section 4.4 the fundamental theorem of calculus the

282 CHAPTER 4 Integration

Section 4.4 The Fundamental Theorem of Calculus

• Evaluate a definite integral using the Fundamental Theorem of Calculus.• Understand and use the Mean Value Theorem for Integrals.• Find the average value of a function over a closed interval.• Understand and use the Second Fundamental Theorem of Calculus.

The Fundamental Theorem of Calculus

You have now been introduced to the two major branches of calculus: differentialcalculus (introduced with the tangent line problem) and integral calculus (introducedwith the area problem). At this point, these two problems might seem unrelated—butthere is a very close connection. The connection was discovered independently byIsaac Newton and Gottfried Leibniz and is stated in a theorem that is appropriatelycalled the Fundamental Theorem of Calculus.

Informally, the theorem states that differentiation and (definite) integration areinverse operations, in the same sense that division and multiplication are inverseoperations. To see how Newton and Leibniz might have anticipated this relationship,consider the approximations shown in Figure 4.27. The slope of the tangent line wasdefined using the (the slope of the secant line). Similarly, the area ofa region under a curve was defined using the (the area of a rectangle).So, at least in the primitive approximation stage, the operations of differentiation anddefinite integration appear to have an inverse relationship in the same sense thatdivision and multiplication are inverse operations. The Fundamental Theorem ofCalculus states that the limit processes (used to define the derivative and definiteintegral) preserve this inverse relationship.

�y�xproduct�y��xquotient

∆x ∆x

∆y

∆y ∆y

Secantline

Tangentline

Slope = Slope ≈

∆x

∆y

Area = ∆y∆x Area ≈ ∆y∆x

Area ofrectangle

Area ofregionundercurve

∆x

(a) Differentiation

Differentiation and definite integration have an “inverse”relationship.Figure 4.27

(b) Definite integration

THEOREM 4.9 The Fundamental Theorem of Calculus

If a function is continuous on the closed interval and is an antideriva-tive of on the interval then

�b

a

f �x� dx � F�b� � F�a�.

�a, b�,fF�a, b�f

E X P L O R A T I O N

Integration and AntidifferentiationThroughout this chapter, you have beenusing the integral sign to denote anantiderivative (a family of functions)and a definite integral (a number).

Antidifferentiation:

Definite integration:

The use of this same symbol for bothoperations makes it appear that theyare related. In the early work withcalculus, however, it was not knownthat the two operations were related.Do you think the symbol was firstapplied to antidifferentiation or todefinite integration? Explain yourreasoning. (Hint: The symbol wasfirst used by Leibniz and was derivedfrom the letter )S.

�

�b

a

f �x� dx

� f �x� dx

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SECTION 4.4 The Fundamental Theorem of Calculus 283

Proof The key to the proof is in writing the difference in a convenientform. Let be the following partition of

By pairwise subtraction and addition of like terms, you can write

By the Mean Value Theorem, you know that there exists a number in the th subin-terval such that

Because you can let and obtain

This important equation tells you that by applying the Mean Value Theorem you canalways find a collection of ’s such that the is a Riemann sumof on Taking the limit produces

The following guidelines can help you understand the use of the FundamentalTheorem of Calculus.

F�b� � F�a� � �b

a

f �x� dx.

�as �� → 0��a, b�.fF�b� � F�a�constantci

F�b� � F�a� � n

i�1 f�ci� �xi .

� xi � xi � xi�1F��ci� � f �ci�,

F��ci� �F�xi� � F�xi�1�

xi � xi�1.

ici

� n

i�1�F�xi� � F�xi�1��.

F�b� � F�a� � F�xn� � F�xn�1� � F�xn�1� � . . . � F�x1� � F�x1� � F�x0�

a � x0 < x1 < x2 < . . . < xn�1 < xn � b

�a, b�.�F�b� � F�a�

Guidelines for Using the Fundamental Theorem of Calculus

1. Provided you can find an antiderivative of you now have a way to evaluatea definite integral without having to use the limit of a sum.

2. When applying the Fundamental Theorem of Calculus, the following notationis convenient.

For instance, to evaluate you can write

3. It is not necessary to include a constant of integration in the antiderivativebecause

� F�b� � F�a�. � �F�b� � C� � �F�a� � C�

�b

a

f �x� dx � F�x� � C�b

a

C

�3

1 x3 dx �

x4

4 �3

1�

34

4�

14

4�

814

�14

� 20.

�31 x3 dx,

� F�b� � F�a�

�b

a

f�x� dx � F�x��b

a

f,

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EXAMPLE 1 Evaluating a Definite Integral

Evaluate each definite integral.

a. b. c.

Solution

a.

b.

c.

EXAMPLE 2 A Definite Integral Involving Absolute Value

Evaluate

Solution Using Figure 4.28 and the definition of absolute value, you can rewrite theintegrand as shown.

From this, you can rewrite the integral in two parts.

EXAMPLE 3 Using the Fundamental Theorem to Find Area

Find the area of the region bounded by the graph of the -axis,and the vertical lines and as shown in Figure 4.29.

Solution Note that on the interval

Integrate between and

Find antiderivative.

Apply Fundamental Theorem.

Simplify. �103

� �163

� 6 � 4 � �0 � 0 � 0�

� 2x3

3�

3x2

2� 2x�

2

0

x � 2.x � 0 Area � �2

0 �2x2 � 3x � 2� dx

�0, 2�.y > 0

x � 2,x � 0xy � 2x2 � 3x � 2,

�52

� ��14

�12 � �0 � 0� � �4 � 2� � �1

4�

12

� �x2 � x�1�2

0 � x2 � x�

2

1�2

�2

0 �2x � 1� dx � �1�2

0 ��2x � 1� dx � �2

1�2 �2x � 1� dx

�2x � 1� � ��2x � 1�,2x � 1,

x < 12

x ≥ 12

�2

0 �2x � 1� dx.

��4

0 sec2 x dx � tan x�

��4

0 � 1 � 0 � 1

�4

1 3�x dx � 3�4

1x1�2 dx � 3x3�2

3�2�4

1� 2�4�3�2 � 2�1�3�2 � 14

�2

1 �x2 � 3� dx � x3

3� 3x�

2

1� �8

3� 6 � �1

3� 3 � �

23

��4

0sec2 x dx�4

13�x dx�2

1�x2 � 3� dx

x−1 1 2

3

2

1

y = 2x − 1y = −(2x − 1)

y = 2x − 1 y

The definite integral of on is Figure 4.28

52.�0, 2�y

x1 2 3 4

4

3

2

1

y = 2x2 − 3x + 2y

The area of the region bounded by the graphof the -axis, and is Figure 4.29

103 .x � 2x � 0,xy,

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The Mean Value Theorem for Integrals

In Section 4.2, you saw that the area of a region under a curve is greater than the areaof an inscribed rectangle and less than the area of a circumscribed rectangle. TheMean Value Theorem for Integrals states that somewhere “between” the inscribed andcircumscribed rectangles there is a rectangle whose area is precisely equal to the areaof the region under the curve, as shown in Figure 4.30.

Proof

Case 1: If is constant on the interval the theorem is clearly valid because can be any point in

Case 2: If is not constant on then, by the Extreme Value Theorem, you canchoose and to be the minimum and maximum values of on Because for all in you can apply Theorem 4.8 to writethe following.

See Figure 4.31.

From the third inequality, you can apply the Intermediate Value Theorem to concludethat there exists some in such that

or

NOTE Notice that Theorem 4.10 does not specify how to determine It merely guaranteesthe existence of at least one number in the interval.c

c.

f �c��b � a� � �b

a

f �x� dx.f �c� �1

b � a �b

a

f �x� dx

�a, b�c

f �M�1b � a�

b

a

f �x� dx ≤ f �m� ≤

f �M��b � a�≤�b

a

f �x� dxf �m��b � a� ≤

�b

a

f �M� dx≤�b

a

f �x� dx �b

a

f �m� dx ≤

�a, b�,xf �m� ≤ f �x� ≤ f �M��a, b�.ff �M�f �m�

�a, b�,f

�a, b�.c�a, b�,f

x

f(c)f

a c b

y

Mean value rectangle:

Figure 4.30

f �c��b � a� � �b

a f �x� dx

f

a b

f(m)

f

a b

f

a b

f(M)

Inscribed rectangle(less than actual area)

Figure 4.31

�b

a f �m� dx � f �m��b � a�

Mean value rectangle(equal to actual area)

�b

a f �x� dx

Circumscribed rectangle(greater than actual area)

�b

a f �M� dx � f �M��b � a�

THEOREM 4.10 Mean Value Theorem for Integrals

If is continuous on the closed interval then there exists a number inthe closed interval such that

�b

a

f �x� dx � f �c��b � a�.

�a, b�c�a, b�,f

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Average Value of a Function

The value of given in the Mean Value Theorem for Integrals is called the averagevalue of on the interval

NOTE Notice in Figure 4.32 that the area of the region under the graph of is equal to thearea of the rectangle whose height is the average value.

To see why the average value of is defined in this way, suppose that youpartition into subintervals of equal width If is any point inthe th subinterval, the arithmetic average (or mean) of the function values at the ’sis given by

Average of

By multiplying and dividing by you can write the average as

Finally, taking the limit as produces the average value of on the intervalas given in the definition above.

This development of the average value of a function on an interval is only oneof many practical uses of definite integrals to represent summation processes. InChapter 7, you will study other applications, such as volume, arc length, centers ofmass, and work.

EXAMPLE 4 Finding the Average Value of a Function

Find the average value of on the interval

Solution The average value is given by

(See Figure 4.33.)

�13

�64 � 16 � �1 � 1�� 483

� 16.

�13 x3 � x2�

4

1

1

b � a �b

a

f �x� dx �13

�4

1 �3x2 � 2x� dx

�1, 4�.f �x� � 3x2 � 2x

�a, b�,fn →�

�1

b � a

n

i�1 f �ci� �x.

an �1n

n

i�1 f �ci��b � a

b � a �1

b � a

n

i�1 f �ci��b � a

n �b � a�,

f �c1�, . . . , f �cn�an �1n

� f �c1� � f �c2� � . . . � f �cn�� .

ciici�x � �b � a��n.n�a, b�

f

f

�a, b�.ff �c�

x

f

a b

Average value

y

Average value

Figure 4.32

�1

b � a �b

a f �x� dx

x1 2 3 4

40

30

20

10 Averagevalue = 16

(4, 40)

(1, 1)

f(x) = 3x2 − 2x

y

Figure 4.33

Definition of the Average Value of a Function on an Interval

If is integrable on the closed interval then the average value of onthe interval is

1b � a�

b

a

f �x� dx.

f�a, b�,f

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EXAMPLE 5 The Speed of Sound

At different altitudes in Earth’s atmosphere, sound travels at different speeds. Thespeed of sound (in meters per second) can be modeled by

where is the altitude in kilometers (see Figure 4.34). What is the average speed ofsound over the interval

Solution Begin by integrating over the interval To do this, you canbreak the integral into five parts.

By adding the values of the five integrals, you have

So, the average speed of sound from an altitude of 0 kilometers to an altitude of80 kilometers is

Average speed �180

�80

0 s�x� dx �

24,64080

� 308 meters per second.

�80

0 s�x� dx � 24,640.

�80

50 s�x� dx � �80

50 ��3

2x � 404.5� dx � �34x2 � 404.5x�

80

50� 9210

�50

32 s�x� dx � �50

32 �3

2x � 254.5� dx � 34x2 � 254.5x�

50

32� 5688

�32

22 s�x� dx � �32

22 �3

4x � 278.5� dx � 38x2 � 278.5x�

32

22� 2987.5

�22

11.5 s�x� dx � �22

11.5 �295� dx � 295x�

22

11.5� 3097.5

�11.5

0 s�x� dx � �11.5

0 ��4x � 341� dx � �2x2 � 341x�

11.5

0� 3657

�0, 80�.s�x�

�0, 80�?x

s�x� � ��4x � 341,295,34x � 278.5,32x � 254.5,

�32x � 404.5,

0 ≤ x < 11.5 11.5 ≤ x < 22 22 ≤ x < 32 32 ≤ x < 50 50 ≤ x ≤ 80

s�x�

The first person to fly at a speed greaterthan the speed of sound was CharlesYeager. On October 14, 1947, Yeager wasclocked at 295.9 meters per second at analtitude of 12.2 kilometers. If Yeager hadbeen flying at an altitude below 11.275kilometers, this speed would not have“broken the sound barrier.” The photoabove shows an F-14 Tomcat, a supersonic,twin-engine strike fighter. Currently, the Tomcat can reach heights of 15.24kilometers and speeds up to 2 mach(707.78 meters per second).

Geo

rge

Hal

l/Cor

bis

Spee

d of

sou

nd (

in m

/sec

)

Altitude (in km)

x10 20 30 40 50 60 70 80 90

350

340

330

320

310

300

290

280

s

Speed of sound depends on altitude.Figure 4.34

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The Second Fundamental Theorem of Calculus

Earlier you saw that the definite integral of on the interval was defined usingthe constant as the upper limit of integration and as the variable of integration.However, a slightly different situation may arise in which the variable is used as theupper limit of integration. To avoid the confusion of using in two different ways,is temporarily used as the variable of integration. (Remember that the definite integralis a function of its variable of integration.)

EXAMPLE 6 The Definite Integral as a Function

Evaluate the function

at

Solution You could evaluate five different definite integrals, one for each of thegiven upper limits. However, it is much simpler to fix (as a constant) temporarilyto obtain

Now, using you can obtain the results shown in Figure 4.35.

You can think of the function as the area under the curvefrom to For the area is 0 and For gives the accumulated area under the cosine curve on the entire interval

This interpretation of an integral as an is used oftenin applications of integration.

functionaccumulation�0, ��2�.F��2� � 1

x � ��2,F�0� � 0.x � 0,t � x.t � 0f �t� � cos taccumulatingF�x�

F�x� � sin x,

� sin x.� sin x � sin 0 �x

0 cos t dt � sin t�

x

0

x

x � 0, ��6, ��4, ��3, and ��2.

F�x� � �x

0 cos t dt

F�x� � �x

a

f �t� dt�b

a

f �x� dx

The Definite Integral as a Function of xThe Definite Integral as a Number

not

txx

xb�a, b�f

t

F(0) = 0

x = 0

y

t

F =

x =

π

π

6 21

6

( )

y

t

F =

x =

π

π

4 22

4

( )

y

t

F =

x =

π

π

3 23

3

( )

y

t

F = 1

x =

π

π

2

2

( )

y

cos is the area under the curve cos from 0 to

Figure 4.35

x.tf�t� �t dtF�x� � �x

0

Constant

Constant Constant

is a function of x.F

is a function of x.

f is a function of t.

f

E X P L O R A T I O N

Use a graphing utility to graph thefunction

for Do you recognize thisgraph? Explain.

0 ≤ x ≤ �.

F�x� � �x

0 cos t dt

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In Example 6, note that the derivative of is the original integrand (with only thevariable changed). That is,

This result is generalized in the following theorem, called the Second FundamentalTheorem of Calculus.

Proof Begin by defining as

Then, by the definition of the derivative, you can write

From the Mean Value Theorem for Integrals you know thereexists a number in the interval such that the integral in the expressionabove is equal to Moreover, because it follows that as So, you obtain

A similar argument can be made for

NOTE Using the area model for definite integrals, you can view the approximation

as saying that the area of the rectangle of height and width is approximately equal tothe area of the region lying between the graph of and the -axis on the interval as shown in Figure 4.36.

�x, x � �x�,xf�xf �x�

f �x� �x � �x��x

x

f �t� dt

�x < 0.

� f �x�.

� lim�x→0

f �c�

F��x� � lim�x→0

1�x

f �c� �x��x → 0.

c → xx ≤ c ≤ x � �x,f �c� �x.�x, x � �x�c

�assuming �x > 0�,

� lim�x→0

1

�x �x��x

x

f �t� dt�.

� lim�x→0

1

�x �x��x

a

f �t� dt � �a

x

f �t� dt�

� lim�x→0

1

�x �x��x

a

f �t� dt � �x

a

f �t� dt�

F��x� � lim�x→0

F�x � �x� � F�x�

�x

F�x� � �x

a

f �t� dt.

F

ddx

�F�x�� ddx

�sin x� �ddx �

x

0 cos t dt� � cos x.

F

tx x + ∆x

f(x)

∆x

f(t)

Figure 4.36

f �x� �x � �x� �x

x f �t� dt

THEOREM 4.11 The Second Fundamental Theorem of Calculus

If is continuous on an open interval containing then, for every in theinterval,

ddx �

x

a

f �t� dt� � f �x�.

xa,If

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Note that the Second Fundamental Theorem of Calculus tells you that if a func-tion is continuous, you can be sure that it has an antiderivative. This antiderivativeneed not, however, be an elementary function. (Recall the discussion of elementaryfunctions in Section P.3.)

EXAMPLE 7 Using the Second Fundamental Theorem of Calculus

Evaluate

Solution Note that is continuous on the entire real line. So, usingthe Second Fundamental Theorem of Calculus, you can write

The differentiation shown in Example 7 is a straightforward application of theSecond Fundamental Theorem of Calculus. The next example shows how this theoremcan be combined with the Chain Rule to find the derivative of a function.

EXAMPLE 8 Using the Second Fundamental Theorem of Calculus

Find the derivative of

Solution Using you can apply the Second Fundamental Theorem ofCalculus with the Chain Rule as shown.

Chain Rule

Definition of

Substitute for

Substitute for

Apply Second Fundamental Theorem of Calculus.

Rewrite as function of

Because the integrand in Example 8 is easily integrated, you can verify thederivative as follows.

In this form, you can apply the Power Rule to verify that the derivative is the same asthat obtained in Example 8.

F��x� � �cos x3��3x2�

� �sin x3� � 1� sin x3 � sin �

2� sin t�

x3

��2 F�x� � �x3

��2 cos t dt

x. � �cos x3��3x2� � �cos u��3x2�

x3.u �d

du �u

��2 cos t dt�du

dx

F�x�.�x3

��2cos t dt �

ddu�

x3

��2cos t dt�du

dx

dFdu

�ddu

�F�x�� dudx

F��x� �dFdu

dudx

u � x3,

F�x� � �x3

��2 cos t dt.

ddx

�x

0 �t2 � 1 dt� � �x2 � 1.

f �t� � �t2 � 1

ddx

�x

0 �t2 � 1 dt�.

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Graphical Reasoning In Exercises 1– 4, use a graphing utilityto graph the integrand. Use the graph to determine whether thedefinite integral is positive, negative, or zero.

1. 2.

3. 4.

In Exercises 5 –26, evaluate the definite integral of the algebraicfunction. Use a graphing utility to verify your result.

5. 6.

7. 8.

9. 10.

11. 12.

13. 14.

15. 16.

17. 18.

19. 20.

21. 22.

23. 24.

25. 26.

In Exercises 27–32, evaluate the definite integral of the trigono-metric function. Use a graphing utility to verify your result.

27.

28.

29.

30.

31.

32.

In Exercises 33–38, determine the area of the given region.

33. 34.

35. 36.

37. 38.

In Exercises 39–42, find the area of the region bounded by thegraphs of the equations.

39.

40.

41. 42.

In Exercises 43–46, find the value(s) of guaranteed by theMean Value Theorem for Integrals for the function over thegiven interval.

43. 44.

45.

46.

In Exercises 47–50, find the average value of the function overthe given interval and all values of in the interval for which thefunction equals its average value.

47. 48.

49. 50. �0, ��2�f �x� � cos x,�0, ��f �x� � sin x,

�1, 3�f �x� �4�x2 � 1�

x2 ,��2, 2�f �x� � 4 � x2,

x

��3, ��3�f �x� � cos x,

��4, ��4�f �x� � 2 sec2 x,

�1, 3�f �x� �9x3,�0, 2�f �x� � x � 2�x,

c

y � 0y � �x2 � 3x,y � 0x � 2,y � x3 � x,

y � 0x � 8,x � 0,y � 1 � 3�x,

y � 0x � 2,x � 0,y � 3x2 � 1,

xπ

2

3

4

1

π2

y

xππ24

1

y

y � x � sin xy � cos x

x1 2

1

y

x1 2 3

3

2

1

y

y �1x2y � �3 � x��x

x−1 1

2

y

x1

14

y

y � 1 � x4y � x � x2

��2

��2 �2t � cos t� dt

��3

��3 4 sec � tan � d�

��2

��4 �2 � csc2 x� dx

��6

��6 sec2 x dx

��4

0 1 � sin2 �

cos2 � d�

��

0 �1 � sin x� dx

�4

0 �x2 � 4x � 3� dx�3

0�x2 � 4� dx

�4

1�3 � �x � 3�� dx�3

0 �2x � 3� dx

��1

�8 x � x2

2 3�x dx�0

�1 �t 1�3 � t 2�3� dt

�2

0 �2 � t��t dt�1

0 x � �x

3 dx

�8

1�2

x dx�1

�1 �3�t � 2� dt

�3

�3 v1�3 dv�4

1 u � 2�u

du

��1

�2 �u �

1u2 du�2

1 � 3

x2 � 1 dx

�1

�1 �t3 � 9t� dt�1

0 �2t � 1�2 dt

�3

1�3x2 � 5x � 4� dx�1

�1 �t2 � 2� dt

�5

2 ��3v � 4� dv�0

�1 �x � 2� dx

�7

2 3 dv�1

0 2x dx

�2

�2 x�2 � x dx�2

�2 x�x2 � 1 dx

��

0 cos x dx��

0

4x2 � 1

dx

E x e r c i s e s f o r S e c t i o n 4 . 4 See www.CalcChat.com for worked-out solutions to odd-numbered exercises.

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51. Velocity The graph shows the velocity, in feet per second, ofa car accelerating from rest. Use the graph to estimate thedistance the car travels in 8 seconds.

Figure for 51 Figure for 52

52. Velocity The graph shows the velocity of a car as soon as thedriver applies the brakes. Use the graph to estimate how far thecar travels before it comes to a stop.

61. Force The force (in newtons) of a hydraulic cylinder in apress is proportional to the square of where is thedistance (in meters) that the cylinder is extended in its cycle.The domain of is and

(a) Find as a function of

(b) Find the average force exerted by the press over the interval

62. Blood Flow The velocity of the flow of blood at a distancefrom the central axis of an artery of radius is

where is the constant of proportionality. Find the average rateof flow of blood along a radius of the artery. (Use 0 and asthe limits of integration.)

63. Respiratory Cycle The volume in liters of air in the lungsduring a five-second respiratory cycle is approximated by themodel

where is the time in seconds. Approximate the average volumeof air in the lungs during one cycle.

64. Average Sales A company fits a model to the monthly salesdata of a seasonal product. The model is

where is sales (in thousands) and is time in months.

(a) Use a graphing utility to graph forUse the graph to explain why the average

value of is 0 over the interval.

(b) Use a graphing utility to graph and the linein the same viewing window. Use the

graph and the result of part (a) to explain why is calledthe trend line.

65. Modeling Data An experimental vehicle is tested on astraight track. It starts from rest, and its velocity (meters persecond) is recorded in the table every 10 seconds for 1 minute.

(a) Use a graphing utility to find a model of the formfor the data.

(b) Use a graphing utility to plot the data and graph the model.

(c) Use the Fundamental Theorem of Calculus to approximatethe distance traveled by the vehicle during the test.

v � at3 � bt2 � ct � d

v

gg�t� � t�4 � 1.8

S�t�f �t�

0 ≤ t ≤ 24.f �t� � 0.5 sin��t�6�

tS

0 ≤ t ≤ 24S�t� �t4

� 1.8 � 0.5 sin�� t6 ,

t

V � 0.1729t � 0.1522t2 � 0.0374t3

V

Rk

v � k�R2 � r 2�

Rrv

�0, ��3�.

x.F

F�0� � 500.�0, ��3�,F

xsec x,F

t

v

1 2 3

Time (in seconds)

4 5

20

40

60

Vel

ocity

(in

fee

t per

sec

ond)

80

100

t

v

4 8 12

Time (in seconds)

16 20

30

60

90

Vel

ocity

(in

fee

t per

sec

ond)

120

150

Writing About Concepts53. State the Fundamental Theorem of Calculus.

54. The graph of is shown in the figure.

(a) Evaluate

(b) Determine the average value of on the interval

(c) Determine the answers to parts (a) and (b) if the graphis translated two units upward.

Figure for 54 Figure for 55–60

In Exercises 55–60, use the graph of shown in the figure.The shaded region has an area of 1.5, and Use this information to fill in the blanks.

55. 56.

57. 58.

59.

60. The average value of over the interval is �.�0, 6�f

�6

0 �2 � f �x�� dx � �

�2

0 �2 f �x� dx � ��6

0 � f �x�� dx � �

�6

2 f �x� dx � ��2

0 f �x� dx � �

�60 f �x� dx � 3.5.A

f

x2 3 4 5 6

AB

y

f

x1 2 3 4 5 6 7

1

2

3

4

y

f

�1, 7�.f

�71 f �x� dx.

f

t 0 10 20 30 40 50 60

v 0 5 21 40 62 78 83

332460_0404.qxd 11/23/04 4:17 PM Page 292


66. Modeling Data A department store manager wants toestimate the number of customers that enter the store from noonuntil closing at 9 P.M. The table shows the number of customers

entering the store during a randomly selected minute eachhour from to with corresponding to noon.

(a) Draw a histogram of the data.

(b) Estimate the total number of customers entering the storebetween noon and 9 P.M.

(c) Use the regression capabilities of a graphing utility to finda model of the form for thedata.

(d) Use a graphing utility to plot the data and graph the model.

(e) Use a graphing utility to evaluate and use theresult to estimate the number of customers entering thestore between noon and 9 P.M. Compare this with youranswer in part (b).

(f ) Estimate the average number of customers entering thestore per minute between 3 P.M. and 7 P.M.

In Exercises 67–72, find as a function of and evaluate it atand

67. 68.

69. 70.

71. 72.

73. Let where is a function whose graph isshown.

(a) Estimate and

(b) Find the largest open interval on which is increasing. Findthe largest open interval on which is decreasing.

(c) Identify any extrema of

(d) Sketch a rough graph of

Figure for 73 Figure for 74

74. Let where is a function whose graph isshown.

(a) Estimate and

(b) Find the largest open interval on which is increasing. Findthe largest open interval on which is decreasing.

(c) Identify any extrema of

(d) Sketch a rough graph of

In Exercises 75– 80, (a) integrate to find as a function of and(b) demonstrate the Second Fundamental Theorem of Calculusby differentiating the result in part (a).

75. 76.

77. 78.

79. 80.

In Exercises 81–86, use the Second Fundamental Theorem ofCalculus to find

81. 82.

83. 84.

85. 86.

In Exercises 87–92, find

87. 88.

89. 90.

91. 92.

93. Graphical Analysis Approximate the graph of on theinterval where Identify the -coordinate of an extremum of To print an enlarged copy of

the graph, go to the website www.mathgraphs.com.

94. Use the graph of the function shown in the figure on the nextpage and the function defined by

(a) Complete the table.

g�x� � �x0 f �t� dt.g

f

t

f

42

2

1

−2

−1

y

g.xg�x� � �x

0 f �t� dt.0 ≤ x ≤ 4,g

F�x� � �x2

0sin �2 d�F�x� � �x3

0sin t2 dt

F�x� � �x2

2 1t 3 dt F�x� � �sin x

0 �t dt

F�x� � �x

�x

t3 dtF�x� � �x�2

x

�4t � 1� dt

F��x�.

F�x� � �x

0 sec3 t dtF�x� � �x

0 t cos t dt

F�x� � �x

1

4�t dtF�x� � �x

�1 �t 4 � 1 dt

F�x� � �x

1

t2

t2 � 1 dtF�x� � �x

�2 �t2 � 2t� dt

F��x�.

F�x� � �x

��3 sec t tan t dtF�x� � �x

��4 sec2 t dt

F�x� � �x

4 �t dtF�x� � �x

8

3�t dt

F�x� � �x

0 t�t2 � 1� dtF�x� � �x

0 �t � 2� dt

xF

g.

g.

gg

g�8�.g�6�,g�4�,g�2�,g�0�,

fg�x� � �x0 f �t� dt,

y

t21 3 7 84 5 6

−2−3−4

−1

4321

f

y

t21 3 7 84

−2−1

654321

f

g.

g.

gg

g�8�.g�6�,g�4�,g�2�,g�0�,

fg�x� � �x0 f �t� dt,

F�x� � �x

0sin � d�F�x� � �x

1 cos � d�

F�x� � �x

2�

2t3 dtF�x� � �x

1 10v2 dv

F�x� � �x

2 �t3 � 2t � 2� dtF�x� � �x

0 �t � 5� dt

x � 8.x � 5,x � 2,xF

�90 N�t� dt,

N�t� � at3 � bt2 � ct � d

t � 0t,t � 1N

1 2 3 4 5 6 7 8 9 10

g�x�

x

t 1 2 3 4 5 6 7 8 9

N 6 7 9 12 15 14 11 7 2

332460_0404.qxd 11/23/04 4:17 PM Page 293


(b) Plot the points from the table in part (a) and graph

(c) Where does have its minimum? Explain.

(d) Where does have a maximum? Explain.

(e) On what interval does increase at the greatest rate?Explain.

(f) Identify the zeros of

95. Cost The total cost (in dollars) of purchasing and main-taining a piece of equipment for years is

(a) Perform the integration to write as a function of

(b) Find and

96. Area The area between the graph of the functionand the axis over the interval is

(a) Find the horizontal asymptote of the graph of

(b) Integrate to find as a function of Does the graph of have a horizontal asymptote? Explain.

Rectilinear Motion In Exercises 97–99, consider a particlemoving along the -axis where is the position of the particleat time is its velocity, and is the distance theparticle travels in the interval of time.

97. The position function is given by Find the total distance the particle travels in 5 units

of time.

98. Repeat Exercise 97 for the position function given by

99. A particle moves along the -axis with velocity At time its position is Find the total

distance traveled by the particle on the interval

100. Buffon’s Needle Experiment A horizontal plane is ruledwith parallel lines 2 inches apart. A two-inch needle is tossedrandomly onto the plane. The probability that the needle willtouch a line is

where is the acute angle between the needle and any one ofthe parallel lines. Find this probability.

True or False? In Exercises 101 and 102, determine whetherthe statement is true or false. If it is false, explain why or givean example that shows it is false.

101. If on the interval then

102. If is continuous on then is integrable on

103. Find the Error Describe why the statement is incorrect.

104. Prove that

105. Show that the function

is constant for

106. Let where is continuous for all

real Find (a) (b) (c) and (d) G�0�.G �x�,G��0�,G�0�,t.

fG�x� � �x

0 s�s

0 f �t�dt� ds,

x > 0.

f �x� � �1�x

0

1t 2 � 1

dt � �x

0

1t 2 � 1

dt

ddx �

v�x�

u�x� f �t� dt� � f �v�x��v��x� � f �u�x��u��x�.

�1

�1 x�2 dx � ��x�1� 1

�1 � ��1� � 1 � �2

�a, b�.f�a, b�,f

G�b� � G�a�.F�b� � F�a� ��a, b�,F��x� � G��x�

θ

�

P �2�

��2

0 sin � d�

1 ≤ t ≤ 4.x � 4.t � 1,t > 0.

v�t� � 1��t,x

0 ≤ t ≤ 5.x�t� � �t � 1��t � 3�2,

0 ≤ t ≤ 5.x�t� � t3 � 6t2 � 9t � 2,

�ba �x��t��dtx��t�t,

x�t�x

Ax.A

g.

A�x� � �x

1 �4 �

4t2 dt.

�1, x�t-4 � 4�t 2g�t� �A

C�10�.C�1�, C�5�,x.C

C�x� � 5000�25 � 3�x

0 t1�4 dt .

xC

t

4

2

−2

−4

2 4 6 8 10

f

y

g.

g

g

g

g.

Section Project: Demonstrating the Fundamental Theorem

Use a graphing utility to graph the function on theinterval Let be the following function of

(a) Complete the table. Explain why the values of are increasing.

(b) Use the integration capabilities of a graphing utility to graph

(c) Use the differentiation capabilities of a graphing utility to graphHow is this graph related to the graph in part (b)?

(d) Verify that the derivative of is Graph and write a short paragraph about how this graph isrelated to those in parts (b) and (c).

ysin2 t.y � �1�2�t � �sin 2t��4

F��x�.

F.

F

F�x� � �x

0sin2 t dt

x.F�x�0 ≤ t ≤ �.y1 � sin2 t

0

F�x�

�5��62��3��2��3��6x

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section 4.4 the fundamental theorem of calculus the

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