yields the rate at which F(x) changes with time; that is,

Figure 5.3 Graph of the developmental rate of stinkbugs. The red dots represent the actual data, and the line is the best-fitting line.

Thus, if we know T(x) and want to know how long it takes the stinkbug to complete development, we need to “solve” for F(x). More generally, if we are given that f(x) is the developmental rate at time x, then F(x) must satisfy

Understanding solutions of this equation is the main goal of this section.

For example, x³ is an antiderivative of 3x² since x³ = 3x². Is x³ the only antiderivative of 3x²? The answer is no. For example x³, x³ + 1, and x³ + π all have the same derivative 3x². Consequently, all are antiderivatives of 3x². Luckily, all antiderivatives of a function are related. Suppose F(x) and G(x) are antiderivatives of f(x) on some interval. Since F′(x) = f(x) = G′(x), the function

has derivative

on this interval. What functions have a derivative equal to zero on an interval? The mean value theorem implies only the constant function. Hence, there must be a constant C such that F(x) = G(x) + C, and we have the following result.

Because of this general form, finding the general form of an antiderivative amounts to finding an antiderivative of f and adding an arbitrary constant C.

Warning! What we did in part c, namely, divide by 6 because we were off by a factor of 6, worked only because 6 is a constant. It doesn't work in general. For example, suppose we wanted to find an antiderivative of e^x2. It would be incorrect to argue that since e^x2 = 2xe^x2, we are off by a factor of 2x and the antiderivative is ex². Indeed, e^x2 does not equal e^x2 as you should verify for yourself.

Corresponding to the many rules of differentiation are rules of antidifferentiation. For instance, if F(x) and G(x) are antiderivatives of f(x) and g(x), respectively, then H(x) = F(x) + G(x) is an antiderivative of h(x) = f(x) + g(x). Indeed, since the derivative of a sum is the sum of the derivatives, we obtain

In a similar manner, we can show that antiderivatives have the following properties.

Combining the antidifferentiation properties and formulas allows us to compute even more antiderivatives, as the following examples illustrate.

To find a particular antiderivative F(x) of f(x) on an interval, we need to know a value of F(x) at a particular value of x to determine the particular value of the arbitrary constant C. If we have this information, then finding the antiderivative is known as an initial value problem.

Differential equations and slope fields

An equation that involves derivatives is called a differential equation. Consider a function y = F(x). Then any equation of the form y′ = f(x), or in Leibniz's notation

is a differential equation, and solving for the antiderivative y = F(x) of f(x) (i.e., F′(x) = f(x)) corresponds to solving this differential equation.

In the next chapter, we discuss differential equations in greater detail. Here, we introduce the topic by considering a physiological phenomenon known as the Weber-Fechner law. This law describes the expected response of an animal or human subject to a stimulus, such as light or sound. More particularly, the Weber-Fechner law in physiological psychology asserts that when a subject is exposed to a stimulus, y, the rate of change of the response x with respect to y is inversely proportional to x. This statement can be written mathematically as

where k a positive constant to be determined through experiment. One can interpret this equation as saying if the stimulus x is small, then small changes in stimulus cause large changes in the response. Alternatively, if stimulus x is large, then small changes in the stimulus do not change the response much.

A particular example of the Weber-Fechner law is the logarithmic decibel scale for measuring the intensity of sound.

For some functions it is impossible to come up with an explicit expression for the antiderivative: for example, f(x) = e^−x2 and f(x) = sin x². In such cases, numerical or graphical methods can be used. One graphical approach involves slope fields, where we can use the fact that the slope of a function y = F(x) at any point (x, y) on its graph is given by the derivative F′(x). We exploit this fact to obtain a “picture” of all slopes F′(x) on the (x, y)-plane, as illustrated in the next example.

Rectilinear motion

We can use antiderivatives to understand the motion of an object along a straight line. We have previously defined velocity v(t) at time t to be the rate of change of position s(t) of an object—that is, s′(t) = v(t), and acceleration a(t) to be the rate of change of velocity v(t) of an object—that is, v′(t) = a(t). Thus, it follows that position is the antiderivative of velocity which, in turn, is the antiderivative of acceleration.

These definitions allow us to study the stooping behavior of the peregrine falcon shown in Figure 5.1. The peregrine falcon is arguably the fastest animal in the world. This long-winged raptor favors direct pursuit of other birds, and its level speed exceeds the speed of most birds upon which it preys. Peregrines gain speed by launching attacks from high and then stooping (steep diving with feet back against the tail and wings close to the body) to attain speeds of well over 300 km/h.

Recently, scientists have accurately measured speeds achieved by peregrine falcons during stooping. One falcon was logged by radar at 183 km/h ≈ (114 mph) after a dive of 305 m ≈ (1,000 ft). This is considerably slower than the 300 km/h our current model would predict. One reason for this discrepancy is that we ignored air resistance in our calculations. This shortcoming can be addressed by formulating a differential equation model that includes the effects of air resistance.

Level 1 DRILL PROBLEMS

Find the general antiderivative of the functions f shown in Problems 1 to 22.

17. f(x) = 3e^x

18. f(θ) = sec²θ for −π/2 < x < π/2

19. f(x) = x^3/2 + x^1/2 + x⁻¹ for x > 0

20. f(u) = u³ − 2u +

21. f(u) = 6u + 3 cos u

22. f(x) = 5x − 4 sin x

Find the antiderivative F(x) of the functions shown in Problems 23 to 28 satisfying the indicated initial condition.

23. f(x) = 2 with F(0) = 1

24. f(x) = 4 with F(1) = −1

25. f(x) = 2x + 3 with F(−3) = 0

26. f(x) = 4 − 5x with F(0) = 4

27. f(x) = 6x⁴ with F(1) = −2

28. f(x) = 2x⁻⁴ for x > 0 with F(2) = 0

29. a. If F′(x) = 1 − 4x, find F so that F(1) = 0.

b. Sketch the graphs of y = F(x), y = F(x) − 2, and y = F(x) + 4.

c. Find a constant C so that the largest value of G(x) = F(x) + C is 0.

30. a. If F′(x) = 2x − 1, find F so that F(2) = 0.

b. Sketch the graphs of y = F(x), y = F(x) − 2, and y = F(x) + 4.

c. Find a constant C so that the smallest value of G(x) = F(x) + C is 0.

The slope F′(x) at each point on a graph is given in Problems 31 to 34 along with one point (x₀, y₀) on the graph. Use this information to find F graphically.

35. Sketch a slope field for

for −2 ≤ x ≤ 2 and 0 ≤ y ≤ 5. Over this slope field, sketch the antiderivative of F(x) of x which satisfies F(0) = 1.

36. Sketch a slope field for

for −5 ≤ x ≤ 5 and −5 ≤ y ≤ 5. Over this slope field, sketch the antiderivative F(x) of x which satisfies F(0) = 0.

37. Sketch a slope field for

for −π ≤ x ≤ π and −2 ≤ y ≤ 2. Over this slope field, sketch the antiderivative F(x) of x which satisfies F(0) = 1.

38. Sketch a slope field for

for −1 ≤ x ≤ 1 and −2 ≤ y ≤ 2. Over this slope field, sketch the antiderivative F(x) of x which satisfies F(−1) = 0.

39. Find the general antiderivative of sin(ax) where a ≠ 0.

40. Find the general antiderivative of e^kx where k ≠ 0.

Level 2 APPLIED AND THEORY PROBLEMS

41. As discussed in Example 5, the developmental rate of a stinkbug as a function of temperature T is −0.06075 + 0.00112T. Assume that the temperature of a typical spring in Davis, California, x days after the start of the stinkbug development period is adequately modeled by the function

1. Find the function F(x) describing the amount of development completed by day x assuming that F(0) = 0.
2. Estimate at what time a stinkbug has completed development.

42. Recall that the developmental rate of a stinkbug as a function of temperature T is −0.06075 + 0.00112T. Assume that the temperature of an atypical day in Davis, California, x days after the start of the stinkbug development period is adequately modeled by the function

1. Find the function F(x) describing the amount of development completed by day x assuming that F(0) = 0.
2. Estimate at what time a stinkbug has completed development.

43. Entomologists Godfrey and Anderson* studied the developmental rates of the hydrilla tuber weevil, which is a species that consumes a weed found in ponds and waterways. As illustrated in Figure 5.5, the developmental rate as a function of temperature (in degrees Celsius) is given by

Figure 5.5 Developmental rate as a function of temperature

1. Suppose that the temperature in C° is given by the function T(t) = 30 + 10 sin(2πt) where t is measured in days. Find the fraction of development F(t) that has been completed at time t for eggs laid at time 0.
2. Estimate how many days it takes the weevil to develop to adulthood.

44. Assume that the temperature in Problem 43 is given by the function T(t) = 50 + . Using the developmental rate for the tuber weevil presented in that problem, estimate how many days it takes the weevil to complete half of its development.

45. A peregrine falcon stoops for 305 meters. Assuming a constant acceleration of 9.8 m/s², find its speed at the end of the stoop.

46. Suppose a food package is dropped out of a balloon that is 100 ft above the ground and ascending at a rate of 10 ft/s. Determine how long it takes the package to hit the ground, assuming a constant gravitational acceleration of 9.8 m/s².

47. Apollo 15 astronaut David Scott dropped a hammer and a feather on the moon to demonstrate that in a vacuum all objects fall at the same rate. He dropped both items from a height of approximately 4 ft. How long did it take each object to hit the ground? (Acceleration on the moon due to gravity is −5.2ft/s².) How long would it take for a hammer to hit the ground on Earth if dropped from a height of 4 ft? (Gravitational acceleration on Earth is −32 ft/s².)

48. Assume the brakes of a certain automobile produce a constant deceleration of 22 ft/s². If the car is traveling at 60 mi/h (88 ft/s) when the brakes are applied, how far will it travel before coming to a complete stop?

49. It is estimated that t months from now, the population of Ferndale, California, will be changing at the rate of 4 + 5t^2/3 people per month. If the current population is 2,000, what will the population be eight months from now?

50. A hypothetical study of a community suggests that t years from now the level of carbon monoxide in the air will be changing at the rate of 0.1t + 0.1 ppm/year. If the current level of carbon monoxide in the air is 3.4 ppm, what will be the level three years from now?

51. One of Poiseuille's laws for the flow of blood in an artery says that if v(r) is the velocity of flow r centimeters from the central axis of the artery, then the velocity decreases at a rate proportional to r. That is, v′(r) = ar where a is a negative constant. Find an expression for v(r) assuming that v(R) = 0, where R is the radius of the artery.

52. Suppose that a silviculturist finds that a certain type of tree grows in such a way that its height h(t)t years after planting is changing at the rate of

If the tree was two feet tall when it was planted, how tall will it be in twenty-seven years?

53. Suppose that a woman driving a sports car down a straight road at 60 mi/h (88 ft/s) sees a cow start to cross the road 200 feet ahead. She takes 0.7 seconds to react to the situation before hitting the brakes, which decelerates the car at the rate of 28 ft/s². Does she stop in time to avoid hitting the cow?

54. A hypothetical population, N, grows in such a way that at time t (years), the growth rate is given by

where N(t) is measured in thousands of individuals and N(0) = 5.

1. Find N(t).
2. What is the minimum population size? When does it occur?

5.2 Accumulated Change and Area under a Curve

In this section, we deal with the problem of finding accumulated change, which can be interpreted as the area under a curve. The early Greeks, particularly Archimedes (287–212 BC), estimated the areas of geometrical objects using the “method of exhaustion,” a precursor to integral calculus. They found increasingly better approximations by filling in areas with increasingly smaller elements of known area (much as we do later in this section in taking Riemann sums).

In elementary school you learned formulas to find areas of squares, triangles, and other polygons. You also are familiar with the formula for the area of a circle with radius r: A = πr². The Egyptians were the first to use this formula more than 5,000 years ago, but the Greeks derived the area of a circle by drawing inscribed polygons or circumscribed polygons, as shown in Figure 5.6, and then using triangles to find the area of those polygons as an approximation. This method, called the method of exhaustion, involves finding the area of a circle by inscribing polygons with increasing numbers of sides (Archimedes stopped at a ninety-six-sided polygon). The area of the circle is the limit of the areas of the inscribed polygons as the number of polygonal sides increases.

Figure 5.6 Using the limit of a sequence of inscribed polygons to find the area of a circle

Figure 5.7 Area under a curve y = f(x)

In this section, we focus on estimating the area under a curve y = f(x) over an interval a to b. As illustrated in Figure 5.7, this means estimating the area defined by the region bounded by the curves y = f(x) (with f(x) ≥ 0 on [a, b]), x = a, x = b, and y = 0. Similar to the method of exhaustion, we find these areas by approximating them with collections of finer and finer rectangles.

To motivate finding area under a curve, we show that area under a curve corresponds to accumulated change. We do this in the context of two biological processes: physiological time for insects and plants and disease incidence as studied by epidemiologists.

Physiological time: degree-days

Plants and insects often require a certain amount of heat to develop from one stage in their life cycle to another stage in their life cycle. This measure of accumulated heat is known as physiological time, and the units used are called degree-days—the accumulated product of time and temperature between the organism's lower and upper developmental thresholds. The lower developmental threshold is the temperature below which the insect or plant cannot develop, while the upper development threshold is the temperature above which it cannot develop.

To simplify the presentation, we initially assume that the temperature remains between the lower and upper developmental thresholds. The more general case is considered later. One degree-day is one day (24 hours) with the temperature one degree above the lower developmental threshold. For example, if the lower developmental threshold of the organism is 47°F and the temperature remains at 48°F for one day or 47.2°F for five days, then in each case, one degree-day is accumulated (i.e., 1 × (48 − 47) = 5 × (47.2 − 47) = 1).

The concept of degree-days is used widely in agriculture and developmental biology. In agricultural publications we may come across the following types of statements: “Around 1539 degree-days are required for sweet corn to ripen, assuming a lower development threshold of 50°F”; or, “corn earworms (pests of corn) have a lower developmental threshold of 54.7°F and require 760 degree-days to develop from egg to adult”. These statements allow us to estimate the time it takes sweet corn to mature or corn earworms to develop in geographical regions with different temperature profiles, as well as to anticipate how these times may change in response to global warming.

Unlike the preceding example, temperatures in fields vary over time. Consequently, computing the accumulation of degree-days, as we shall see, requires finding the area of an appropriate region defined by the temperature curve and the lower developmental threshold.

For example, the temperature in Lincoln, Nebraska, on June 23, 2006 is given by the function f(t)°F, illustrated in Figure 5.8, where t going from 0 to 1 represents one day from midnight to midnight. If an organism of interest (say, sweet corn) has a lower developmental threshold of 50°F, then, as we explore in the next example, the total accumulated degree-days is given by the area between the curves y = f(t) and y = 50 from t = 0 to t = 1. This area corresponds to the shaded area in Figure 5.8.

Figure 5.8 Temperature for Lincoln, Nebraska, on June 23, 2006. The shaded area corresponds to the accumulated degree-days for an organism with a lower developmental threshold of 50°F.

The Bombay plague epidemic

In epidemiology, scientists keep track of various rates associated with disease, including the incidence rate, which measures the number of new disease cases per unit of time (e.g., day or week), and the mortality rate, which reports the number of deaths due to the disease per unit of time. For instance, during the outbreak of the plague in Bombay in 1905–1906, the weekly mortality rates due to the plague were recorded, and the values obtained are plotted in Figure 5.10.

Figure 5.10 Mortality rate from plague in Bombay (now called Mumbai) from December 17, 1905 to July 21, 1906. Data shown in red and the fitted function in blue.

In a landmark paper,* two mathematicians, W. O. Kermack and A. G. McKendrick, showed that these data could be reasonably well fitted by the function

where t is measured in weeks and the hyperbolic secant function sech x equals . If we want to estimate the total number of deaths using this function, what do we need to compute? Consider a small interval of time, from t to t + Δt. Since the mortality rate over this interval is given approximately by f(t), the number of deaths over this time interval is approximately f(t)Δt, that is, the area of a rectangle of width Δt and height f(t). Notice how the units work out in this product: f(t) has units of deaths/week and Δt has units of weeks. The product f(t)Δt has units of deaths. These arguments suggest that the area under the curve f(t) from t = 0 to t = 30 equals the total number of deaths, as discussed in the next example. Since we know the actual number of deaths—we simply add up all the weekly data—the next example examines how the area under f(t) approximates the actual number of deaths.

In Figure 5.11 notice that when the curve is on the rise, as in the first three rectangles, the area is overestimated (the green area above the curve), and when the curve is on the decline, the area is underestimated (the white area below the curve). This is a result of the height of the rectangles being defined by the value of the function on the right side of each interval. The reverse would be true if the height of the rectangles were defined by the value of the function on the left side of each interval.

The area problem

The previous examples illustrate the importance of finding areas under curves. These examples also show that we can approximate areas by approximating the region with rectangles, computing the area of each rectangle, and summing up the areas. This observation is the key that unlocks the area problem. We pursue this approach further in the following example.

Example 4 suggests that area under x² over the interval [0, 1] is . But how can we really be sure that these numbers converge to ? We address this question in the next example.

Examples 4 and 5 provide the core idea of how to define area under a nonnegative function y = f(x) from x = a to x = b. First, we divide the interval [a, b] into n equally spaced subintervals of width Δx = . Let

To approximate the height of f over a subinterval [a_i, a_i+1], choose a point x_i on the interval [a_i, a_i+1]. The points x_i are called sample points. In our examples, we chose left or right endpoints as our sample points, but we could have picked any point in each interval. The height of f over [a_i, a_i+1] is approximately f(x_i). The area of f over [a_i, a_i+1] is approximately f(x_i) Δx. Adding all these rectangular areas up yields

This sum is known as a Riemann sum after the brilliant mathematician Georg Friedrich Bernhard Riemann (1826–1866; see the in Problem Set 5.2). Now we write this sum succinctly, using the summation notation presented in the regression subsection of Section 4.3.

We have seen how area can be approximated by a Riemann sum and how this approximation improves as n become large, approaching the true area as n → ∞. Therefore, we write

We cannot know that the method really works unless we have a theorem that tells us that a limit exists and that this limit is independent of the way we choose the sample points in the subintervals. For continuous functions such a theorem does exist, but its proof is a topic for a course in real analysis. (The “real” refers to real-valued functions in contrast to the “complex” of complex numbers and complex-valued functions.)

In Problem Set 5.2, some of the problems require one or more of the following summation formulas. These formulas can be verified using mathematical induction.

Level 1 DRILL PROBLEMS

First sketch the region under the graph of y = f(x) on the interval [a, b] in Problems 1 to 12. Then approximate the area of each region by using right endpoints and the formula

for Δx = and the indicated values of n.

1. f(x) = 2x + 1 on [0, 1] for n = 4

2. f(x) = 4x + 1 on [0, 1] for n = 8

3. f(x) = x² on [0, 2] for n = 4

4. f(x) = x² on [0, 2] for n = 6

5. f(x) = x³ on [1, 3] for n = 4

6. f(x) = 4x² + 2 on [0, 1] for n = 4

7. f(x) = x² + x³ on [0, 1] for n = 4

8. f(x) = e^x on [0, 1] for n = 4

9. f(x) = x⁻¹ on [1, 2] for n = 4

10. f(x) = on [1, 4] for n = 4

11. f(x) = cos x on for n = 4

12. f(x) = x + sin x on for n = 3

Use a calculator to estimate the area under the curve y = f(x) on each interval given in Problems 13 to 18 as a sum of ten terms evaluated at right endpoints.

13. f(x) = 4x on [0, 1]

14. f(x) = x² on [0, 4]

15. f(x) = cos x on

16. f(x) = x + sin x on

17. f(x) = ln(x² + 1) on [0, 3]

18. f(x) = e^−3x2 on [0, 1]

Use a summation formula in Problems 19 to 24.

19. Prove that

for L_n as defined in Example 4.

20. Use Riemann sums and left endpoints to prove that the area under y = x from x = 0 to x = 2 equals 2.

21. Use Riemann sums and right endpoints to prove that the area under y = x from x = 0 to x = 4 equals 8.

22. Use Riemann sums and right endpoints to prove that the area under y = x³ from x = 0 to x = 4 is 64.

23. Use Riemann sums and left endpoints to prove that the area under y = x³ from x = 0 to x = 2 is 4.

24. Use Riemann sums and right endpoints to prove that the area under y = x + 3x² from x = 0 to x = 2 is 10.

Level 2 APPLIED AND THEORY PROBLEMS

25. The lower developmental threshold of sweet corn is 50°F and requires 1,587 degree-days for maturing. If the temperature were to remain a constant 75°F, how long would it take for the corn to mature?

26. The pistachio has a lower developmental threshold of 50°F and requires 1,197 degree-days for shell hardening. If the temperature were to remain a constant 72°F, how long would it take for the pistachio's shell to harden?

27. The black turtle bean has a lower developmental threshold of 41°F and requires 1,365.5 degree-days for 50% anthesis (i.e., until 50% of all the flowers have blossomed). If the temperature were to remain a constant 68.5°F, how long would it take to reach the required 50% anthesis?

28. Estimate mortality due to the plague in Bombay (Mumbai) by approximating the region under

deaths per week from t = 0 to t = 30 with rectangles of width 15 weeks. Would you expect your answer to be more or less accurate than the result of Example 3?

29. Estimate mortality due to the plague by approximating the region under

deaths per week from t = 0 to t = 30 with rectangles of width three weeks and height given by their right endpoints. Would you expect your answer to be more of less accurate than the result of Example 3?

30. The weekly rate of cases of influenza A (strain unknown) studied by WHO/NREVSS during the 2003–2004 season is plotted in Figure 5.15. Estimate the total number of cases (i.e., the area under the curve) over the interval [40, 56] using the right endpoints of two-week intervals. Sketch the corresponding rectangles in the figure.

Figure 5.15 Weekly rate of cases of influenza A

31. Repeat Problem 30 using left endpoints.

32. The daily maximum (max) and minimum (min) temperatures at Westmoreland, California, for the period from April 1 to 9 are given in Table 5.3.

Table 5.3 Early morning minimum and mid-afternoon maximum temperatures at Westmoreland, California, for nine days (minimum on April 10 included for later use)

The lower developmental threshold for cotton is k = 60°F. If M_i and m_i are used to denote the max and min temperatures on day i, respectively, then estimate the degree-day accumulation over the nine-day period using this formula:

degree-days accumulated on day i

Note that the logic behind this formula is that on days for which k ≤ m_i we use the average temperature above the threshold, and on days for which m_i < k < M_i we estimate that the temperature is above threshold for a proportion of the day, and then use the average as applying over this period of time to obtain the middle expression in the equation.

33. The lower developmental threshold for the elm leaf beetle is 52°F. Use the formula in Problem 32 to estimate the degree-day accumulation for the elm leaf beetle in Stockton, California, over a two-week period of time using the data listed in Table 5.4.

Table 5.4 Early morning minimum and mid-afternoon maximum temperatures at Stockton, California (fire station # 4) for 14 days (minimum on March 15 included for later use)

34. In the figure that follows, m_i and m_i+1 are the minimum daily temperatures on consecutive days i and (i + 1), respectively, which are assumed to be exactly one day apart. M_i is the maximum daily temperature on day i and is assumed to occur exactly half way between the two minimum temperatures. The parameter k is the lower developmental threshold for a particular species.

If these values satisfy M_i > m_i > k and k > m_i+1, then show that the degree-day accumulation over the one-day period, as depicted by the green area in the figure, is given by the expression

Alternatively, if M_i > m_i > k and M_i > m_i+1 > k then show that the degree-day accumulation over day i is given by

35. The obliquebanded leafroller, is an agricultural pest with a lower developmental threshold of 43°F. Estimate the accumulated degree-days for this insect growing in Westmoreland, California, over the period from the morning of April 2nd to the morning of April 3rd using the data in Table 5.3 (see Problem 32), and the formula presented in Problem 34.

36. The codling moth, is an agricultural pest with a lower developmental threshold of 50°F. Estimate the accumulated degree-days for this insect growing in Stockton, California, over the period from the morning of April 1st to the morning of April 2nd using the data in Table 5.3 (see Problem 32), and the formula presented in Problem 34.

37. Assume the temperature in degrees Fahrenheit is given by

where t is time in days. Assume the lower development threshold is 40°F and estimate the number of degree-days that accumulate from t = 0 to t = 10 days using time intervals of width two.

38. Use the temperature variation model in Problem 37 to estimate the number of degree-days accumulated from t = 0 to t = 20 for citrus flower which has a lower developmental threshold of 49°F. Use time intervals of width four days.

39. Suppose the velocity v (in meters per second) of a runner during the first few seconds of a race is given by

Plot these points in the t v plane. Sketch the velocity curve. Estimate the distance traveled by the runner by estimating the area under the velocity curve; use rectangles with heights given by a right endpoint approximation.

40. A pneumotachograph is a medical device used to measure the rate at which air is exhaled by a patient's lungs. Suppose Figure 5.16 shows the rate of exhalation for a particular patient. The area under the graph provides a measure of the total volume of air in the lungs during exhalation. Use a Riemann sum with n = 8 and right-endpoint subinterval representatives to estimate the volume.

Figure 5.16 Rate of exhalation

41. An industrial plant spills pollutant into a lake. Suppose that the pollutant spread out to form the pattern shown in Figure 5.17. All distances are in feet.

Figure 5.17 Pollutant spill

Use a Riemann sum with n = 6 and right-endpoint subinterval representatives to estimate the area of the spill.

42.

In this section, we saw that history honored Georg Riemann by naming an important process after him. In his personal life Riemann was frail, bashful, and timid; but in his professional life, he was one of the all-time giants in mathematics. In his book, Space Through the Ages, Cornelius Lanczos wrote, “Although Riemann's collected papers fill only one single volume of 538 pages, this volume weighs tons if measured intellectually. Every one of his many discoveries was destined to change the course of mathematical science.” One of these discoveries is the Riemann zeta function

which had already been considered by Euler. In this function, the sum is over all natural numbers n, and the product is over all prime numbers. While Euler considered the zeta function in the context of a real variable z, Riemann considered the zeta function in the context of a complex variable z. Except for a few trivial exceptions, the roots of ζ(s) all lie between 0 and 1. Riemann conjectured that the zeta function had infinitely many nontrivial roots, all with their real parts equal to 1/2. This is the famous Riemann hypothesis, which remains one of the most important unsolved problems in mathematics. The Clay Mathematics Institute, for example, has offered a million dollar prize for solving this conjecture. (For more information about this institute, visit http://www.claymath.org.) So you can become a millionaire doing mathematics!

Write a paper on Georg Riemann; in particular, discuss this million dollar prize.

5.3 The Definite Integral

Previously we defined area under a nonnegative function as the limit of a Riemann sum. In this section we define this limit for any continuous function (positive or negative) and develop its geometrical meaning as well as its properties.

For a nonnegative continuous function f(x) from x = a to x = b, we defined the area under the curve as

where Δx = and x_i is a point from the interval [a + (i − 1) Δx, a + iΔx]. Theorem 5.1 from the previous section implies that

exists and is independent of the sample points x_i whenever f is continuous. When f takes on negative values, the integral no longer corresponds to the area under the curve, but the signed area as we soon shall see. The existence of the limit is so important that Leibniz (see , page 538) developed the following special notation for it.

In the definition of the definite integral, the function f that is being integrated is called the integrand; the interval [a, b] is the interval of integration; and the endpoints a and b are called, respectively, the lower and the upper limits of integration. The variable x is called the variable of integration. Notice that in taking the limit the Greek letters are supplanted by the Roman letters: Δ becomes a d and Σ becomes an elongated S.

Note that the two expressions obtained for the integrals in Example 1 must be the same for the theory of integration to be consistent. This will be demonstrated in Section 5.5, after we consider how to change the variable of integration.

Geometrical meaning of the definite integral

We saw previously that f(x)dx corresponds to the area under the curve y = f(x) provided that f(x) ≥ 0 from x = a to x = b. The following example uses this fact to evaluate an integral.

What happens if f(x) changes sign on the interval? In this case, f(x)dx is the signed area of the region R determined by the curve y = f(x) and the lines y = 0, x = a, and x = b. More specifically, if f changes sign on the interval [a, b], then the region R breaks up into two pieces: one piece, call it R⁻, that lies below the x axis as illustrated by the red region in Figure 5.19 and another piece, call it R⁺, that lies above the x axis as illustrated by the green region in Figure 5.19.

If A⁺ and A⁻ denote the areas of R⁺ and R⁻, respectively, then

Figure 5.19 Geometry of the definite integral

Properties of definite integrals

Integrals satisfy several useful properties, some of which are summarized in the following box.

These properties can be proved using Riemann sums and limit laws (see Problem Set 5.3).

Combining the properties of integrals with the geometrical interpretation of the integral allows one to quickly compute certain integrals.

We conclude this section with some additional properties of the definite integral.

Nonnegativity can be proved using the definition of a definite integral, and nonnegativity, in turn, can be used to prove dominance and bounding. For example, to prove dominance, suppose that f(x) ≥ g(x) from x = a to x = b. Then, f(x) − g(x) ≥ 0 from x = a to x = b. Applying the property of differences and nonnegativity yields

Hence,

Figure 5.24 Geometrical depiction of the splitting property where the signed area f(x)dx from x = a to x = b equals the signed area f(x)dx from x = a to x = c plus the signed area f(x)dx from x = c to x = b

If we set M and m to be the maximum value and minimum value, respectively, of f on the interval [a, b], then the bounding property provides crude estimates for the value of a definite integral. When working through detailed computations by hand, these crude estimates allow us to see whether our work has resulted in a reasonable answer. Finally, a proof of the splitting property is somewhat subtle, but geometrically intuitive, as illustrated in Figure 5.24.

Level 1 DRILL PROBLEMS

Express the limits in Problems 1 to 6 as definite integrals of the form f(x)dx.

Express the definite integrals in Problems 7 to 12 as limits of Riemann sums.

First sketch the region under the graph of y = f(x) on the interval [a, b]. Then use the interpretation of the definite integral f(x)dx as a signed area to evaluate the integrals in Problems 13 to 16.

Evaluate each of the integrals in Problems 17 to 22 by using the following information together with the sum rule and the splitting property:

Use integral properties to establish the statements in Problems 23 to 26.

27. Use the fact that x²dx = and the geometrical interpretation of the integral to find

28. Use the graph of y = cos x to evaluate

on the indicated interval.

Using right endpoints with n = 5, approximate the definite integrals in Problems 33 to 36. Indicate whether each approximation is greater than or less than the actual definite integral.

Use Riemann sums with right endpoints, along with a summation formula (see Section 5.2) to evaluate the integrals in Problems 37 and 38.

Show that each statement about area in Problems 39 to 42 is true in general, or if not, provide a counterexample. It will probably help to sketch the indicated region for each problem.

39. If C > 0 is a constant, the region under the line y = C on the interval [a, b] has area A = C(b − a).

40. If C > 0 is a constant and b > a ≥ 0, the region under the line y = Cx on the interval [a, b] has area A = C(b − a).

41. Let f be a function that satisfies f(x) ≥ 0 for x in the interval [a, b]. Then the area under the curve y =[f(x)]² on the interval [a, b] must always be greater than the area under y = f(x) on the same interval.

42. A function f is said to be even if f(−x) = f(x). If f is even and f(x) ≥ 0 throughout the interval [−a, a], then the area under the curve y = f(x) on this interval is twice the area under y = f(x) on [0, a].

Level 2 APPLIED AND THEORY PROBLEMS

43. We saw in Example 8 that Thompson seedless grapes have a lower developmental threshold of 50°F and require approximately 3,000 degree-days to ripen. Suppose the temperature in the field is given by

over a 15 day period where x is time in days. Write an expression involving definite integrals that represents the number of degree-days accumulated from x = 0 to x = 15, and evaluate this expression. What percentage of ripening took place during this 15 day period?

44. Assume for a particular variety of grape that the lower developmental threshold is 30°F and that it takes 2,500 degree-days to ripen. If the temperature in degrees Fahrenheit over a particular 15 day period is given by

where x is the time in days, then write an expression involving definite integrals that represents the number of degree-days accumulated from x = 0 to x = 15. Evaluate this expression and calculate the percentage ripening the occurred over this 15 day period.

45. Assume that over a 50 day period the rate at which individuals in a population die from disease is given by the function f(x), where x is days and the units of f(x) are individuals per day. If f(x) has the form

then write an expression involving definite integrals that represents the number of deaths over this 50 day period and evaluate this expressing using your knowledge of how to calculate areas of geometrical objects.

46. Assume that over a 60 day period the rate at which individuals in a population die from disease is given by the function f(x), where x is days and the units of f(x) are individuals per day. If f(x) has the form

then write an expression involving definite integrals that represents the number of deaths over this 60 day period and evaluate this expressing using your knowledge of how to calculate areas.

47. A function f is said to be odd if f(−x) = −f(x). Show that if f is odd on the interval [−a, a], then f(x)dx = 0.

48. Prove the sum rule for integrals using the definition of a definite integral and properties of limits and summations.

49. Generalize the splitting property by showing that for a ≤ c ≤ d ≤ b

whenever all these integrals exist.

50. Prove the bounding rule for definite integrals: If f is continuous on the closed interval [a, b] and m ≤ f(x) ≤ M for constants, m, M, and all x in the closed interval, then

51. Gilles de Roberval (1602–1675) started his study of mathematics at the age of 14 years. He had a distinguished career and was a founding member of the Académie Royale des Sciences. In 1669 he invented the Roberval balance (Figure 5.25), which is still widely used today.

Figure 5.25 Roberval's balance variations

Roberval had a chair position as professor of mathematics at the Collége Royale. Every three years there was a contest by competitive examination (written by the incumbent!) to determine who would occupy this position. It is said for this reason, Roberval kept many of his techniques of integration secret until his death. We do know, however, that he developed powerful methods in the early study of integration. These methods are described in his treatise Traité des indivisibles. For instance, in this treatise, he computed the definite integral of sin(x) using obscure trigonometric identities. It is these identities that the computer algebra system uses to simplify the sum and take the limit. For this quest, you may stand on the shoulders of Roberval and use technology to write a Riemann sum for

using right endpoints. Then use technology to simplify this sum to evaluate this definite integral.

5.4 The Fundamental Theorem of Calculus

In this section, we discuss the evaluation theorem and the fundamental theorem of calculus. These theorems link antiderivatives, which we can compute relatively easily, with definite integrals and Riemann sums. We show that antiderivatives and Riemann sums, when they both exist, are the same thing.

Evaluation theorem and net change

To appreciate the power of the evaluation theorem, compare the work of Example 1 with work required to find the limit of the Riemann sum for each of these functions.

To simplify our work, we introduce the following notation:

Or, when there is no ambiguity,

As illustrated in Example 3 of Section 5.2, an important interpretation of the evaluation theorem is that it relates the accumulated change of a function over an interval to the area under its derivative. Thus, we have the following procedure for calculating accumulated change.

For instance, if F(x) represents the total number of births in the world by year x, then F(b) − F(a) is the number of births that occurred between years a and b. Now suppose that f(x) = F′(x), which by definition implies that F′(x) is the instantaneous birthrate x. The evaluation theorem asserts that the area under the instantaneous birthrate equals the accumulated change in births. Does this make sense? At the level of units it certainly does. The instantaneous birthrate has units births per year. Hence, the area over an interval of time has units births per year multiplied by years. This equals births, which has the same units as the accumulated change. Moreover, if we broke up the interval [a, b] into small subintervals, then the number of births in a given subinterval would be approximately the instantaneous birthrate at some point in the subinterval in question multiplied by the length of the subinterval, that is, the area of a rectangle lying above this subinterval. Adding up all these little rectangular areas would give us simultaneously an approximation for the number of births over [a, b] and the area under F′. More generally, if one integrates the rate of change over an interval [a, b], then one gets the accumulated change over [a, b].

Fundamental theorem of integral calculus

The answer to Example 3 gives us the net increase in the length of the ram's horn over the whole six-year period. Suppose, though, that we want to find the net increase at any age x during the six-year period from age 3 to age 9. The net increase is given by the function

for 3 ≤ x ≤ 9. For example, we found F(9) ≈ 33.8148 cm. In writing the integral defining F(x), we took advantage of the fact that the variable of integration is a dummy variable (it disappears once the integration has been performed). Consequently, to avoid confusion, we choose the variable u of integration to be different from our time variable x. By the evaluation theorem,

Figure 5.27 Estimated growth of a ram's horn in centimeters

Plotting F(x) from x = 3 to x = 9 as shown in Figure 5.27 illustrates how the net increase of the length of the horn changes over this time interval. Notice, as we might expect, the length of the horn is increasing at a decreasing rate.

We can now generalize this idea for any continuous function f defined on the interval [a, b] by considering the function

If we interpret f(x) as a rate, then F(x) describes how the accumulated change varies as a function of x. Alternatively, F(x) describes how the signed area under f confined to the interval [a, x] varies as a function of x.

Using the fundamental theorem, we can easily compute the accumulation of degree-days.

Indefinite integrals

Since the fundamental theorem of calculus ensures the existence of antiderivatives via integrals, it is appropriate to introduce a notation for the general antiderivative.

The fact that the indefinite integral has no upper limit of integration or lower limit of integration distinguishes it from a definite integral. It is important to remember that the indefinite integral represents a family of functions (a particular function plus an arbitrary constant), whereas the definite integral of a specified function represents a value.

Level 1 DRILL PROBLEMS

In Problems 1 to 8, evaluate the definite integral.

Find the indefinite integrals in Problems 9 to 16.

Compute F′(x) for the functions given in Problems 17 to 22.

Figure 5.29 Graph of f

1. For what values of x does F(x) have a local maximum or minimum?
2. For what values of x is F concave up? concave down?
3. At what values of x does F(x) achieve a global maximum? global minimum?
4. Sketch the graph of F(x).

27. Let G(x) = g(u)du where the graph of g is shown in Figure 5.30.

Figure 5.30 Graph of g

1. For what values of x does G(x) have a local maximum or minimum?
2. For what values of x is G concave up? concave down?
3. At what values of x does G(x) achieve a global maximum? global minimum?
4. Sketch the graph of G(x).

Level 2 APPLIED AND THEORY PROBLEMS

28. Use the model for bighorn rams formulated by Jorgenson and colleagues (see Example 3) to find the net increase in length of a ram's horn from x = 3 to x = 7.

29. Use the model for bighorn rams formulated by Jorgenson and colleagues (see Example 3) to find the net increase in length of a ram's horn from x = 5 to x = 9.

30. Citrus flowers in Tulare County, California, have a lower developmental threshold of 49°F and require approximately 767 degree-days to reach petal-fall (i.e., 50% of citrus flowers have lost their petals). Suppose the temperature in the fields is given by

where x is the time in days.

1. Write an expression involving definite integrals that represents the time, x, at which 767 degree-days have accumulated.
2. Use technology to estimate the time x.

31. Sweet corn in western Oregon has a lower developmental threshold of 50°F and requires approximately 1,597 degree-days to reach maturity. Suppose the temperature in the fields is given by

where x is the time in days.

1. Write an expression involving definite integrals that represents the time, x, at which 1,597 degree-days have accumulated.
2. Use technology to estimate the time x.

32. The rate of change of ant diversity along an elevational gradient is given by

where x is elevation above sea level measured in kilometers. If F(1) = 6.9, find an expression for F(x), the number of ant species at an elevation of x km. Compare your answer to Example 4 of Section 2.7.

33. Prove the evaluation theorem (Theorem 5.2) using the fundamental theorem of calculus (Theorem 5.3).

5.5 Substitution

In the next two sections, we discuss three techniques of integration: substitution, integration by parts, and partial fractions. The first two of these techniques are counterparts to rules of differentiation. Unlike differentiation, techniques of integration are incomplete; not every function has an elementary representation of its indefinite integral. Consequently, part of the skill you need to acquire is knowing which integration techniques apply to which functions. Since it is possible to compute integrals using technology, you may be asking: Why bother with these techniques? There are several reasons. First, by learning integration techniques and determining when and in which order to use them, you gain insight into how these techniques work. Second, sometimes technology needs a helping hand; implementing an integration technique (especially substitution) by hand may allow technology to complete a calculation it could not otherwise do. Third, computing integrals builds important mathematical skills: it's like lifting weights to improve your strength for sports.

Substitution for indefinite integrals

Our integration efforts begin with an antidifferentiation form of the chain rule. Consider the integral

Basic rules of antidifferentiation provide no direct way of computing this integral. However, look carefully at this integral, and note that the derivative of the denominator equals the numerator. This observation suggests introducing a new variable, u = x² + 5, that, on differentiation, yields = 2x. In this next step we replace 2xdx = dx in the integrand with du. A mathematical justification for this step is dealt with in more advanced texts. Finally, integrating with respect to the variable u, we obtain

Although we have not justified our steps rigorously, we can verify our answer by differentiating. Since

it follows that ln(x² + 5) + C is the general antiderivative of .

To see why this approach worked, consider the integral

for functions f and g. In our example dx, we have f(x) = 1/x and g(x)= x² + 5. If F is an antiderivative of f, then

Equivalently, if we make the change of variables u = g(x), then

We summarize this procedure in the following box.

This procedure may seem difficult now, because you may not be sure of what to let the variable u represent. Just remember, at least initially, you are looking for one part of the integrand that is the derivative of another part of the integrand. If you practice enough, things will get easier!

As the previous two examples illustrate, after making a substitution and simplifying, there should be no x values in the integrand. Sometimes eliminating all the x's requires additional work.

Some of you will, no doubt, have access to a calculator or a computer program that can assist in the process of integration. Although technology is very useful, there are times when evaluating an integral by hand will result in a simpler form of the answer. In other cases, technology will give an incomplete form that needs to be adjusted.

Lest you think that if you purchase a calculator you will not need to study and master techniques of integration, consider the following example.

Substitution for definite integrals

We have two methods for dealing with definite integrals. The first is to return to the original variable (as we did with indefinite integrals). The second is to keep track of the change of variables in the limits of integration. We illustrate both of these methods by revisiting Example 4.

Method I: Return to the original variable.

Method II: Keep track of the change of variables in the limits of integration.

Consider the general case of this second method. Let u = g(x) and F(x) be an antiderivative of f(x). Then,

We summarize this observation in the following box.

Level 1 DRILL PROBLEMS

Problems 1 to 8 present pairs of integration problems; one of the pair will use substitution and one will not. As you are working these problems, think about when substitution may be appropriate.

Use substitution to find the indefinite integrals in Problems 9 to 16.

Use substitution to evaluate the definite integrals in Problems 17 to 24.

Level 2 APPLIED AND THEORY PROBLEMS

28. In Example 7, U.S. population growth was modeled by

where t is decades after 1790. If each person eats food at a rate of one ration per year, find the total number of rations of food eaten in the United States between 1800 and 1900.

29. Assume that a dust mite population starts with 10 dust mites and grows at a rate of 10e^0.3t dust mites per hour. How many dust mites will there be one day from now?

30. Suppose an environmental study indicates that the ozone level, L, in the air above a major metropolitan center is changing at a rate modeled by the function

parts per million per hour (ppm/h) t hours after 7:00 A.M.

1. Express the ozone level L(t) as a function of t if L is 4 ppm at 7:00 A.M.
2. Use the graphing utility of your calculator to find the time between 7:00 A.M. and 7:00 P.M. when the highest level of ozone occurs. What is the highest level?

31. The Gompertz law of tumor growth is given by the equation

where N is the size of the tumor, t is time (measured in days), b is the asymptotic size of the tumor, and a is a measurement of the tumor growth rate. Assume a = 1 and b = 10. Integrate both sides of the Gompertz equation and solve for N in terms of t. To get rid of the integration constant, assume that N equals 5 at time t = 0.

32. In Example 6 of Section 1.5, we modeled the uptake of glucose by bacterial populations off of the coast of Peru by the function

where x is micrograms of glucose per liter. Suppose the concentration of glucose x is decaying exponentially in time: x(t) = 100e^−0.01t micrograms per liter where t is measured in hours.

1. Write a function U(t) that describes how the uptake rate is changing in time.
2. Determine the net uptake of a cell from t = 0 to t = 6 hours.

33. In Example 5 of Section 2.4, we found that the rate at which wolves kill moose can be modeled by

where x is measured in number of moose per km². Suppose that the density of moose is increasing exponentially according to the function x(t) = 0.1e^0.2t moose per km² where t is measured in hundreds of days. Determine the number of moose killed by a wolf from t = 0 to t = 3.

34. In Problem 42 in Problem Set 2.4, we examined how wolf densities in North America depend on moose densities. We found that the following function provides a good fit to the data:

where x is number of moose per km². Assume the moose density is increasing exponentially according the function x(t) = 0.1e^0.2t moose per km² where t is measured in hundreds of days. Determine the change in the wolf density from t = 0 to t = 3.

5.6 Integration by Parts and Partial Fractions

In this section, we consider two important techniques of integration that will be useful in later chapters.

Integration by parts

Integration by parts is a procedure based on inverting the product rule for differentiation. To derive a formula for this procedure, we begin with the product rule for differentiating functions f(x) and g(x), assuming these derivatives exist.

If we let u = f(x) and v = g(x), then du = f′(x)dx, dv = g′(x)dx, and we obtain the following simplified formula.

To evaluate integrals using integration by parts, we want to choose u and dv so that the new integral is easier to integrate than the original integral.

We noted that there were two possible choices for u and dv in Example 1. The other choice is to let u = e^x and dv = x dx. If we make this choice, and substitute into the formula for integration by parts, then we obtain

(Try this yourself to practice the technique.) In this case, instead of simplifying the problem and solving it, we just made it more complicated! Thus, it is important to try different substitutions to see which works best.

In the next example, it is necessary to apply integration by parts more than once. As you will see, though, when we do so a second time, we return to the original integral.

Sometimes you need to combine techniques to conquer an integral.

Integration by parts extends to definite integrals in a natural way.

Partial fractions

Partial fractions is an integration technique enable us to integrate any rational function; that is, functions of the form

Integration problems involving rational functions arise commonly in enzyme kinetics, evolutionary games, and population dynamics. For example, in describing the growth of a population of size N(t) with a growth that is positively impacted by its own size, we may encounter an integral of the form

The appropriate integration procedure is to write (expand) the rational function into a sum of two simpler functions that we can directly integrate. More specifically, we try to find constants A and B such that

Equivalently, multiplying both sides by N(N − 1) gives

This equation holds for all N only if A = −1 and A + B = 0; in other words, B = −A = −(−1) = 1. Thus, we can write

and

Although it is possible to deal with all rational functions, we confine our discussion to rational functions f(x) = P(x)/Q(x), such that Q(x) can be expressed as a product of n distinct linear factors:

If the degree of P(x) is less than the degree of Q(x) (i.e., n > m), then we can always find constants A₁, A₂,..., A_n such that

For integration methods for more general rational functions, go online or read another calculus text.

Alternatively, if the degree of P(x) is greater than or equal to the degree of Q(x), then we can perform long division and factor the remainder term. When Q(x) can be decomposed into linear factors, there is a simple method to determine the coefficients. This method is the Heaviside “cover-up” method—named after Oliver Heaviside (see Problem 35, , in Problem Set 5.6).

Level 1 DRILL PROBLEMS

Find each integral in Problems 1 to 10.

Find the exact value of the definite integrals in Problems 11 to 16 using integration by parts. Check your answer by using a calculator to find an approximate answer correct to four decimal places.

Find the indicated integrals in Problems 17 to 22.

In Problems 23 to 28, first use an appropriate substitution and then use integration by parts or partial fractions to evaluate the integral. Remember to give your answers in terms of x.

30. a. Evaluate ∫ cos²x dx. Hint: Use the trigonometric identity: cos²x = (1 + cos(2x)).

b. Use part a to evaluate ∫ x cos²x dx using integration by parts.

Level 2 APPLIED AND THEORY PROBLEMS

31. The 1988 film Stand and Deliver, starring Edward James Olmos pictured above, provides an alternative perspective—tabular integration—on integration by parts. This technique involves writing a table with two columns, one labeled D for differentiation and another labeled I for integration. The first row of the D column contains u, the part to be differentiated in the original integral. The second row in the D column contains . The third row in the D column contains . Proceed in this manner until the product of the functions in the last row either equals 0 or is a constant multiple of what you started with. The first row of the I column contains dv, the part to be integrated. For the second, third, and so on rows in the I column, place the successive integrals. For example, rework Example 1, namely ∫ xe^xdx, with u = x and d v = e^x:

Now, draw diagonal lines from the first element of the D column to the second element of the I column, from the second element of D to the third element of I and so forth. Multiply the elements at the ends of each of the diagonal lines, take an alternating sum of these products, and add the integral of the product of terms in the last row. For Example 1, and the table shown here we have

which is the same result shown in Example 1. Use this method to find the integral in Example 2.

32. Use the tabular method from Stand and Deliver (Problem 31) on the integral in Example 3.

33. Use the tabular method from Stand and Deliver (Problem 31) on the integral in Example 4.

34. Contrast the method of integration by parts as illustrated by the examples in the text and the tabular method from Stand and Deliver (Problem 31).

35. * Oliver Heaviside (1850–1925) was born in the same London slums as Charles Dickens. When young, Heaviside had scarlet fever, which left him partly deaf. He finished his schooling in 1865 at age 16. He was a top student but failed geometry. Heaviside went to work as a telegrapher, which drew him into the study of electricity. He read Maxwell's Treatise on Electricity and Magnetism and managed to reduce Maxwell's whole field theory into two equations.

Next Heaviside picked up the new ideas of vector analysis, which inspired him to develop a theory he called operational calculus, a powerful tool. However, its foundation was not rigorous, so his methods were not accepted by many mathematicians of his day. Only people like Kelvin, Rayleigh, and Hertz saw the brilliance that was driving Heaviside faster than method could follow. Heaviside knew what he was doing. He growled at his detractors, “Shall I refuse my dinner because I do not fully understand... digestion?”

Like vector analysis, Heaviside's calculus stood the test of time. So did the rest of his work. For example, he gave us the theory for long distance telephones. Ultimately, Heaviside grew sick of fighting his detractors and retired to Torquay in southwest England. A sketch of Oliver Heaviside is shown in Figure 5.34.

Figure 5.34 Oliver Heaviside (1850–1925)

Now let us consider Heaviside's so-called cover-up method for determining coefficients with partial fractions. Consider the antiderivative from Example 8. Find A₁, A₂, and A₃ such that

The coefficients are found, one at a time, by “covering” that factor and evaluating the remaining expression by the value that causes the “covered” factor to be zero. That is, first cover x:

The covered factor is 0 when x = 0, so evaluate the noncovered portion at x = 0:

Thus, A₁ = −2. Next, cover the factor under the A₂ term:

Evaluate for x = −1:

Finally, cover the factor under the A₃ term:

Evaluate for x = 1:

Thus,

Explain why this cover-up method of Heaviside works.

36. Assume that after t hours on the job, a factory worker can produce 100te^−0.5t units per hour. How many units does the worker produce during the first three hours?

37. After t weeks, suppose that contributions in response to a fund-raising campaign were coming in at the rate of 2,000te^−0.2t dollars per week. How much money was raised during the first five weeks?

38. An actuary measures the probability that a person in a certain population will die at age x by the formula

where λ is a parameter such that 0 < λ < e.

1. For a given λ, find the maximum value of P(x).
2. Sketch the graph of P(x).
3. Find the area under the probability curve y = P(x) for 0 ≤ x ≤ 10, and interpret your result.

39. A population P, grows at the rate

thousand individuals per year at time t (in years). By how much does the population change during the eighth year?

40. Suppose that a drug is assimilated into a patient's bloodstream at a rate modeled by

where t is the number of minutes since the drug was taken. Find the total amount of drug assimilated into the patient's bloodstream during the second minute.

41. Recovering from an environmental perturbation, a (hypothetical) population exhibits dampened oscillations of the form

where t is measured in days. As a part of a sampling effort, a scientist captures and releases individuals from this population at a rate of 0.1N(t) individuals per acre per day. If the scientist is sampling one acre, determine the number of individuals she captures and releases in seven days.

42. Recovering from an environmental perturbation, a (hypothetical) population exhibits dampened oscillations of the form

where t is measured in days. As a part of a sampling effort, a scientist captures and releases individuals from this population at a rate of 0.01N(t) individuals per acre per day. If the scientist is sampling one acre, determine the number of individuals she captures and releases in ten days.

5.7 Numerical Integration

We have seen that integration is, in general, a more difficult task than differentiation. In differentiation, knowing the derivatives of several elementary functions (e.g., sin x, e^x, and xⁿ) and a set of basic rules (e.g., product rule, chain rule, and quotient rule) allows us to differentiate rather complex-looking functions. In contrast, integration is more complicated. The number of rules and special cases, and uncertainty about which rule to apply, make integration more of an art than a science. An optimist would argue that armed with enough rules, and a great deal of practice, we could express the integral of any reasonable continuous function in terms of familiar functions. Unfortunately, this is not true.

You may have encountered some of these functions. For instance, some calculators return the same integral when asked to compute ∫ e^−x2dx, while others return the answer . This answer becomes circular when you find that the definition of the erf function, which is short for error function, is

Why is technology of no help when it comes to integrating this function analytically?

To answer this question, recall that an “elementary function” is a function that can be expressed in terms of power functions, exponentials, sines, cosines, and logarithms via the usual algebraic processes, including the solving (with or without radicals) of polynomials. Thus, elementary functions are all the “precalculus functions,” including polynomials and trigonometric and logarithmic functions. A theorem in mathematics asserts that there are an infinite number of elementary functions without an elementary antiderivative. Here are some examples of these functions:

The more common ones get their own names. For instance, we saw that up to some scaling factors, “erf” is the antiderivative of e^−x2. We can also find out using technology that “Si” is the antiderivative of sin x/x. Unfortunately, these functions are not exceptional.

What can we do when we need to integrate functions that are integrable but do not have elementary derivatives? If they are definite integrals, we could approximate them using Riemann sums, or we could use one of several other approximation schemes. In this section, we discuss five numerical schemes for approximating definite integrals. Three of these schemes, left endpoint rule, right endpoint rule, and midpoint rule, differ only in the manner that the sample points x_i are chosen. A fourth scheme, the trapezoidal rule, and a fifth scheme, Simpson's rule, involve approximating the functions with piecewise linear and piecewise quadratic segments, respectively. These five schemes, three of which are depicted in Figure 5.35, differ in how rapidly they converge (as n → ∞) to the true value of the definite integral. These rates of convergence are described via error estimates.

Figure 5.35 Numerical estimates of the area under a curve on the interval [a, b] using the left endpoint, midpoint, and trapezoidal rules, as labeled.

Left and right endpoint approximations

We begin with the simplest approximation schemes, left endpoint approximation and right endpoint approximation. For presentation purposes, our discussion focuses on left endpoint approximation (see first panel in Figure 5.35). By analogy, similar statements apply to the right endpoint approximation.

Suppose f is a continuous function from x = a to x = b and we want to estimate f(x)dx. By definition, f(x)dx, is a limit of Riemann sums. Consequently, given n, we partition the interval [a, b] into n equal subintervals with endpoints:

where a₁ = a + Δx, a₂ = a + 2Δx,..., a_n = a +nΔx, and Δx = . Taking the left endpoints, x₁ = a₀, x₂ = a₁,..., x_n = a_n−1 as our sample points, we have

As a first example, we begin with a simple function, so that we can examine the error generated by taking an approximation.

We might have added another column to our answer for Example 1. What is the error of the approximation? That is,

We calculate the error for each of the n values in the 2nd table of Example 1 and find that the approximations are underestimates by 0.461, 0.228, 0.090, 0.045, and 0.023 for the respective entries in the table. As we would expect, the errors tend to decrease as n increases. This leads to two questions: First, how quickly do the errors decrease with n? Second, if we don't know the true value of the definite integral, how can we estimate the error?

To answer both questions requires introducing error bounds: upper bounds for the magnitude of the error. These upper bounds often involve understanding the derivatives of the integrand. For example, suppose we know for some constant K₁ > 0 that |f′(x)| ≤ K₁ for all x between a and b. The evaluation theorem implies that

for any point x in [a, a + Δx]. Since f′(u) ≤ K₁, the dominance property of integrals implies that

for x between a and a + Δx. Equivalently,

Similarly, since f′(u) ≥ −K₁, the dominance property of integrals implies

A graphical interpretation of these two inequalities is shown in Figure 5.36. The graph of f(x) above the interval [a, a + Δx] lies in a triangular wedge with area K₁(Δx)². To estimate the error, we can consider three cases: the graph lies entirely in the upper half of triangular wedge (as illustrated in Figure 5.36), lies entirely in the lower half of the triangular wedge, or pass through both halves of the triangular wedge. When the graph lies entirely within one half of the wedge, it lies in a region with area K(Δx)²/2. Hence, the error in approximating f(x)dx with f(x₁)Δx is less than or equal to K₁(Δx)²/2. On the other hand, if the graph of f(x) passes through both the upper and lower half of the triangular wedge, then over- and under-estimates partially cancel one another, and the error is less than if the graph lied entirely in either the upper or lower half of the triangular wedge.

Figure 5.36 Estimating errors for the left endpoint rule

Similarly, over any of the subintervals, the error in approximating the actual value with f(x_k)Δx is at most K₁(Δx)²/2. Let E_L be the error by using a left endpoint approximation. Then, summing the error estimates over the n subintervals yields

where

is the error of the left endpoint approximation.

The preceding example illustrates several important points. First, even though we initially chose a reasonably large n (say, 10), the error bound was so large that we could not be certain about any of the digits. Second, an extremely large n is needed to ensure an estimate with accuracy to 0.001. Third, the approximation for the large n value was not very different from the estimate for the smaller n value of 10.

Midpoint rule

An alternative numerical approximation scheme is the midpoint rule in which the sample points chosen are the midpoints of each of the subintervals (see center panel in Figure 5.35). We begin, as we did with the left endpoints, by partitioning the interval [a, b] into n subintervals with endpoints:

where a₁ = a + Δx, a₂ = a + 2Δx,..., a_n = a + nΔx, and Δx = . This time, we take the midpoints,

as our sample points to get

Associated with the midpoint rule is an error estimate. If |f″(x)| ≤ K₂ for x on [a, b], then the midpoint error bound is

where

If the midpoint rule differs from the left and right endpoint rules only by shifting the sample points by Δx/2, why does it do so much better? First, observe that both the midpoint rule and left endpoint rule integrate constant functions perfectly. Indeed, L_n and M_n for c dx equal c(b − a). Second, for the case of linear functions f(x) = cx + d on an interval [a, b], it is simple to show (verify this for yourselves!) that E_M = 0 while E_L = |c|(b − a)/2. Hence, the midpoint rule integrates linear functions perfectly; consequently, it introduces error only for functions with nonzero, second-order derivatives. This explains why the error bound for the midpoint rule involves second-order terms (i.e., |f″(x)|and n²). In contrast, since the left endpoint rule introduces errors for functions whose derivatives are nonzero, the error bound for the left endpoint rule involves first-order terms (i.e.,|f′(x)|and n).

Another rule that perfectly estimates linear functions is the trapezoidal rule. This rule also has error bounds in terms of second-order rather than first-order derivatives. This rule is presented in the definition box at the end of this section and can be studied in more detail in Problem 38 of Problem Set 5.7.

Simpson's rule

As Example 3 illustrated, the midpoint rule was a significant improvement over the left endpoint rule because the error bound decreased like 1/n² instead of like 1/n. The small price we had to pay for this improvement was that bounds are calculated in terms of the second rather than first derivative. With this sweet taste of success, can we do better? As it turns out, yes! There is another rule, Simpson's rule, for approximating integrals using parabolas to approximate the curve. This method requires breaking the interval into an even number n of subintervals. Let a = x₀ < x₁ = a + Δx < ··· < x_n = b be the endpoints of these subintervals of width Δx = (b − a)/n. The approximation is given by

Given |f⁽⁴⁾(x)| ≤ K₄ for x on [a, b], the Simpson error bound is

where

Note that throughout this section we used the notation K_n to represent the bound of the nth derivative of f(x) for the cases n = 1, 2, and 4; that is,

This convention stresses the fact that each derivative has its own bound.

As the error estimate for Simpson's rule suggests from the foregoing example, Simpson's rule has much better convergence properties (i.e., the rate at which the error decreases with increasing n) than the midpoint rule. Why? Well, the Simpson's rule integrates a cubic function (i.e., third-order polynomials) perfectly. Hence, only nonzero fourth-order derivatives result in errors, and the error bound involves fourth-order terms (i.e., |f⁽⁴⁾(x)| and n⁴).

For convenience, we next summarize the four methods of numerical integration described in this section. We include the trapezoidal rule (see the third panel in Figure 5.35 and Problem 38 in Problem Set 5.7).

A simple example reinforces how efficient Simpson's rule is compared with the left endpoint and midpoint rules in converging to a solution.

Level 1 DRILL PROBLEMS

Approximate the integrals in Problems 1 to 12 using these four rules:

1. left endpoint rule
2. right endpoint rule
3. midpoint rule
4. Simpson's rule

Estimate the value of the integrals in Problems 13 to 22 to within the prescribed accuracy; use the rule indicated by the subscript on the error bound.

In Problems 23 to 28, determine how many subintervals are required to guarantee accuracy to within 0.00005 using these two rules:

1. midpoint rule
2. Simpson's rule

29. Estimate the area in the graph in Figure 5.38 using the left endpoint rule, right endpoint rule, and Simpson's rule with n = 6.

Figure 5.38 Estimate shaded area

30. Estimate the area in the graph in Figure 5.39 using the left endpoint rule, right endpoint rule, and Simpson's rule with n = 4.

Figure 5.39 Estimate shaded area

Level 2 APPLIED AND THEORY PROBLEMS

31. Governmental health agencies throughout the world monitor cases of human diseases. For example, the Office of Population Censuses and Surveys in the United Kingdom published weekly case reports about measles in England and Whales. Theoretical ecologist Ben Bolker published the data from 1948 to 1966 on the Web at http://ms.mcmaster.ca/bolker/measdata/ewmeas.dat The first 89 weeks at four week intervals is shown in the table below.

Use Simpson's rule and right endpoint rule to estimate the total number of cases over these 89 weeks.

32. Using inverse trigonometric functions, it can be shown that

Use this result to estimate π correct to four decimal places by applying Simpson's rule to this integral and using the appropriate error estimate.

33. Black Plague revisited Recall that for the outbreak of the plague in Bombay in 1905–1906, the mortality rate due to the plague was approximated by Kermack and McKendrick with the function

deaths per week.

1. Write a definite integral that represents the number of deaths that accumulated from t = 0 to t = 30.
2. Estimate the definite integral using Simpson's rule with n = 10.

34. The data set for 80 hours of the discharge (in m³/s) for the Raging River is shown here:

The data set for the first 24 hours is summarized in the following table:

Use Simpson's rule to estimate the total amount of discharge in the first 24 hours.

35. Sweet corn has a lower development threshold of 50°F and requires 1587 degree-days to complete development. On July 3, 2006, the temperatures in northern Illinois were as follows (measurements performed by the Northern Illinois Agronomy Research Center).

1. Using the right endpoint rule, estimate the number of degree-days that elapsed on this summer day.
2. If the temperatures on July 3rd typified the temperatures throughout the summer, estimate how many days it would take sweet corn to mature.

36. Repeat Problem 35 using Simpson's rule. Do you expect your answer to be more or less accurate than the answer to Problem 35?

37. The weekly rate of cases of influenza A (strain unknown) studied by WHO/NREVSS during the 2003–2004 season is plotted here:

Estimate the total number of cases (i.e., the area under the curve) from week 40 to week 56 using Simpson's rule on two-week intervals.

38. Trapezoidal rule for numerical integration: Assume that the area under a curve f(x) on an interval [a, b] is approximated by n trapezoids made up from rectangles topped by right-angle triangles so that the left- and right-hand sides of the (k + 1)^th trapezoid over the interval [x_k, x_k+1], are of heights f(x_k) and f(x_k+1), respectively, as illustrated in Figure 5.35. Let a = x₀ < x₁ = a + Δx < ··· < x_n = b and Δx =(b − a)/n. Show that the approximate area is given by

For completeness, we note that the error

satisfies

where K₂ satisfies |f″(x)| ≤ K₂ on the interval [a, b].

39.

Takakazu Seki Kōwa was born in Fujioka, Japan, the son of a samurai, but was adopted by a patriarch of the Seki family. Seki invented and used an early form of determinants for solving systems of equations. He also invented a method for approximating areas that is very similar to the rectangular method introduced in this section. His method, known as the yenri (circle principle), found the area of a circle by dividing the circle into small rectangles, as shown in Figure 5.40 (drawn by a student of Seki).

Figure 5.40 Early Asian calculus

Data Source: From Sawguchi Kazuyuki kokon Sampoki 1670. A History of Japanese Mathematics David Eugene Smith and Yoshio Mikami

Draw a circle with radius of 10 cm. Draw vertical chords through each centimeter on a diameter (you should have 18 rectangles). Measure the heights of the rectangles and approximate the area of the circle by adding the areas of the rectangles. Compare this method with the formula for the area of this circle.

40.

Isaac Newton invented a preliminary version of Simpson's rule. In 1779, Newton wrote an article to an addendum to Methodus Differentials (1711) in which he gave the following example: If there are four ordinates at equal intervals, let A be the sum of the first and fourth, B the sum of the second and third, and R the interval between the first and fourth; then the area between the first and fourth ordinates is approximated by (A + 3B)R. This is known today as the Newton-Cotes three-eighths rule, which can be expressed in the form

Roger Cotes and James Stirling (1692–1770) both knew this formula, as well as what we call in this section Simpson's rule. In 1743, this rule was rediscovered by Thomas Simpson (1710–1761).

using the Newton-Cotes three-eights rule, then compare with an approximation made using left endpoints (rectangles) with n = 8. Which of the rules gives the most accurate estimate?

5.8 Applications of Integration

In the preceding sections of this chapter, we motivated definite integration with area under a curve and accumulated change. In this section, we consider some additional applications that use Riemann sums to formulate integrals for survival and renewal processes, estimate cardiac output, and compute the amount of work required to perform a task.

Survival and renewal

Survival and renewal is the study of a population, or group of individuals, with the goal of predicting the size of the group at some future time. In the following example, a survival function gives the fraction of individuals in a group, or population, that can be expected to remain in the group for any specified period of time. In addition, a renewal function gives the rate at which new members arrive. Survival and renewal problems arise in many areas of study, including sociology, ecology, demography, and even finance—where the “population” is the number of dollars in an investment account, and “survival and renewal” refer to results of an investment strategy.

This example provides a guide to developing a more general formulation for survival and renewal processes. Suppose a population initially has N₀ individuals and receives new individuals (renews) at a rate r(t), and suppose the fraction of individuals remaining (surviving) in the population after t units of time after entering the population is s(t). If we want to determine the number of individuals in the population at time T, we can divide the interval [0, T] into subintervals of width Δt = T/n. The number of individuals arriving into the population during the kth time interval is approximately r(t_k)Δt. The fraction of these r(t_k)Δt individuals surviving to time T is approximately s(T − t_k). Hence, the number of individuals entering during the kth time interval and surviving to time T is approximately s(T − t_k)r(t_k)Δt. Summing up over all these time intervals yields

Taking the limit as n → ∞ yields

Of the N₀ individuals who were initially present, s(T)N₀ of them survive to time T. Hence, the number of individuals in the population at time T is given by the following survival and renewal function.

Another application of the renewal equation is in the area of finances, where we are concerned with the growth of economies or our personal fortunes. The key difference in this application is that instead of calculating how capital (population of dollars) decays, we are interested in how capital grows.

Although it is not common practice in banking systems to compound continuously, it is a reasonable approximation for daily compounding. Recall from Example 6 of Section 1.4 that compounding continuously at a rate of c% per year implies that if you put N dollars into an account, then t years later there will be e^ct/100N dollars in the account.

The solution to Example 3 shows that if money is being added to an account at a rate of r(t), the account is continually compounded at an interest rate of c%, and there is initially N₀ dollars in the account, then the total amount in the account T years from now is

This is just another survival-renewal equation with s(t) = e^cT/100.

Cardiac output

Cardiac output is the volume of blood pumped by the heart in a specified interval of time. Estimating cardiac output is important, because it is an indicator of certain heart diseases.

Figure 5.42 Schematic of the heart

Cardiac output can be measured using dye dilution. A known quantity of dye, say D milligrams (mg), is injected into a main vein near the heart. This dye circulates with the blood through the body (from the right ventricle of the heart to the lungs to the left ventricle and into the arteries) and returns to the left ventricle (see schematic in Figure 5.42). The concentration of the dye, c(t) milligrams per liter (mg/l), passing through an artery is monitored. To compute cardiac output from these recorded concentrations, assume that the cardiac output (i.e., blood flow) remains at a constant rate, F liters per second (l/s), during the experiment. The rate at which dye is passing through the artery at time t seconds is given by Fc(t) milligrams per second (mg/s). Notice how the units work out here: c(t) has units mg/L and F has units L/s. Hence c(t)F has units

Assume that the entire amount of dye passes through the artery between time t = 0 and t = T. The net amount of dye passing through the artery over the time interval from 0 to T is

By conservation of mass, the net amount of dye observed must equal the initial amount of dye, D; that is,

Solving for the cardiac output, F, yields

Work

How much pasta should a person eat to dig a posthole? How much energy should a grizzly bear expend to dig out a pocket gopher? To answer these questions, we need to understand the relationship between work and energy. To complete any work, we need energy. A standard unit of energy is a calorie. A calorie is the amount of energy required to heat one gram of water one degree Celsius. What does this mean?

In the article “What is Energy?” in Exploring Food Magazine (vol. 14 #4 1990), Paul Doherty writes about his undergraduate days as a biophysics student at MIT and describes an experiment where the professor set a peanut on fire. The energy from this peanut was sufficient to heat a test tube filled with 10 grams of water from room temperature to 100° Celsius, and then boil away 2 grams of water.

Doherty calculated that amount of energy required to heat and then boil the water. To heat 10 grams of water from 20° C to boiling (100° C) requires (100 − 20) × 10 = 800 calories. Further, since 540 calories are needed to boil away a gram of water (at sea level), another 1080 calories were needed to vaporize 2 grams of water. Thus, in the demonstration, the burning peanut delivered 1880 calories of heat flow to the test tube of water. This might seem like a lot of calories for one peanut. The reason for this surprising number is that a single dietary calorie (i.e., the type of calorie that is reported with food), which is abbreviated Cal or simply C, is actually one kilocalorie, abbreviated kcal. Hence, a peanut typically represents 1.5 to 2 Cal of energy.

What can we do with all this energy once ingested? We can turn it into work.

From Newtonian physics we know that

which implies that units of force are mass × distance × time⁻². Hence, the units of work are mass × distance² × time⁻². A standard work unit, called a Joule(J), is defined as follows:

The conversion between joules and dietary calories is given by the formula

Figure 5.45 El Capitan, Yosemite

Often people try to achieve great things by moving huge loads of rocks and dirt, such as digging canals, cutting through mountains to build roads and railways, or hollowing out deep cellars to build skyscrapers. Calculus can be used to compute the amount of work required to accomplish such feats, as evidenced in the following example.

The next example computes the amount of work required to remove soil from a geometrically more complicated region.

Level 1 DRILL PROBLEMS

After reviewing the mental health clinic in Example 1, for Problems 1 to 6 calculate the number of patients in the clinic after 15 months if the patient survival rate s(t) and the renewal rate r(t) are as given.

1. s(t) = e^−t/20 and r(t) = 20 per month. How did doubling the renewal rate change the answer from what we found in Example 1?

2. s(t) = e^−t/40 and r(t) = 10 per month. How did halving the survival rate change the answer from what we found in Example 1?

3. s(t) = e^−t/10 and r(t) = 20 per month

4. s(t) = e^−t/40 and r(t) = 20 per month

5. s(t) = e^−t/20 and r(t) = 10 + t per month

6. s(t) = and r(t) = 10 per month

After reviewing the fire ants in Example 2, for Problems 7 to 12 calculate the number of workers after five years if the worker survival rate s(t) is as given. Use technology to numerically evaluate the integrals.

7. s(t) = e^−0.625t, that is, survival rate is doubled

8. s(t) = e^−2.5t, that is, survival rate is halved

9. s(t) =

10. s(t) = e^−1.25t and, in addition, the proportion of queens alive at time t is q(t) = e^−0.1t

11. s(t) = e^−1.25t and, in addition, the proportion of queens alive at time t is q(t) = e^−0.2t

12. s(t) = e^−1.25t and, in addition, the proportion of queens alive at time t is q(t) = e^−0.05t

For Problems 13 to 16, reconsider Example 3. Calculate the amount of money Peggy Sue will have in her account by age 60 if she adds A dollars per year to her account, she opens her account at age B years, and the annual interest rate is 10%.

13. A = 4,000 and B = 20

14. A = 4,000 and B = 30

15. A = 1,000 and B = 10 (she starts really young!)

16. A = 10,000 and B = 40 (she starts very late!)

Answer Problems 17 to 19 by finding the work done; express your answer using the unit of foot-pounds (ft-lb).

17. Lifting a 90 lb bag of concrete 3 ft

18. Lifting a 50 lb bag of salt 5 ft

19. Lifting a 850 lb billiard table 15 ft

Level 2 APPLIED AND THEORY PROBLEMS

20. Analysts speculate that patients will enter a new clinic at a rate of 300 + 100 sin individuals per month. Moreover, the likelihood an individual is in the clinic t months later is e^−t. Find the number of patients in the clinic one year from now.

21. A patient receives a continuous drug infusion at a rate of 10 mg/h. Studies have shown that t hours after injection, the fraction of drug remaining in a patient's body is e^−2t. If the patient initially has 5 mg of drug in her bloodstream, then what is the amount of drug in the patient's bloodstream 24 hours later?

22. Consider a mental health clinic that initially has 300 patients, and accepts 100 new patients per month; the fraction of patients receiving treatment for t or more months is given by f(t).

Using Riemann sums with left endpoints, estimate the number of patients in the clinic after 12 months.

23. Consider the following two scenarios involving an IRA account that yields 9% continuous interest.

1. You graduate from college at age 22, get a job, and open an IRA account. You deposit $1,000 per year until age 65. How much money is the account at age 65? How much money did you pay into this account?
2. You graduate from college at age 22 and do not bother to start an IRA account until you reach 32. Then you deposit $2,000 per year into the IRA account until you reach age 65. How much money is in your IRA account at age 65? How much money did you pay into this account?

24. The administrators of a town estimate that the fraction of people who will still be residing in the town t years from now is given by the function S(t) = e^−0.04t. The current population is 20,000 people and new people are arriving at a rate of 500 per year.

1. What will be the population size 10 years from now?
2. What will be the population size 100 years from now?

25. After 5 mg of dye is injected into a vein, we obtain the concentration levels shown in the following table. The variable t is in seconds and c(t) is in mg/liter. Using Simpson's rule, compute the cardiac output.

26. Sediment flow.* Methods used in the measuring of cardiac output can be applied to other situations. For example, ecologists and scientists are interested in how much sediment is moved by a river. Data on the water flow and suspended sediment in the Des Moines River near Saylorville Lake, Iowa, is given in the following table. Using Simpson's rule, compute the total amount (kilograms) of suspended sediment that passed the measurement point for the period ending December 15, 1993.

Des Moines River Basin Water Discharge Records (December 1993)

27. Kety-Schmidt technique.* Seymour Kety and Carl Schmidt described a widely acknowledged and accurate method for the determination of cerebral blood flow and cerebral physiological activity (e.g., metabolic rate of oxygen). For example, a patient breathes 15% nitrous oxide (N₂O). After the start of administration, the arterial concentration, A, is measured in the radial artery. This is the concentration before the blood enters the brain. The venous concentration, V, is measured at the base of the skull in the superior bulb of the internal jugular vein (at the point of exit of the jugular vein from the brain). This process for measuring the blood flow of cerebral physiological activity is commonly referred to as the Kety-Schmidt technique. A sample table is shown below.

1. Initially, the rate of N₂O flow into the brain is greater than the flow out of the brain. Moreover, after approximately ten minutes, the concentrations flowing into the brain and from the brain are approximately equal. The brain has become saturated with N₂O. Assuming a constant cerebral blood flow rate, F, use Simpson's rule to estimate the total amount of N₂O accumulated in the brain during ten minutes. Your answer will depend on F.
2. Through other means, the maximum amount of N₂O in the brain can be measured. Suppose that the maximum amount is determined to be 58.8 cc. Determine F.

28. A bucket weighing 75 lb when filled and 10 lb when empty is pulled up the side of a 100 ft building. How much more work in foot-pounds is done in pulling up the full bucket than the empty bucket?

29. A 20 ft rope weighing 0.4 lb/ft hangs over the edge of a building 100 ft high. How much work in foot-pounds is done in pulling the rope to the top of the building? Assume that the top of the rope is flush with the top of the building and that the lower end of the rope is swinging freely.

30. How much ice water do you need to ingest to burn off 300 calories? Assume your body temperature is 37°C and the energy required to digest ice water is the energy needed to raise the ice to body temperature.

31. A children's book, a steam shovel claims that she can do as much work in one day as 100 men can do in seven days. In a modern society we assume that the work needed to dig the cellar considered in Example 8 is done by a machine. But how many calories would 100 men produce in seven days, if we assume that each ate two pounds of pasta a day and worked with 10% efficiency? How does this compare with the work done (i.e., calories needed) to dig the cellar considered in Example 8? Assume a serving of pasta is two ounces and contains 200 Cal.

32. Determine the length of a rectangular trench you can dig with the energy gained from eating one Milky Way bar (270 Cal). Assume that you convert the energy gained from the food with 10% efficiency and that the trench is 1 meter wide and 1 meter deep. Assume the density of soil is 1,000 kg/m³.

In the next two problems, use the fact that a serving of pasta contains 200 Cal and the density of soil is 1,000 kg/m³.

33. How much work does it take to dig up a conical hole of depth 5 meters and diameter 6 meters? How many servings of pasta are required to complete this work, assuming the energy from the pasta is converted with 5% efficiency to work?

34. How much work does it take to dig a hemispherical pit with radius 10 meters? How many servings of pasta are required to complete this work, assuming the energy from the pasta is converted with 5% efficiency to work?

1. Find the general antiderivative of f(x) = .

   Evaluate the definite integrals in Problems 2 to 5.

2. 
3. 
4. 
5. 
6. Find the area under the curve y = over [1, 2].
7. The slope F′(x) = at each point is shown in Figure 5.49.

   

   Figure 5.49 Slope field

   Find F passing through (1, −2) both graphically and analytically.

8. The Royalty rose has a lower developmental threshold of 41.4°F and requires 473 degree-days for harvesting time. If the temperature were to remain a constant 72°F, how long would it take for this rose to mature?
9. Find where

   

10. Evaluate the following integral

    

11. Consider a mental health clinic that initially has 450 patients and accepts 150 new patients per month; the fraction of patients receiving treatment for t or more months is given by f(t):

    

    Using Riemann sums with left endpoints, estimate the number of patients in the clinic after 12 months.

12. Find an upper bound for

    

13. The rate of infection of a disease in a population of 10,000 is given by the function

    

    where t is the time in months since the disease broke out.

    1. Use technology to plot R(t). Why is this a reasonable description of a disease spreading in a population?
    2. Compute the number of people infected by the disease by time T.
    3. Use technology to approximate the time when 50% of the population have the disease.
14. In a wild week of temperature fluctuations, the temperature in Corvallis, Oregon, is given by

    

    where t is measured in days. Find the number of degree-days that have elapsed for a beet armyworm over the first week. Note: The lower developmental threshold of a beet armyworm is 54°F.

15. Express tan x dx as the limit of a Riemann sum using right endpoints.
16. Express as a definite integral.
17. Find
18. Suppose
19. Use the geometrical interpretation of the definite integral to find (1 − |x|)dx. Be sure to provide a sketch.
20. A stone was dropped off a tower and hit the ground at a speed of 200 ft/s. What was the height of the tower?