# Mathematics

9. Oil production An oil refinery produces oil at a variable rate given by

Q′1t2 = c 800 if 0 … t 6 302600 – 60t if 30 … t 6 40 200 if t Ú 40,

where t is measured in days and Q is measured in barrels.

a. How many barrels are produced in the first 35 days? b. How many barrels are produced in the first 50 days? c. Without using integration, determine the number of barrels

produced over the interval 360, 804. 40–43. Population growth 40. Starting with an initial value of P102 = 55, the population of

a prairie dog community grows at a rate of P′1t2 = 20 – t>5 (prairie dogs>month), for 0 … t … 200, where t is measured in months.

a. What is the population 6 months later? b. Find the population P1t2, for 0 … t … 200.

41. When records were first kept 1t = 02, the population of a rural town was 250 people. During the following years, the population grew at a rate of P′1t2 = 3011 + 1t2, where t is measured in years. a. What is the population after 20 years? b. Find the population P1t2 at any time t Ú 0.

42. The population of a community of foxes is observed to fluctuate on a 10-year cycle due to variations in the availability of prey. When population measurements began 1t = 02, the population was 35 foxes. The growth rate in units of foxes>year was observed to be

P′1t2 = 5 + 10 sin pt 5

.

a. What is the population 15 years later? 35 years later? b. Find the population P1t2 at any time t Ú 0.

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6.1 Velocity and Net Change 409

43. A culture of bacteria in a Petri dish has an initial popula- tion of 1500 cells and grows at a rate 1in cells>day2 of N′1t2 = 100e-0.25t. Assume t is measured in days. a. What is the population after 20 days? After 40 days? b. Find the population N1t2 at any time t Ú 0.

44. Flow rates in the Spokane River The daily discharge of the Spo- kane River as it flows through Spokane, Washington, in April and June is modeled by the functions

r11t2 = 0.25t2 + 37.46t + 722.47 1April2 and r21t2 = 0.90t2 – 69.06t + 2053.12 1June2,

where the discharge is measured in millions of cubic feet per day and t = 1 corresponds to the first day of the month (see figure).

t

r

0 15 25205 100

500

1000

1500

2000 r2(t) ! 0.90t

2 ” 69.06t # 2053.12

r1(t) ! 0.25t 2 # 37.46t # 722.47

a. Determine the total amount of water that flows through Spokane in April (30 days).

b. Determine the total amount of water that flows through Spokane in June (30 days).

c. The Spokane River flows out of Lake Coeur d’Alene, which contains approximately 0.67 mi3 of water. Determine the per- centage of Lake Coeur d’Alene’s volume that flows through Spokane in April and June.

45–48. Marginal cost Consider the following marginal cost functions.

a. Find the additional cost incurred in dollars when production is increased from 100 units to 150 units.

b. Find the additional cost incurred in dollars when production is increased from 500 units to 550 units.

45. C′1×2 = 2000 – 0.5x 46. C′1×2 = 200 – 0.05x 47. C′1×2 = 300 + 10x – 0.01×2 48. C′1×2 = 3000 – x – 0.001×2 Further Explorations 49. Explain why or why not Determine whether the following

statements are true and give an explanation or counterexample.

a. The distance traveled by an object moving along a line is the same as the displacement of the object.

b. When the velocity is positive on an interval, the displacement and the distance traveled on that interval are equal.

c. Consider a tank that is filled and drained at a flow rate of V′1t2 = 1 – t2>100 1gal>min2, for t Ú 0, where t is measured in minutes. It follows that the volume of water in the tank increases for 10 min and then decreases until the tank is empty.

d. A particular marginal cost function has the property that it is positive and decreasing. The cost of increasing production from A units to 2A units is greater than the cost of increasing production from 2A units to 3A units.

50–51. Velocity graphs The figures show velocity functions for motion along a straight line. Assume the motion begins with an initial position of s102 = 0. Determine the following: a. The displacement between t = 0 and t = 5 b. The distance traveled between t = 0 and t = 5 c. The position at t = 5 d. A piecewise function for s1t2 50.

t51

3

1

v

0

51.

t51

3

1

v

0

52–55. Equivalent constant velocity Consider the following veloc- ity functions. In each case, complete the sentence: The same distance could have been traveled over the given time period at a constant velocity of ________.

52. v1t2 = 2t + 6, for 0 … t … 8 53. v1t2 = 1 – t2>16, for 0 … t … 4 54. v1t2 = 2 sin t, for 0 … t … p 55. v1t2 = t125 – t221>2, for 0 … t … 5 56. Where do they meet? Kelly started at noon 1t = 02 riding a

bike from Niwot to Berthoud, a distance of 20 km, with veloc- ity v1t2 = 15>1t + 122 (decreasing because of fatigue). Sandy started at noon 1t = 02 riding a bike in the opposite direction from Berthoud to Niwot with velocity u1t2 = 20>1t + 122 (also decreasing because of fatigue). Assume distance is measured in kilometers and time is measured in hours.

a. Make a graph of Kelly’s distance from Niwot as a function of time.

b. Make a graph of Sandy’s distance from Berthoud as a function of time.

c. When do they meet? How far has each person traveled when they meet?

d. More generally, if the riders’ speeds are v1t2 = A>1t + 122 and u1t2 = B>1t + 122 and the distance between the towns is D, what conditions on A, B, and D must be met to ensure that the riders will pass each other?

e. Looking ahead: With the velocity functions given in part (d), make a conjecture about the maximum distance each person can ride (given unlimited time).

57. Bike race Theo and Sasha start at the same place on a straight road, riding bikes with the following velocities (measured in mi>hr). Assume t is measured in hours.

Theo: vT1t2 = 10, for t Ú 0 Sasha: vS1t2 = 15t, for 0 … t … 1 and vS1t2 = 15, for t 7 1

a. Graph the velocity functions for both riders. b. If the riders ride for 1 hr, who rides farther? Interpret your

answer geometrically using the graphs of part (a). c. If the riders ride for 2 hr, who rides farther? Interpret your

answer geometrically using the graphs of part (a).

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414 Chapter 6 Applications of Integration

In cases such as these, we treat y as the independent variable and divide the interval 3c, d4 into n subintervals of width ∆y = 1d – c2>n (Figure 6.17). On the kth subinterval, a point yk

* is selected, and we construct a rectangle that extends from the left curve to the right curve. The kth rectangle has length f 1yk*2 – g1yk*2, so the area of the kth rectangle is 1 f 1yk*2 – g1yk*22∆y. The area of the region is approximated by the sum of the areas of the rectangles. In the limit as n S ∞ and ∆y S 0, the area of the region is given as the definite integral

A = lim nS∞

a n

k = 1 1 f 1yk*2 – g1yk*22∆y = Ldc 1 f 1y2 – g1y22 dy.

➤ This area formula is analogous to the one given on page 412; it is now expressed with respect to the y-axis. In this case, f 1y2 – g1y2 is the length of a rectangle and dy represents its width. We sum (integrate) the areas of the rectangles 1 f 1y2 – g1y22 dy to obtain the area of the region.

Given these points of intersection, we see that the region R1 is bounded by y = -x2 + 3x + 6 and y = -2x on the interval 3-1, 04. Similarly, region R2 is bounded by y = -x2 + 3x + 6 and y = 2x on 30, 34 (Figure 6.15b). Therefore,

A = L 0

-1 11-x2 + 3x + 62 – 1-2×2 2 dx + L30 11-x2 + 3x + 62 – 2×2 dx

(1++++++)++++++* (++++++)+++++1* area of region R1 area of region R2

= L 0

-1 1-x2 + 5x + 62 dx + L30 1-x2 + x + 62 dx Simplify.

= a – x3 3

+ 5 2

x2 + 6xb ` 0 -1

+ a – x3 3

+ 1 2

x2 + 6xb ` 3 0 Fundamental Theorem

= 0 – a 1 3

+ 5 2

– 6b + a -9 + 9 2

+ 18b – 0 = 50 3

. Simplify.

Related Exercises 15–22

Integrating with Respect to y There are occasions when it is convenient to reverse the roles of x and y. Consider the regions shown in Figure 6.16 that are bounded by the graphs of x = f 1y2 and x = g1y2, where f 1y2 Ú g1y2, for c … y … d (which implies that the graph of f lies to the right of the graph of g). The lower and upper boundaries of the regions are y = c and y = d, respectively.

Figure 6.16

x x

x

d

c

O

y

d

c

O

y

d

c

O

y

x ! g(y) x ! g(y) x ! g(y)

x ! f (y) x ! f (y) x ! f (y)

x

d

c

!y

f (yk*) ” g(yk*)

x # g(y) x # f (y)

O

Area of region: A ! k#1

n

$ ( f (yk*) ” g(yk*)) !y

( f (yk*), yk*)

(g(yk*), yk*)

y

Area of kth rectangle # ( f (yk*) ” g(yk*)) !y

yk*

Figure 6.17

DEFINITION Area of a Region Between Two Curves with Respect to y Suppose that f and g are continuous functions with f 1y2 Ú g1y2 on the interval 3c, d4. The area of the region bounded by the graphs x = f 1y2 and x = g1y2 on 3c, d4 is

A = L d

c 1 f 1y2 – g1y22 dy.

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6.2 Regions Between Curves 415

EXAMPLE 3 Integrating with respect to y Find the area of the region R bounded by the graphs of y = x3, y = x + 6, and the x-axis.

SOLUTION The area of this region could be found by integrating with respect to x. But this approach requires splitting the region into two pieces (Figure 6.18). Alternatively, we can view y as the independent variable, express the bounding curves as functions of y, and make horizontal slices parallel to the x-axis (Figure 6.19).

Figure 6.18

20 x

8

y

y ! x ” 6 y ! x3

(#6, 0)

y ! 0

#6

0

0

2

(2, 8)

R

Area ! ! ((x ” 6) # 0) dx ” ! ((x ” 6) # x3) dx Figure 6.19

20 x

8

x ! y1/3

(“6, 0)

(2, 8)

R

y

0

8

Rectangle length ! y1/3 ” (y ” 6) Rectangle width ! #y

x ! y ” 6

Slice the region using horizontal rectangles from y ! 0 to y ! 8.

y ! 0

Area ! ! (y1/3 ” (y ” 6)) dy

Solving for x in terms of y, the right curve y = x3 becomes x = f 1y2 = y1>3. The left curve y = x + 6 becomes x = g1y2 = y – 6. The intersection point of the curves satisfies the equation y1>3 = y – 6, or y = 1y – 623. Expanding this equation gives the cubic equation

y3 – 18y2 + 107y – 216 = 1y – 821y2 – 10y + 272 = 0, whose only real root is y = 8. As shown in Figure 6.19, the areas of the slices through the region are summed from y = 0 to y = 8. Therefore, the area of the region is given by

L 8

0 1y1>3 – 1y – 622 dy = a 3

4 y4>3 – y2

2 + 6yb ` 8

0 Fundamental Theorem

= a 3 4

# 16 – 32 + 48b – 0 = 28. Simplify. Related Exercises 23–32

QUICK CHECK 3 The region R is bounded by the curve y = 1x, the line y = x – 2, and the x-axis. Express the area of R in terms of (a) integral(s) with respect to x and (b) integral(s) with respect to y.

EXAMPLE 4 Calculus and geometry Find the area of the region R in the first quad- rant bounded by the curves y = x2>3 and y = x – 4 (Figure 6.20). SOLUTION Slicing the region vertically and integrating with respect to x requires two integrals. Slicing the region horizontally requires a single integral with respect to y. The second approach appears to involve less work.

Slicing horizontally, the right bounding curve is x = y + 4 and the left bound- ing curve is x = y3>2. The two curves intersect at 18, 42, so the limits of integration are y = 0 and y = 4. The area of R is

L 4

0 11y + 42 – y3>22 dy = a y2

2 + 4y – 2

5 y5>2b ` 4

0 =

56 5

. s r

right curve left curve

➤ You may use synthetic division or a root finder to factor the cubic polynomial in Example 3. Then the quadratic formula shows that the equation

y2 – 10y + 27 = 0

has no real roots.

Figure 6.20

0 x84

y

4

0

4

Area of region R

! ! ((y ” 4) # y3/2) dy

(8, 4)

y ! x # 4, x ! y ” 4

y ! x2/3, x ! y3/2

R

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416 Chapter 6 Applications of Integration

Basic Skills 5–8. Finding area Determine the area of the shaded region in the following figures. 5.

x

y

y ! x

y ! x2 ” 2

6.

x

y

y ! x y ! x3