# Mathematics

known rate Q′. Then the net change in Q between t = a and t = b 7 a is

Q1b2 – Q1a2 = Lba Q′1t2 dt. net change in Q

Given the initial value Q102, the future value of Q at time t Ú 0 is Q1t2 = Q102 + L t0 Q′1×2 dx.

e The correspondences between velocity–displacement problems and more general prob- lems are shown in Table 6.1.

Table 6.1

Velocity–Displacement Problems General Problems

Position s1t2 Quantity Q1t2 (such as volume or population) Velocity: s′1t2 = v1t2 Rate of change: Q′1t2 Displacement: s1b2 – s1a2 = Lba v1t2 dt Net change: Q1b2 – Q1a2 = Lba Q′1t2 dt Future position: s1t2 = s102 + L t0 v1x2 dx Future value of Q: Q1t2 = Q102 + L t0 Q′1×2 dx EXAMPLE 5 Cell growth A culture of cells in a lab has a population of 100 cells when nutrients are added at time t = 0. Suppose the population N1t2 (in cells/hr) increases at a rate given by

N′1t2 = 90e-0.1t. Find N1t2, for t Ú 0.

➤ Although N is a positive integer (the number of cells), we treat it as a continuous variable in this example.

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410 Chapter 6 Applications of Integration

d. Which rider arrives first at the 10-, 15-, and 20-mile markers of the race? Interpret your answer geometrically using the graphs of part (a).

e. Suppose Sasha gives Theo a head start of 0.2 mi and the riders ride for 20 mi. Who wins the race?

f. Suppose Sasha gives Theo a head start of 0.2 hr and the riders ride for 20 mi. Who wins the race?

58. Two runners At noon 1t = 02, Alicia starts running along a long straight road at 4 mi>hr. Her velocity decreases according to the function v1t2 = 4>1t + 12, for t Ú 0. At noon, Boris also starts running along the same road with a 2-mi head start on Alicia; his velocity is given by u1t2 = 2>1t + 12, for t Ú 0. Assume t is measured in hours.

a. Find the position functions for Alicia and Boris, where s = 0 corresponds to Alicia’s starting point.

b. When, if ever, does Alicia overtake Boris?

59. Running in a wind A strong west wind blows across a circular running track. Abe and Bess start running at the south end of the track, and at the same time, Abe starts running clockwise and Bess starts running counterclockwise. Abe runs with a speed (in units of mi>hr) given by u1w2 = 3 – 2 cos w and Bess runs with a speed given by v1u2 = 3 + 2 cos u, where w and u are the central angles of the runners.

wind

Abe

Bess

start

!”

a. Graph the speed functions u and v, and explain why they describe the runners’ speeds (in light of the wind).

b. Compute the average value of u and v with respect to the central angle.

c. Challenge: If the track has a radius of 110 mi, how long does it take each runner to complete one lap and who wins the race?

Applications 60. Filling a tank A 2000-liter cistern is empty when water be-

gins flowing into it 1at t = 02 at a rate (in L>min) given by Q′1t2 = 31t, where t is measured in minutes. a. How much water flows into the cistern in 1 hour? b. Find and graph the function that gives the amount of water in

the tank at any time t Ú 0. c. When will the tank be full?

61. Depletion of natural resources Suppose that r1t2 = r0 e-kt, with k 7 0, is the rate at which a nation extracts oil, where r0 = 107 barrels>yr is the current rate of extraction. Suppose also that the estimate of the total oil reserve is 2 * 109 barrels. a. Find Q1t2, the total amount of oil extracted by the nation after

t years. b. Evaluate lim

tS∞ Q1t2 and explain the meaning of this limit.

c. Find the minimum decay constant k for which the total oil reserves will last forever.

d. Suppose r0 = 2 * 107 barrels>yr and the decay constant k is the minimum value found in part (c). How long will the total oil reserves last?

62. Snowplow problem With snow on the ground and falling at a constant rate, a snowplow began plowing down a long straight road at noon. The plow traveled twice as far in the first hour as it did in the second hour. At what time did the snow start falling? Assume the plowing rate is inversely proportional to the depth of the snow.

63. Filling a reservoir A reservoir with a capacity of 2500 m3 is filled with a single inflow pipe. The reservoir is empty when the inflow pipe is opened at t = 0. Letting Q1t2 be the amount of water in the reservoir at time t, the flow rate of water into the reservoir 1in m3>hr2 oscillates on a 24-hr cycle (see figure) and is given by

Q′1t2 = 20 a1 + cos pt 12

b .

0 t2412

40

20

Q!

Time (hr)

Fl ow

ra te

o f w

at er

(m 3 /

hr )

a. How much water flows into the reservoir in the first 2 hr? b. Find and graph the function that gives the amount of water in

the reservoir over the interval 30, t4, where t Ú 0. c. When is the reservoir full?

64. Blood flow A typical human heart pumps 70 mL of blood with each stroke (stroke volume). Assuming a heart rate of 60 beats>min 11 beat>s2, a reasonable model for the outflow rate of the heart is V′1t2 = 7011 + sin 2pt2, where V1t2 is the amount of blood (in milliliters) pumped over the interval 30, t4, V102 = 0, and t is measured in seconds. a. Graph the outflow rate function. b. Verify that the amount of blood pumped over a one-second

interval is 70 mL. c. Find the function that gives the total blood pumped between

t = 0 and a future time t 7 0. d. What is the cardiac output over a period of 1 min? (Use calcu-

lus; then check your answer with algebra.)

65. Air flow in the lungs A simple model (with different parameters for different people) for the flow of air in and out of the lungs is

V′1t2 = -p 2

sin pt 2

,

where V1t2 (measured in liters) is the volume of air in the lungs at time t Ú 0, t is measured in seconds, and t = 0 corresponds to a time at which the lungs are full and exhalation begins. Only a fraction of the air in the lungs in exchanged with each breath. The amount that is exchanged is called the tidal volume.

a. Find and graph the volume function V assuming that V102 = 6 L.

b. What is the breathing rate in breaths/min? c. What is the tidal volume and what is the total capacity of the

lungs?

T

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6.1 Velocity and Net Change 411

66. Oscillating growth rates Some species have growth rates that oscillate with an (approximately) constant period P. Consider the growth rate function

N′1t2 = r + A sin 2pt P

where A and r are constants with units of individuals>yr, and t is measured in years. A species becomes extinct if its population ever reaches 0 after t = 0. a. Suppose P = 10, A = 20, and r = 0. If the initial popula-

tion is N102 = 10, does the population ever become extinct? Explain.

b. Suppose P = 10, A = 20, and r = 0. If the initial popula- tion is N102 = 100, does the population ever become extinct? Explain.

c. Suppose P = 10, A = 50, and r = 5. If the initial popula- tion is N102 = 10, does the population ever become extinct? Explain.

d. Suppose P = 10, A = 50, and r = -5. Find the initial population N102 needed to ensure that the population never becomes extinct.

67. Power and energy Power and energy are often used interchange- ably, but they are quite different. Energy is what makes matter move or heat up and is measured in units of joules (J) or Calories (Cal), where 1 Cal = 4184 J. One hour of walking consumes roughly 106 J, or 250 Cal. On the other hand, power is the rate at which energy is used and is measured in watts 1W; 1 W = 1 J>s2. Other useful units of power are kilowatts 11 kW = 103 W2 and megawatts 11 MW = 106 W2. If energy is used at a rate of 1 kW for 1 hr, the total amount of energy used is 1 kilowatt-hour 1kWh2, which is 3.6 * 106 J.

Suppose the power function of a large city over a 24-hr period is given by

P1t2 = E′1t2 = 300 – 200 sin pt 12

,

where P is measured in megawatts and t = 0 corresponds to 6:00 p.m. (see figure).

0 t2412

400

600

200

P

Time (hr)

Po w

er (M

W )

!t 12

P(t) ! E “(t) ! 300 # 200 sin

a. How much energy is consumed by this city in a typical 24-hr period? Express the answer in megawatt-hours and in joules.

b. Burning 1 kg of coal produces about 450 kWh of energy. How many kg of coal are required to meet the energy needs of the city for 1 day? For 1 year?

c. Fission of 1 g of uranium-235 1U-2352 produces about 16,000 kWh of energy. How many grams of uranium are needed to meet the energy needs of the city for 1 day? For 1 year?

d. A typical wind turbine can generate electrical power at a rate of about 200 kW. Approximately how many wind turbines are needed to meet the average energy needs of the city?

68. Variable gravity At Earth’s surface, the acceleration due to gravity is approximately g = 9.8 m>s2 (with local variations). However, the acceleration decreases with distance from the sur- face according to Newton’s law of gravitation. At a distance of y meters from Earth’s surface, the acceleration is given by

a1y2 = – g11 + y>R22, where R = 6.4 * 106 m is the radius of Earth.

a. Suppose a projectile is launched upward with an initial velocity of v0 m>s. Let v1t2 be its velocity and y1t2 its height (in meters) above the surface t seconds after the launch. Neglecting forces

such as air resistance, explain why dv dt

= a1y2 and dy dt

= v1t2. b. Use the Chain Rule to show that

dv dt

= 1 2

d dy

1v22. c. Show that the equation of motion for the projectile is

1 2

d dy

1v22 = a1y2, where a1y2 is given previously. d. Integrate both sides of the equation in part (c) with respect to y

using the fact that when y = 0, v = v0. Show that

1 2

1v2 – v202 = g R a 11 + y>R – 1b . e. When the projectile reaches its maximum height, v = 0.

Use this fact to determine that the maximum height is

ymax = Rv0

2

2gR – v20 .

f. Graph ymax as a function of v0. What is the maximum height when v0 = 500 m>s, 1500 m>s, and 5 km>s?

g. Show that the value of v0 needed to put the projectile into orbit (called the escape velocity) is 12gR.

Additional Exercises 69–72. Another look at the Fundamental Theorem

69. Suppose that f and g have continuous derivatives on an interval 3a, b4. Prove that if f 1a2 = g1a2 and f1b2 = g1b2, then 1ba f ′1×2 dx = 1ba g′1×2 dx.

70. Use Exercise 69 to prove that if two runners start and finish at the same time and place, then regardless of the velocities at which they run, their displacements are equal.

71. Use Exercise 69 to prove that if two trails start at the same place and finish at the same place, then regardless of the ups and downs of the trails, they have the same net change in elevation.

72. Without evaluating integrals, prove that

L 2

0

d dx

112 sin px22 dx = L20 ddx1x1012 – x232 dx. QUICK CHECK ANSWERS 1. Displacement = -20 mi (20 mi south); distance traveled = 100 mi 2. Suppose the object moves in the posi- tive direction, for 0 … t … 3, and then moves in the negative direction, for 3 6 t … 5. 3. A function; a number 4. Displacement = 0; distance traveled = 1 5. 1720 m 6. The production cost would increase more between 9000 and 12,000 books than between 12,000 and 15,000 books. Graph C′ and look at the area under the curve.

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412 Chapter 6 Applications of Integration

6.2 Regions Between Curves In this section, the method for finding the area of a region bounded by a single curve is generalized to regions bounded by two or more curves. Consider two functions f and g continuous on an interval 3a, b4 on which f 1×2 Ú g1x2 (Figure 6.11). The goal is to find the area A of the region bounded by the two curves and the vertical lines x = a and x = b.

Once again, we rely on the slice-and-sum strategy (Section 5.2) for finding areas by Riemann sums. The interval 3a, b4 is partitioned into n subintervals using uniformly spaced grid points separated by a distance ∆x = 1b – a2>n (Figure 6.12). On each sub- interval, we build a rectangle extending from the lower curve to the upper curve. On the kth subinterval, a point xk

* is chosen, and the height of the corresponding rectangle is taken to be f 1xk*2 – g1xk*2. Therefore, the area of the kth rectangle is 1 f 1xk*2 – g1xk*22 ∆x (Figure 6.13). Summing the areas of the n rectangles gives an approximation to the area of the region between the curves:

A ≈ a n

k = 1 1 f 1xk*2 – g1xk*22 ∆x.

Figure 6.11

Lower curve y ! g(x)

ba

Upper curve y ! f (x)

y

x

Area ! A

Figure 6.12

y

y ! f (x)

y ! g(x)

a ! x0

“x ! width of each rectangle

! b # a

n

b ! xn xn#1

x2x1

“x x

Figure 6.13

ba x

y

f (xk) ! g(xk)

“x

Area of region: A ! k#1

n

$ ( f (xk) ! g(xk)) “x

Area of kth rectangle # ( f (xk) ! g(xk)) “x

(xk, f (xk))*

*

* *

*

(xk, g(xk))

xk

*

*

* *

*

*

y # g(x)

y # f (x)

DEFINITION Area of a Region Between Two Curves Suppose that f and g are continuous functions with f 1×2 Ú g1x2 on the interval 3a, b4. The area of the region bounded by the graphs of f and g on 3a, b4 is

A = L b

a 1 f 1×2 – g1x22 dx.

As the number of grid points increases, ∆x approaches zero and these sums approach the area of the region between the curves; that is,

A = lim nS∞

a n

k = 1 1 f 1xk*2 – g1xk*22∆x.

The limit of these Riemann sums is a definite integral of the function f – g.

➤ It is helpful to interpret the area formula: f 1×2 – g1x2 is the length of a rectangle and dx represents its width. We sum (integrate) the areas of the rectangles 1 f 1×2 – g1x22 dx to obtain the area of the region.

QUICK CHECK 1 In the area formula for a region between two curves, verify that if the lower curve is g1x2 = 0, the formula becomes the usual formula for the area of the region bounded by y = f 1×2 and the x-axis.

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6.2 Regions Between Curves 413

EXAMPLE 1 Area between curves Find the area of the region bounded by the graphs of f 1×2 = 5 – x2 and g1x2 = x2 – 3 (Figure 6.14). SOLUTION A key step in the solution of many area problems is finding the intersec- tion points of the boundary curves, which often determine the limits of integration. The intersection points of these two curves satisfy the equation 5 – x2 = x2 – 3. The solutions of this equation are x = -2 and x = 2, which become the lower and upper limits of integration, respectively. The graph of f is the upper curve and the graph of g is the lower curve on the interval 3-2, 24. Therefore, the area of the region is

A = L 2

-2 115 – x22 – 1×2 – 322 dx Substitute for f and g.

s s f 1×2 g1x2

= 2L 2

0 18 – 2×22 dx Simplify and use symmetry.

= 2a8x – 2 3

x3b ` 2 0 Fundamental Theorem

= 64 3

. Simplify.

Notice how the symmetry of the problem simplifies the integration. Additionally, note that the area formula A = 1ba 1 f 1×2 – g1x22 dx is valid even if one or both curves lie below the x-axis, as long as f 1×2 Ú g1x2 on 3a, b4.

Related Exercises 5–14

QUICK CHECK 2 Interpret the area formula when written in the form A = 1ba f 1×2 dx – 1ba g1x2 dx, where f 1×2 Ú g1x2 Ú 0 on 3a, b4. EXAMPLE 2 Compound region Find the area of the region bounded by the graphs of f 1×2 = -x2 + 3x + 6 and g1x2 = ! 2x ! (Figure 6.15a). SOLUTION The lower boundary of the region is bounded by two different branches of the absolute value function. In situations like this, the region is divided into two (or more) subregions whose areas are found independently and then summed; these subregions are labeled R1 and R2 (Figure 6.15b). By the definition of absolute value,

g1x2 = ! 2x ! = b2x if x Ú 0 -2x if x 6 0.

Figure 6.14

Slice the region using vertical rectangles from x ! “2 to x ! 2.

Rectangle height ! f(x) ” g(x) Rectangle width !#x

g(x) ! x2 ” 3

f(x) ! 5 ” x24

3

2

1

“2

“1

(“2, 1) (2, 1)

“2 “1 21 x

(x, f(x))

(x, g(x))

y

#x

Figure 6.15 (a)

!1 321 x

6

4

2

y

g(x) ” !2x!

f (x) ” !x2 # 3x # 6

0

(b)

!1 321

g(x) ” !2x!

f (x) ” !x2 # 3x # 6

0

(3, 6)

(!1, 2)

0

Area ” ” ((!x2 # 3x # 6) ! (!2x) dx !1

R1

R2

4

2 0

3

Area ” ” ((!x2 # 3x # 6) ! 2x) dx

6

y

x

The left intersection point of f and g satisfies -2x = -x2 + 3x + 6, or x2 – 5x – 6 = 0. Solving for x, we find that 1x + 121x – 62 = 0, which implies x = -1 or x = 6; only the first solution is relevant. The right intersection point of f and g satisfies 2x = -x2 + 3x + 6; you should verify that the relevant solution in this case is x = 3.

➤ The solution x = 6 corresponds to the intersection of the parabola y = -x2 + 3x + 6 and the line y = -2x in the fourth quadrant, not shown in Figure 6.15 because g1x2 = -2x only when x 6 0.

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418 Chapter 6 Applications of Integration

40–43. Regions between curves Sketch the region and find its area.

40. The region bounded by y = sin x and y = x1x – p2, for 0 … x … p

41. The region bounded by y = 1x – 122 and y = 7x – 19 42. The region bounded by y = 2 and y =

121 – x2 43. The region bounded by y = x2 – 2x + 1 and y = 5x – 9

44–50. Either method Use the most efficient strategy for computing the area of the following regions.

44. The region bounded by x = y1y – 12 and x = -y1y – 12 45. The region bounded by x = y1y – 12 and y = x>3 46. The region bounded by y = x3, y = -x3, and

3y – 7x – 10 = 0

47. The region bounded by y = 1x, y = 2x – 15, and y = 0 48. The region bounded by y = x2 – 4, 4y – 5x – 5 = 0, and

y = 0, for y Ú 0

49. The region in the first quadrant bounded by y = 5 2

– 1 x

and y = x

50. The region in the first quadrant bounded by y = x -1, y = 4x, and y = x>4

51. Comparing areas Let f 1×2 = xp and g1x2 = x1>q, where p 7 1 and q 7 1 are positive integers. Let R1 be the region in the first quadrant between y = f 1×2 and y = x and let R2 be the region in the first quadrant between y = g1x2 and y = x. a. Find the area of R1 and R2 when p = q, and determine which

region has the greater area. b. Find the area of R1 and R2 when p 7 q, and determine which

region has the greater area. c. Find the area of R1 and R2 when p 6 q, and determine which

region has the greater area.

52–55. Complicated regions Find the area of the regions shown in the following figures.

52. y

O

y ! 4!2x

y ! 2×2

y ! “4x # 6

x

53.

y ! 9 ” x2

y ! “x

y ! 8x

x

y

5 2