Stewart J. Calculus

Подождите немного. Документ загружается.

1. (a) Draw the line and use geometry to ﬁnd the area under this line, above the

-axis, and between the vertical lines and .

(b) If , let be the area of the region that lies under the line between

and . Sketch this region and use geometry to ﬁnd an expression for .

2. (a) If , let

represents the area of a region. Sketch that region.

A!x"

A!x" !

!1 % t

" dt

x & $1

A!x"

A!x"t ! xt ! 1

y ! 2t % 1A!x"x " 1

t ! 3t ! 1t

y ! 2t % 1

AREA FUNCTIONS

D I S C O V E R Y

P R O J E C T

312

|| ||

CHAPTER 5 INTEGRALS

61– 62 Use properties of integrals, together with Exercises 27 and

28, to prove the inequality.

61. 62.

63. Prove Property 3 of integrals.

64. Prove Property 6 of integrals.

65. If is continuous on , show that

[Hint: .]

66. Use the result of Exercise 65 to show that

67. Let if is any rational number and if is

any irrational number. Show that is not integrable on .

68. Let and if . Show that is not

integrable on . [Hint: Show that the ﬁrst term in the

Riemann sum, , can be made arbitrarily large.]

69– 70 Express the limit as a deﬁnite integral.

69. [Hint: Consider .]

70.

71. Find . Hint: Choose to be the geometric mean of

and

(

that is,

)

and use the identity

m!m % 1"

m % 1

i$1

lim

n l )

i!1

1 % !i%n"

f !x" ! x

lim

n l )

i!1

f !x

" *x

#0, 1$

x ! 1f !x" ! 1%xf !0" ! 0

#0, 1$f

xf !x" ! 1xf !x" ! 0

(

f !x" sin 2x dx

(

f !x"

! f !x" !

f !x"

f !x" dx

f !x"

#a, b$f

(

x sin x dx !

(

% 1 dx &

Write as a single integral in the form :

48. If and , ﬁnd .

If and , ﬁnd

50. Find if

51. Suppose has absolute minimum value and absolute max-

imum value . Between what two values must

lie? Which property of integrals allows you to make your

conclusion?

52–54 Use the properties of integrals to verify the inequality with-

out evaluating the integrals.

52.

54.

55 – 60 Use Property 8 to estimate the value of the integral.

55. 56.

57. 58.

59. 60.

(

!x $ 2 sin x" dx

1 % x

$ 3x % 3" dx

(

tan x dx

1 % x

x dx

(

cos x dx !

(

2 !

1 % x

dx ! 2

53.

1 % x

dx !

1 % x

f !x" dxM

f !x" !

(

3 for x

x for x & 3

f !x" dx

#2 f !x" % 3t!x"$ dx

t!x" dx ! 16x

f !x" dx ! 37

49.

f !x" dxx

f !x" dx ! 3.6x

f !x" dx ! 12

f !x" dx %

f !x" dx $

f !x" dx

47.

SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS

|| ||

313

THE FUNDAMENTAL THEOREM OF CALCULUS

The Fundamental Theorem of Calculus is appropriately named because it establishes a

connection between the two branches of calculus: differential calculus and integral calcu-

lus. Differential calculus arose from the tangent problem, whereas integral calculus arose

from a seemingly unrelated problem, the area problem. Newton’s mentor at Cambridge,

Isaac Barrow (1630–1677), discovered that these two problems are actually closely

related. In fact, he realized that differentiation and integration are inverse processes. The

Fundamental Theorem of Calculus gives the precise inverse relationship between the

derivative and the integral. It was Newton and Leibniz who exploited this relationship and

used it to develop calculus into a systematic mathematical method. In particular, they saw

that the Fundamental Theorem enabled them to compute areas and integrals very easily

without having to compute them as limits of sums as we did in Sections 5.1 and 5.2.

The ﬁrst part of the Fundamental Theorem deals with functions deﬁned by an equation

of the form

where is a continuous function on and varies between and . Observe that tbax#a, b$f

t!x" !

f !t" dt

5.3

(b) Use the result of Exercise 28 in Section 5.2 to ﬁnd an expression for .

(d) If and h is a small positive number, then represents the area

of a region. Describe and sketch the region.

(e) Draw a rectangle that approximates the region in part (d). By comparing the areas of

these two regions, show that

(f) Use part (e) to give an intuitive explanation for the result of part (c).

;

3. (a) Draw the graph of the function in the viewing rectangle

by .

(b) If we deﬁne a new function by

then is the area under the graph of from 0 to [until becomes negative, at

which point becomes a difference of areas]. Use part (a) to determine the value of

at which starts to decrease. [Unlike the integral in Problem 2, it is impossible to

evaluate the integral deﬁning to obtain an explicit expression for .]

, . . . , , . Then use these values to sketch a graph of .

(d) Use your graph of from part (c) to sketch the graph of using the interpretation of

as the slope of a tangent line. How does the graph of compare with the graph

of ?

4. Suppose is a continuous function on the interval and we deﬁne a new function

by the equation

Based on your results in Problems 1–3, conjecture an expression for .

t+!x"

t!x" !

f !t" dt

t#a, b$f

t+t+!x"

t+t

tt!2"t!1.8"t!0.6"

t!0.4"t!0.2"

t!x"t

t!x"x

t!x"

f !x"xft!x"

t!x" !

cos!t

" dt

#$1.25, 1.25$

#0, 2$f !x" ! cos!x

A!x % h" $ A!x"

) 1 % x

A!x % h" $ A!x"x & $1

A+!x"

A!x"

depends only on , which appears as the variable upper limit in the integral. If is a ﬁxed

number, then the integral is a deﬁnite number. If we then let vary, the number

also varies and deﬁnes a function of denoted by .

If happens to be a positive function, then can be interpreted as the area under the

graph of from to , where can vary from to . (Think of as the “area so far” func-

tion; see Figure 1.)

EXAMPLE 1 If is the function whose graph is shown in Figure 2 and

, ﬁnd the values of , , , , , and . Then sketch a

rough graph of .

SOLUTION First we notice that . From Figure 3 we see that is the

area of a triangle:

To ﬁnd we add to the area of a rectangle:

We estimate that the area under from 2 to 3 is about 1.3, so

For , is negative and so we start subtracting areas:

We use these values to sketch the graph of in Figure 4. Notice that, because is

positive for , we keep adding area for and so is increasing up to ,

where it attains a maximum value. For , decreases because is negative. M

If we take and , then, using Exercise 27 in Section 5.2, we have

t!x" !

t dt !

a ! 0f !t" ! t

f !t"tx " 3

x ! 3tt

f !t"t

t!5" ! t!4" %

f !t" dt ) 3 % !$1.3" ! 1.7

t!4" ! t!3" %

f !t" dt ) 4.3 % !$1.3" ! 3.0

f !t"t " 3

FIG URE 3

(

)

(

)

(

)

Å4.3

(

)

Å3

(

)

Å1.7

t!3" ! t!2" %

f !t" dt ) 3 % 1.3 ! 4.3

t!2" !

f !t" dt !

f !t" dt

f !t" dt ! 1 % !1 ! 2" ! 3

t!1"t!2"

t!1" !

f !t" dt !

!1 ! 2" ! 1

t!1"t!0" ! x

f !t" dt ! 0

t!5"t!4"t!3"t!2"t!1"t!0"t!x" ! x

f !t" dt

tbaxxaf

t!x"f

t!x"xx

f !t" dt

314

|| ||

CHAPTER 5 INTEGRALS

a bx

area=©

y=f(t)

FIG URE 1

y=f(t)

FIG URE 2

FIG URE 4

f(t)dt

Notice that , that is, . In other words, if is deﬁned as the integral of by

Equation 1, then turns out to be an antiderivative of , at least in this case. And if we

sketch the derivative of the function shown in Figure 4 by estimating slopes of tangents,

we get a graph like that of in Figure 2. So we suspect that in Example 1 too.

To see why this might be generally true we consider any continuous function with

. Then can be interpreted as the area under the graph of from

to , as in Figure 1.

In order to compute from the deﬁnition of derivative we ﬁrst observe that,

for is obtained by subtracting areas, so it is the area under the

graph of from to (the gold area in Figure 5). For small you can see from the

ﬁgure that this area is approximately equal to the area of the rectangle with height and

width :

Intuitively, we therefore expect that

The fact that this is true, even when is not necessarily positive, is the ﬁrst part of the Fun-

damental Theorem of Calculus.

THE FUNDAMENTAL THEOREM OF CALCULUS, PART 1 If is continuous on ,

then the function deﬁned by

is continuous on and differentiable on , and .

PROOF If and are in , then

(by Property 5)

and so, for ,

For now let us assume that . Since is continuous on , the Extreme

Value Theorem says that there are numbers and in such that f !u" ! m#x, x % h$

#x, x % h$fh " 0

t!x % h" $ t!x"

x%h

f !t" dt

h " 0

x%h

f !t" dt

f !t" dt %

x%h

f !t" dt

t!x % h" $ t!x" !

x%h

f !t" dt $

f !t" dt

!a, b"x % hx

t+!x" ! f !x"!a, b"#a, b$

a ! x ! bt!x" !

f !t" dt

#a, b$f

t+!x" ! lim

t!x % h" $ t!x"

! f !x"

t!x % h" $ t!x"

) f !x"

t!x % h" $ t!x" ) hf !x"

f !x"

hx % hxf

h " 0, t!x % h" $ t!x"

t+!x"

ft!x" !

f !t" dtf !x" & 0

t+ ! ff

ftt+ ! ft+!x" ! x

SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS

|| ||

315

a b

x x+h

FIG URE 5

N We abbreviate the name of this theorem as

FTC1. In words, it says that the derivative of a

deﬁnite integral with respect to its upper limit is

the integrand evaluated at the upper limit.

and , where and are the absolute minimum and maximum values of

on . (See Figure 6.)

By Property 8 of integrals, we have

that is,

Since , we can divide this inequality by :

Now we use Equation 2 to replace the middle part of this inequality:

Inequality 3 can be proved in a similar manner for the case

(See Exercise 59.)

Now we let . Then and , since and lie between and .

Therefore

and

because is continuous at . We conclude, from (3) and the Squeeze Theorem, that

If or , then Equation 4 can be interpreted as a one-sided limit. Then Theo-

rem 3.2.4 (modiﬁed for one-sided limits) shows that is continuous on . M

Using Leibniz notation for derivatives, we can write FTC1 as

when is continuous. Roughly speaking, Equation 5 says that if we ﬁrst integrate and

then differentiate the result, we get back to the original function .

EXAMPLE 2 Find the derivative of the function .

SOLUTION Since is continuous, Part 1 of the Fundamental Theorem of

Calculus gives

t+!x" !

1 % x

f !t" !

1 % t

t!x" !

1 % t

f !t" dt ! f !x"

#a, b$t

bx ! a

t+!x" ! lim

h l 0

t!x % h" $ t!x"

! f !x"

lim

h l 0

f !v" ! lim

v l x

f !v" ! f !x"

lim

h l 0

f !u" ! lim

u l x

f !u" ! f !x"

x % hxvuv l xu l xh l 0

f !u" !

t!x % h" $ t!x"

! f !v"

f !u" !

x%h

f !t" dt ! f !v"

hh " 0

f !u"h !

x%h

f !t" dt ! f !v"h

mh !

x%h

f !t" dt ! Mh

#x, x % h$

fMmf !v" ! M

316

|| ||

CHAPTER 5 INTEGRALS

FIG URE 6

x u

√=x+h

y=ƒ

Module 5.3 provides visual

evidence for FTC1.

TE C

EXAMPLE 3 Although a formula of the form may seem like a strange

way of deﬁning a function, books on physics, chemistry, and statistics are full of such

functions. For instance, the Fresnel function

is named after the French physicist Augustin Fresnel (1788–1827), who is famous for his

works in optics. This function ﬁrst appeared in Fresnel’s theory of the diffraction of light

waves, but more recently it has been applied to the design of highways.

Part 1 of the Fundamental Theorem tells us how to differentiate the Fresnel function:

This means that we can apply all the methods of differential calculus to analyze (see

Exercise 53).

Figure 7 shows the graphs of and the Fresnel function

. A computer was used to graph by computing the value of this

integral for many values of . It does indeed look as if is the area under the graph

of from to [until , when becomes a difference of areas]. Figure 8

shows a larger part of the graph of .

If we now start with the graph of in Figure 7 and think about what its derivative

should look like, it seems reasonable that . [For instance, is increasing

when and decreasing when .] So this gives a visual conﬁrmation of

Part 1 of the Fundamental Theorem of Calculus. M

EXAMPLE 4 Find .

SOLUTION Here we have to be careful to use the Chain Rule in conjunction with FTC1.

Let . Then

(by the Chain Rule)

(by FTC1)

In Section 5.2 we computed integrals from the deﬁnition as a limit of Riemann sums

and we saw that this procedure is sometimes long and difﬁcult. The second part of the

Fundamental Theorem of Calculus, which follows easily from the ﬁrst part, provides us

with a much simpler method for the evaluation of integrals.

! sec!x

" ! 4x

! sec u

sec t dt

sec t dt !

sec t dt

u ! x

sec t dt

f !x"

0f !x" " 0

SS+!x" ! f !x"

S!x"x ) 1.4x0f

S!x"x

SS!x" ! x

f !t" dt

f !x" ! sin!

(

%2"

S+!x" ! sin!

(

%2"

S!x" !

sin!

(

%2" dt

t!x" ! x

f !t" dt

SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS

|| ||

317

FIG URE 7

ƒ=sin(π≈/2)

S(x)=

sin(πt@/2)dt

FIG URE 8

The Fresnel function

S(x)=

sin(πt@/2)dt

0.5

THE FUNDAMENTAL THEOREM OF CALCULUS, PART 2 If is continuous on

, then

where is any antiderivative of , that is, a function such that .

PROOF Let . We know from Part 1 that ; that is, is an anti-

derivative of . If is any other antiderivative of on , then we know from Corol-

lary 4.2.7 that and differ by a constant:

for . But both and are continuous on and so, by taking limits of both

sides of Equation 6 (as and ), we see that it also holds when and

If we put in the formula for , we get

So, using Equation 6 with and , we have

Part 2 of the Fundamental Theorem states that if we know an antiderivative of , then

we can evaluate simply by subtracting the values of at the endpoints of the

interval . It’s very surprising that , which was deﬁned by a complicated pro-

cedure involving all of the values of for , can be found by knowing the val-

ues of at only two points, and .

Although the theorem may be surprising at ﬁrst glance, it becomes plausible if we inter-

pret it in physical terms. If is the velocity of an object and is its position at time t,

then , so s is an antiderivative of . In Section 5.1 we considered an object that

always moves in the positive direction and made the guess that the area under the velocity

curve is equal to the distance traveled. In symbols:

That is exactly what FTC2 says in this context.

EXAMPLE 5 Evaluate the integral .

SOLUTION The function is continuous on and we know from Sec-

tion 4.9 that an antiderivative is , so Part 2 of the Fundamental Theorem

gives

dx ! F!1" ! F!!2" !

!1"

!!2"

! !

F!x" !

#!2, 1$f !x" ! x

v!t" dt ! s!b" ! s!a"

vv!t" ! s"!t"

s!t"

v!t"

baF!x"

a # x # bf !x"

f !x" dx#a, b$

f !x" dx

f !t" dt

! t!b" ! t!a" ! t!b"

F!b" ! F!a" ! #t!b" $ C $ ! #t!a" $ C $

x ! ax ! b

t!a" !

f !t" dt ! 0

t!x"x ! a

x ! b

x ! ax l b

x l a

#a, b$tFa

F!x" ! t!x" $ C

#a, b$fFf

tt"!x" ! f !x"t!x" ! x

f !t" dt

F" ! ffF

f !x" dx ! F!b" ! F!a"

#a, b$

318

|| ||

CHAPTER 5 INTEGRALS

N We abbreviate this theorem as FTC2.

Notice that FTC2 says we can use any antiderivative F of f. So we may as well use

the simplest one, namely , instead of or . M

We often use the notation

So the equation of FTC2 can be written as

Other common notations are and .

EXAMPLE 6 Find the area under the parabola from 0 to 1.

SOLUTION An antiderivative of is . The required area is found using

Part 2 of the Fundamental Theorem:

If you compare the calculation in Example 6 with the one in Example 2 in Section 5.1,

you will see that the Fundamental Theorem gives a much shorter method.

EXAMPLE 7 Find the area under the cosine curve from to , where .

SOLUTION Since an antiderivative of is , we have

In particular, taking , we have proved that the area under the cosine curve

from 0 to is . (See Figure 9.) M

When the French mathematician Gilles de Roberval ﬁrst found the area under the sine

and cosine curves in 1635, this was a very challenging problem that required a great deal

of ingenuity. If we didn’t have the beneﬁt of the Fundamental Theorem, we would have to

compute a difﬁcult limit of sums using obscure trigonometric identities (or a computer

algebra system as in Exercise 25 in Section 5.1). It was even more difﬁcult for Roberval

because the apparatus of limits had not been invented in 1635. But in the 1660s and 1670s,

when the Fundamental Theorem was discovered by Barrow and exploited by Newton and

Leibniz, such problems became very easy, as you can see from Example 7.

EXAMPLE 8 What is wrong with the following calculation?

SOLUTION To start, we notice that this calculation must be wrong because the answer is

negative but and Property 6 of integrals says that when

f !x" dx & 0f !x" ! 1%x

& 0

dx !

! !

! 1 ! !

sin!

%2" ! 1

b !

A !

cos x dx ! sin x

]

! sin b ! sin 0 ! sin b

F!x" ! sin xf !x" ! cos x

0 # b #

%2b0

A !

dx !

AF!x" !

f !x" ! x

y ! x

#F!x"$

F!x"

F"! fwhere

f !x" dx ! F!x"

]

F!x"

]

! F!b" ! F!a"

$ C

$ 7F!x" !

SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS

|| ||

319

N In applying the Fundamental Theorem we use

a particular antiderivative of . It is not neces-

sary to use the most general antiderivative.

FIG URE 9

y=cosx

area=1

. The Fundamental Theorem of Calculus applies to continuous functions. It can’t

be applied here because is not continuous on . In fact, f has an inﬁ-

nite discontinuity at , so

DIFFEREN T I ATI O N A N D I N T E G R ATION A S INV E R S E P ROC E S S E S

We end this section by bringing together the two parts of the Fundamental Theorem.

THE FUNDAMENTAL THEOREM OF CALCULUS

Suppose is continuous on .

If , then .

, where is any antiderivative of , that is, .

We noted that Part 1 can be rewritten as

which says that if is integrated and then the result is differentiated, we arrive back at the

original function . Since , Part 2 can be rewritten as

This version says that if we take a function , ﬁrst differentiate it, and then integrate the

result, we arrive back at the original function , but in the form . Taken

together, the two parts of the Fundamental Theorem of Calculus say that differentiation

and integration are inverse processes. Each undoes what the other does.

The Fundamental Theorem of Calculus is unquestionably the most important theorem

in calculus and, indeed, it ranks as one of the great accomplishments of the human mind.

Before it was discovered, from the time of Eudoxus and Archimedes to the time of Galileo

and Fermat, problems of ﬁnding areas, volumes, and lengths of curves were so difﬁcult

that only a genius could meet the challenge. But now, armed with the systematic method

that Newton and Leibniz fashioned out of the Fundamental Theorem, we will see in the

chapters to come that these challenging problems are accessible to all of us.

F!b" ! F!a"F

F"!x" dx ! F!b" ! F!a"

F"!x" ! f !x"f

f !t" dt ! f !x"

F" ! ffF

f !x" dx ! F!b" ! F!a"

t"!x" ! f !x"t!x" !

f !t" dt

#a, b$f

does not exist

x ! 0

#!1, 3$f !x" ! 1%x

f & 0

320

|| ||

CHAPTER 5 INTEGRALS

Openmirrors.com

SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS

|| ||

321

5–6 Sketch the area represented by . Then ﬁnd in two

ways: (a) by using Part 1 of the Fundamental Theorem and (b) by

evaluating the integral using Part 2 and then differentiating.

5. 6.

7–18 Use Part 1 of the Fundamental Theorem of Calculus to ﬁnd

the derivative of the function.

7. 8.

10.

11.

12.

14.

15. 16.

18.

19–36 Evaluate the integral.

19. 20.

21. 22.

23. 24.

25. 26.

27. 28.

29. 30.

31. 32.

33. 34.

$ 1

!1 $ 2y"

sec

(

tan

(

sec

t dt

!y ! 1"!2y $ 1" dy

x ! 1

(

3 $ x

)

x!2 $ x

" dx

cos

(

4%5

(

1 $

)

!5 ! 2t $ 3t

" dt

6 dx

! 2x" dx

y !

1%x

sin

t dty !

1!3x

1 $ u

17.

y !

cos x

!1 $ v

dvy !

tan x

t $

h!x" !

1 $ r

drh!x" !

1%x

sin

t dt

13.

G!x" !

cos

t dt

(

Hint:

1 $ sec t

dt ! !

1 $ sec t dt

F!x" !

1 $ sec t dt

t!r" !

$ 4

dxt!y" !

sin t dt

t!x" !

!2 $ t

dtt!x" !

$ 1

t!x" !

(

1 $

)

dtt!x" !

t"!x"t!x"

1. Explain exactly what is meant by the statement that “differenti-

ation and integration are inverse processes.”

2. Let , where is the function whose graph

is shown.

(a) Evaluate for and 6.

(b) Estimate .

minimum value?

(d) Sketch a rough graph of .

Let , where is the function whose graph

is shown.

(a) Evaluate , , , , and .

(b) On what interval is increasing?

(d) Sketch a rough graph of .

4. Let , where is the function whose graph is

shown.

(a) Evaluate and .

(b) Estimate , and .

(d) Where does have a maximum value?

(e) Sketch a rough graph of .

(f) Use the graph in part (e) to sketch the graph of .

Compare with the graph of .

t"!x"

t!0"t!!2", t!!1"

t!3"t!!3"

ft!x" !

f !t" dt

1 5

t!6"t!3"t!2"t!1"t!0"

ft!x" !

f !t" dt

t!7"

x ! 0, 1, 2, 3, 4, 5,t!x"

4 6

ft!x" ! x

f !t" dt

E X E R C I S E S

5.3