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5-15 SECTION 5.2 INTEGRATION BY SUBSTITUTION 385

SECTION 5.2 Integration by Substitution

The majority of functions that occur in practical situations can be differentiated by
applying rules and formulas such as those you learned in Chapter 2. Integration, how-
ever, is at least as much an art as a science, and many integrals that appear decep-
tively simple may actually require a special technique or clever insight.
For example, we easily find that

! 1
x 7 dx " x 8 ! C
8
by applying the power rule, but suppose we wish to compute

! (3x ! 5)7 dx

We could proceed by expanding the integrand (3x ! 5)7 and then integrating term by
term, but the algebra involved in this approach is daunting. Instead, we make the
change of variable
1
u " 3x ! 5 so that du " 3 dx or dx " du
3
Just-In-Time REVIEW Then, by substituting these quantities into the given integral, we get

Recall that the differential of

y " f(x) is dy " f#(x) dx. !
(3x ! 5)7 dx " u 7 !
$13 du%
" $ u %!C" u
1 1 1
8 8
! C power rule
3 8 24
1 since u " 3x ! 5
" (3x ! 5)8 ! C
24

We can check this computation by differentiating using the chain rule (Section 2.4):

" #
d 1 1
(3x ! 5)8 " [8(3x ! 5)7(3)] " (3x ! 5)7
dx 24 24

1
which verifies that (3x ! 5)8 is indeed an antiderivative of (3x ! 5)7.
24
The change of variable procedure we have just demonstrated is called integration
by substitution, and it amounts to reversing the chain rule for differentiation. To see
why, consider an integral that can be written as

! !
f(x) dx " g(u(x))u#(x) dx

Suppose G is an antiderivative of g, so that G# " g. Then, according to the chain rule

d
[G(u(x))] " G#(u(x)) u#(x)
dx
" g(u(x)) u#(x) since G# " g
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386 CHAPTER 5 Integration 5-16

Therefore, by integrating both sides of this equation with respect to x, we find that

! !
f(x) dx " g(u(x))u#(x) dx

" !$ d
dx %
[G(u(x))] dx

" G(u(x)) ! C since &G# " G

In other words, once we have an antiderivative for g(u), we also have one for f (x).
A useful device for remembering the substitution procedure is to think of
u " u(x) as a change of variable whose differential du " u#(x) dx can be manipulated
algebraically. Then

! !
f(x) dx " g(u(x)) u#(x) dx

!
" g(u) du substitute du for u#(x) dx

" G(u) ! C where G is an antiderivative of g

" G(u(x)) ! C substitute u(x) for u
Here is a step-by-step procedure for integrating by substitution.

Using Substitution to Integrate ! f (x) dx

Step 1. Choose a substitution u " u(x) that “simplifies” the integrand f(x).
Step 2. Express the entire integral in terms of u and du " u#(x) dx. This means
that all terms involving x and dx must be transformed to terms involv-
ing u and du.
Step 3. When step 2 is complete, the given integral should have the form

! !
f(x) dx " g(u) du

If possible, evaluate this transformed integral by finding an

antiderivative G(u) for g(u).
Step 4. Replace u by u(x) in G(u) to obtain an antiderivative G(u(x)) for f (x),
so that

! f(x) dx " G(u(x)) ! C

An old saying goes, “The first step in making rabbit stew is to catch a rabbit.”
Likewise, the first step in integrating by substitution is to find a suitable change of
variable u " u(x) that simplifies the integrand of the given integral &f(x) dx without
adding undesired complexity when dx is replaced by du " u#(x) dx. Here are a few
guidelines for choosing u(x):
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5-17 SECTION 5.2 INTEGRATION BY SUBSTITUTION 387

1. If possible, try to choose u so that u#(x) is part of the integrand f(x).

2. Consider choosing u as the part of the integrand that makes f(x) difficult to inte-
grate directly, such as the quantity inside a radical, the denominator of a fraction,
or the exponent of an exponential function.
3. Don’t “oversubstitute.” For instance, in our introductory example &(3x ! 5)7 dx,
a common mistake is to use u " (3x ! 5)7. This certainly simplifies the integrand,
but then du " 7(3x ! 5)6(3) dx, and you are left with a transformed integral that
is more complicated than the original.
4. Persevere. If you try a substitution that does not result in a transformed integral
you can evaluate, try a different substitution.
Examples 5.2.1 through 5.2.6 illustrate how substitutions are chosen and used in var-
ious kinds of integrals.

EXAMPLE 5.2.1

Find !( 2x ! 7 dx.

Solution
We choose u " 2x ! 7 and obtain
1
du " 2 dx so that dx " du
2
Then the integral becomes

!( !
2x ! 7 dx " (u $12 du%
" !
1 1'2
2
u du since (u " u1'2

1 u 3'2 1
" ! C " u 3'2 ! C power rule
2 3'2 3
1
" (2x ! 7)3'2 ! C substitute 2x ! 7 for u
3

EXAMPLE 5.2.2

Find ! 8x(4x2 $ 3)5 dx.

Solution
First, note that the integrand 8x(4x2 $ 3)5 is a product in which one of the factors,
8x, is the derivative of an expression, 4x2 $ 3, that appears in the other factor. This
suggests that you make the substitution
u " 4x2 $ 3 with du " 4(2x dx) " 8x dx
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388 CHAPTER 5 Integration 5-18

to obtain

! !
8x(4x 2 $ 3)5 dx " (4x 2 $ 3)5(8x dx)

!
" u 5 dx

1
" u6 ! C power rule
6
1 2
" (4x $ 3)6 ! C substitute 4x2 $ 3 for u
6

EXAMPLE 5.2.3

Find ! 4
x 3e x !2 dx.

Solution
If the integrand of an integral contains an exponential function, it is often useful to
substitute for the exponent. In this case, we choose
u " x4 ! 2 so that du " 4x3 dx
and

! x 3e x
4
!2
!
dx " e x
4
(x 3 dx)
!2

!
" eu $14 du% since du " 4x3 dx

1
" eu ! C exponential rule
4
1 x4 !2
" e !C substitute x4 ! 2 for u
4

EXAMPLE 5.2.4

Find ! x
x$1
dx.

Solution
Following our guidelines, we substitute for the denominator of the integrand, so that
u " x $ 1 and du " dx. Since u " x $ 1, we also have x " u ! 1. Thus,

! x
x$1
dx "
u
!
u!1
du

" !" 1!
1
u
du # divide

" u ! ln |u| ! C constant and logarithmic rules

" x $ 1 ! ln |x $ 1| ! C substitute x $ 1 for u
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5-19 SECTION 5.2 INTEGRATION BY SUBSTITUTION 389

EXAMPLE 5.2.5

Find ! 3x ! 6
(2x 2 ! 8x ! 3
dx.

Solution
This time, our guidelines suggest substituting for the quantity inside the radical in the
denominator; that is,
u " 2x2 ! 8x ! 3 du " (4x ! 8) dx

At first glance, it may seem that this substitution fails, since du " (4x ! 8) dx
appears quite different from the term (3x ! 6) dx in the integral. However, note
that

3
(3x ! 6) dx " 3(x ! 2) dx " (4)[(x ! 2) dx]
4
3 3
" [(4x ! 8) dx] " du
4 4
Substituting, we find that

! 3x ! 6
(2x ! 8x ! 3
2
dx " !1
(2x ! 8x ! 3
2
[(3x ! 6) dx]

" !$ % !
(
1 3
u 4
du "
3
4
u $1'2 du

" $ %
1'2
3 u 3
! C " (u ! C
4 1'2 2
3 substitute
" (2x 2 ! 8x ! 3 ! C u " 2x2 ! 8x ! 3
2

EXAMPLE 5.2.6

Find ! (ln x)2

x
dx.

Solution
Because

d 1
(ln x) "
dx x
the integrand

(ln x)2
x
" (ln x)2
1
x $%
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390 CHAPTER 5 Integration 5-20

1
is a product in which one factor is the derivative of an expression ln x that appears
x
1
in the other factor. This suggests substituting u " ln x with du " dx so that
x

! (ln x)2
x
! 1
$ %
dx " (ln x)2 dx
x

! 1
" u 2 du " u 3 ! C
3
1
" (ln x)3 ! C substitute ln x for u
3

Sometimes an integral “looks” like it should be evaluated using a substitution but

closer examination reveals a more direct approach. Consider Example 5.2.7.

EXAMPLE 5.2.7

Find ! e 5x!2 dx.

Solution
You can certainly handle this integral using the substitution
u " 5x ! 2 du " 5 dx
but it is not really necessary since e5x!2 " e5xe2, and e2 is a constant. Thus,

! !
e 5x!2 dx " e 5xe 2 dx

!
" e 2 e 5x dx factor constant e2 outside integral

e 5x
" e2 " #
5
!C exponential rule

1
" e 5x!2 ! C since e2e5x " e5x!2
5

In Example 5.2.7, we used algebra to put the integrand into a form where sub-
stitution was not necessary. In Examples 5.2.8 and 5.2.9, we use algebra as a first
step, before making a substitution.

EXAMPLE 5.2.8

Find ! x 2 ! 3x ! 5
x!1
dx.
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5-21 SECTION 5.2 INTEGRATION BY SUBSTITUTION 391

Solution
There is no easy way to approach this integral as it stands (remember, there is no
“quotient rule” for integration). However, suppose we simply divide the denominator
into the numerator:

x!2
x!1 )x 2 ! 3x ! 5
$x(x ! 1)
2x ! 5
$2(x ! 1)
3
that is,

x 2 ! 3x ! 5 3
"x!2!
x!1 x!1

3
We can integrate x ! 2 directly using the power rule. For the term , we use the
x!1
substitution u " x ! 1; du " dx:

! x2 ! 3x ! 5
x!1
dx " !"
x!2!
3
x!1 #
dx

! !
" x dx ! 2 dx ! ! 3
u
du
u"x!1
du " dx
1
" x2 ! 2x ! 3 ln |u| ! C
2
1 2
" x ! 2x ! 3 ln |x ! 1| ! C substitute x ! 1 for u
2

EXAMPLE 5.2.9

Find ! 1
1 ! e $x
dx.

Solution
You may try to substitute w " 1 ! e$x. However, this is a dead end because
dw " $e$x dx but there is no e$x term in the numerator of the integrand. Instead,
note that

1 1 1
" "
1 ! e $x 1 ex ! 1
1! x
e ex
x
e
" x
e !1
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392 CHAPTER 5 Integration 5-22

Now, if you substitute u " ex ! 1 with du " ex dx into the given integral, you get

! 1
1!e $x dx " x
ex
e !1
!
dx " x
1
e !1
!
(e x dx)

" !1
u
du

" ln |u| ! C
" ln |e x ! 1| ! C substitute ex ! 1 for u

When Substitution The method of substitution does not always succeed. In Example 5.2.10, we consider
Fails an integral very similar to the one in Example 5.2.3 but just enough different so no
substitution will work.

EXAMPLE 5.2.10

Evaluate ! 4
x 4e x !2
dx.

Just-In-Time REVIEW Solution

Notice that if u " x4 ! 2, then The natural substitution is u " x4 ! 2, as in Example 5.2.3. As before, you find
x4 " u $ 2, so 1
du " 4x3 dx, so x3 dx " du, but this integrand involves x4, not x3. The “extra” factor
4
x " (u $ 2)1'4 " (u $ 2 4
4
of x satisfies x " ( u $ 2, so when the substitution is made, you have

! x4ex
4
!2
!
dx " xex
4
!2
!
(x3 dx) " (
4 1
u $ 2 eu ( du)
4
which is hardly an improvement on the original integral! Try a few other possible
substitutions (say, u " x2 or u " x3) to convince yourself that nothing works.

An Application EXAMPLE 5.2.11

Involving Substitution The price p (dollars) of each unit of a particular commodity is estimated to be chang-
ing at the rate
dp $135x
"
dx (9 ! x 2
where x (hundred) units is the consumer demand (the number of units purchased at
that price). Suppose 400 units (x " 4) are demanded when the price is $30 per unit.
a. Find the demand function p(x).
b. At what price will 300 units be demanded? At what price will no units be demanded?
c. How many units are demanded at a price of $20 per unit?
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5-23 SECTION 5.2 INTEGRATION BY SUBSTITUTION 393

Solution
a. The price per unit p(x) is found by integrating p#(x) with respect to x. To per-
form this integration, use the substitution

1
u " 9 ! x2, du " 2x dx, x dx " du
2
to get

p(x) "!( $135x

9!x 2 dx " !
$135 1
u 1/2 2 $%
du

"
$135
2
!
u $1/2 du

$ %
$135 u 1'2
" !C
2 1'2
" $135(9 ! x 2 ! C substitute 9 ! x 2 for u

Since p " 30 when x " 4, you find that

30 " $135(9 ! 42 ! C
C " 30 ! 135(25 " 705

so

p(x) " $135(9 ! x 2 ! 705

b. When 300 units are demanded, x " 3 and the corresponding price is

p(3) " $135(9 ! 32 ! 705 " $132.24 per unit

No units are demanded when x " 0 and the corresponding price is

p(0) " $135(9 ! 0 ! 705 " $300 per unit

c. To determine the number of units demanded at a unit price of $20 per unit, you
need to solve the equation

$135(9 ! x 2 ! 705 " 20

135(9 ! x 2 " 685
685
(9 ! x 2 "
135
2
9 ! x * 25.75 square both sides
x 2 * 16.75
x * 4.09

That is, roughly 409 units will be demanded when the price is $20 per unit.
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394 CHAPTER 5 Integration 5-24

EXERCISES 5.2
In Exercises 1 and 2, fill in the table by specifying the
substitution you would choose to find each of the four 8. !
[(x $ 1)5 ! 3(x $ 1)2 ! 5] dx

!
given integrals.
2
1. 9. xe x dx

a. !
(3x ! 4)
Integral

5'2
dx
Substitution u
10. !
2xe x
2
$1
dx

b. ! 4
dx
11. !
t(t 2 ! 1)5 dt

!(
3$x

c. ! 2
te 2$t dt
12. 3t t2 ! 8 dt

d. !
t(2 ! t 2)3 dt
13. !
x 2(x 3 ! 1)3'4 dx

2. 14. !
x 5e 1$x dx
6

Integral Substitution u
15. !2y 4
dy
!
5
3
y !1

!
a. dx
(2x $ 5)4 y2
16. dy
! (y ! 5)2
3
3
b. x 2e $x dx

c. !t
et
e !1
dt
17. !
(x ! 1)(x 2 ! 2x ! 5)12 dx

d. ! 3
(t
t!3
2 ! 6t ! 5
dt 18. !
(3x 2 $ 1)e x
3
$x
dx

In Exercises 3 through 36, find the indicated integral

19. ! 3x 4 ! 12x 3 ! 6
x ! 5x 4 ! 10x ! 12
5 dx

and check your answer by differentiation.

!10x 3 $ 5x

!
20. dx
(x 4 $ x 2 ! 6
(2x ! 6)5 dx
!
3.
3u $ 3

!
21. du
(u 2 $ 2u ! 6)2
4. e5x!3 dx

5. ! (4x $ 1 dx
22. ! 6u $ 3
4u 2 $ 4u ! 1
du

6. ! 1
dx
23. !
ln 5x
x
dx

7. !
3x ! 5

e 1$x dx
24. ! 1
x ln x
dx
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5-25 SECTION 5.2 INTEGRATION BY SUBSTITUTION 395

25. ! 1
x(ln x)2
dx
In Exercises 43 through 46, the slope f #(x) at each
point (x, y) on a curve y " f(x) is given, along with a

!
particular point (a, b) on the curve. Use this
ln x 2 information to find f(x).
26. dx
x
43. f #(x) " (1 $ 2x)3/2; (0, 0)
27. ! x2 ! 1
2
2x ln (x ! 1)
dx 44. f #(x) " x(x 2 ! 5; (2, 10)
2
45. f #(x) " xe 4$x ; ($2, 1)
28. ! e (x
(x
dx
46. f #(x) "
2x
1 ! 3x 2
; (0, 5)

29. ! e x ! e $x
e x $ e $x
dx In Exercises 47 through 50, the velocity v(t) " x#(t) at
time t of an object moving along the x axis is given,
30. ! e $x(1 ! e 2x) dx
along with the initial position x(0) of the object. In
each case, find:

!
(a) The position x(t) at time t.
x
31. dx (b) The position of the object at time t " 4.
2x ! 1 (c) The time when the object is at x " 3.

32. ! t$1
t!1
dt 47. x#(t) " $2(3t ! 1)1/2 ; x(0) " 4
$1

!(
48. x#(t) " ; x(0) " 5
1 ! 0.5t
33. x 2x ! 1 dx
1

!
49. x#(t) " ; x(0) " 0
x (2t ! 1
34. 3 dx
(4 $ 3x $2t

!
50. x#(t) " ; x(0) " 4
1 (1 ! t2)3'2
35. dx
( (x ! 1)
x ( 51. MARGINAL COST At a certain factory, the
[Hint: Let u " (x ! 1.] marginal cost is 3(q $ 4)2 dollars per unit when

!
the level of production is q units.
$ %
2'3
1 1
36. $1 dx a. Express the total production cost in terms of the
x2 x overhead (the cost of producing 0 units) and the

"Hint: Let u " 1x $1.#

number of units produced.
b. What is the cost of producing 14 units if the
overhead is $436?
In Exercises 37 through 42, solve the given initial 52. DEPRECIATION The resale value of a certain
value problem for y = f(x). industrial machine decreases at a rate that depends
dy on its age. When the machine is t years old, the
37. " (3 $ 2x)2 where y " 0 when x " 0 rate at which its value is changing is $960e$t/5
dx
dy dollars per year.
38. " (4x ! 5 where y " 3 when x " 1 a. Express the value of the machine in terms of its
dx
age and initial value.
dy 1 b. If the machine was originally worth $5,200,
39. " where y " 1 when x " 0
dx x ! 1 how much will it be worth when it is 10 years
dy old?
40. " e 2$x where y " 0 when x " 2
dx 53. TREE GROWTH A tree has been transplanted
dy x!2 and after x years is growing at the rate of
41. " 2 where y " 3 when x " $1 1
dx x ! 4x ! 5 1! meters per year. After 2 years, it has
(x ! 1)2
dy ln(x
42. " where y " 2 when x " 1 reached a height of 5 meters. How tall was it
dx x when it was transplanted?
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396 CHAPTER 5 Integration 5-26

54. RETAIL PRICES In a certain section of the a. Find V(x).

country, it is estimated that t weeks from now, the b. How much will the land be worth in 10 years?
price of chicken will be increasing at the rate of c. Use the graphing utility of your calculator with
p#(t) " 3(t ! 1 cents per kilogram per week. If TRACE and ZOOM to determine how long it
chicken currently costs $2.30 per kilogram, what will take for the land to be worth $1,000 per
will it cost 8 weeks from now? acre.
55. REVENUE The marginal revenue from the sale 59. AIR POLLUTION In a certain suburb of Los
of x units of a particular commodity is estimated Angeles, the level of ozone L(t) at 7:00 A.M. is
to be 0.25 parts per million (ppm). A 12-hour weather
2
R#(x) " 50 ! 3.5xe $0.01x dollars per unit forecast predicts that the ozone level t hours later
where R(x) is revenue in dollars. will be changing at the rate of
a. Find R(x), assuming that R(0) " 0. 0.24 $ 0.03t
L#(t) "
b. What revenue should be expected from the sale (36 ! 16t $ t 2
of 1,000 units? parts per million per hour (ppm/hr).
56. WATER POLLUTION An oil spill in the ocean a. Express the ozone level L(t) as a function of t.
is roughly circular in shape, with radius R(t) feet When does the peak ozone level occur? What is
t minutes after the spill begins. The radius is the peak level?
increasing at the rate b. Use the graphing utility of your calculator to
sketch the graph of L(t) and use TRACE and
21 ZOOM to answer the questions in part (a).
R#(t) " ft'min
0.07t ! 5 Then determine at what other time the ozone
a. Find an expression for the radius R(t), assuming level will be the same as it is at 11:00 A.M.
that R " 0 when t " 0. 60. SUPPLY The owner of a fast-food chain
b. What is the area A " %R2 of the spill after 1 hour? determines that if x thousand units of a new meal
57. DRUG CONCENTRATION The concentration item are supplied, then the marginal price at that
C(t) in milligrams per cubic centimeter (mg/cm3) level of supply is given by
of a drug in a patient’s bloodstream is 0.5 mg/cm3 x
p#(x) " dollars per meal
immediately after an injection and t minutes later (x ! 3)2
is decreasing at the rate where p(x) is the price (in dollars) per unit at
$0.01e 0.01t which all x meal units will be sold. Currently,
C#(t) " 0.01t mg'cm3 per minute 5,000 units are being supplied at a price of $2.20
(e ! 1)2
A new injection is given when the concentration per unit.
drops below 0.05 mg/cm3. a. Find the supply (price) function p(x).
a. Find an expression for C(t). b. If 10,000 meal units are supplied to restaurants
b. What is the concentration after 1 hour? After in the chain, what unit price should be charged
3 hours? so that all the units will be sold?
c. Use the graphing utility of your calculator with 61. DEMAND The manager of a shoe store
TRACE and ZOOM to determine how much determines that the price p (dollars) for each pair of
time passes before the next injection is given. a popular brand of sports sneakers is changing at
58. LAND VALUE It is estimated that x years from the rate of
now, the value V(x) of an acre of farmland will be $300x
p#(x) " 2
increasing at the rate of (x ! 9)3'2
0.4x 3 when x (hundred) pairs are demanded by
V#(x) " consumers. When the price is $75 per pair, 400
(0.2x 4 ! 8,000
dollars per year. The land is currently worth $500 pairs (x " 4) are demanded by consumers.
per acre.
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5-27 SECTION 5.3 THE DEFINITE INTEGRAL AND THE FUNDAMENTAL THEOREM OF CALCULUS 397

a. Find the demand (price) function p(x). 64. MARGINAL PROFIT Repeat Exercise 63 for
b. At what price will 500 pairs of sneakers be 11 $ x
demanded? At what price will no sneakers be marginal revenue R#(x) " and for the
(14 $ x
demanded? 2
marginal cost C#(x) " 2 ! x ! x .

!
c. How many pairs will be demanded at a price of
$90 per pair? 65. Find x 1'3(x 2'3 ! 1)3'2 dx. [Hint: Substitute
62. SUPPLY The price p (dollars per unit) of a
u " x2/3 ! 1 and use x2/3 " u $ 1.]
particular commodity is increasing at thc rate

p#(x) "
20x
(7 $ x)2
66. Find ! x 3(4 $ x 2) $1'2 dx. [Hint: Substitute

when x hundred units of the commodity are u " 4 $ x2 and use the fact that x2 " 4 $ u.]
supplied to the market. The manufacturer supplies
200 units (x " 2) when the price is $2 per unit.
a. Find the supply firnction p(x).
67. Find ! e 2x
1 ! ex
dx. [Hint: Let u " 1 ! ex.]

b. What price corresponds to a supply of

500 units?
68. Find ! e $x(1 ! e x)2 dx. [Hint: Is it better to set

63. MARGINAL PROFIT A company determines u " 1 ! ex or u " ex? Or is it better to not
that the marginal revenue from the production of even use the method of substitution?]
x units is R#(x) " 7 $ 3x $ 4x2 hundred dollars
per unit, and the corresponding marginal cost is
C#(x) " 5 ! 2x hundred dollars per unit. By how
much does the profit change when the level of
production is raised from 5 to 9 units?

SECTION 5.3 The Definite Integral and the

Fundamental Theorem of Calculus
Suppose a real estate agent wants to evaluate an unimproved parcel of land that is
100 feet wide and is bounded by streets on three sides and by a stream on the fourth
side. The agent determines that if a coordinate system is set up as shown in Figure 5.2,
the stream can be described by the curve y " x3 ! 1, where x and y are measured in
hundreds of feet. If the area of the parcel is A square feet and the agent estimates its
land is worth $12 per square foot, then the total value of the parcel is 12A dollars. If
the parcel were rectangular in shape or triangular or even trapezoidal, its area A could
be found by substituting into a well-known formula, but the upper boundary of the
parcel is curved, so how can the agent find the area and hence the total value of
the parcel?
Our goal in this section is to show how area under a curve, such as the area A
in our real estate example, can be expressed as a limit of a sum of terms called a
definite integral. We will then introduce a result called the fundamental theorem of
calculus that allows us to compute definite integrals and thus find area and other quan-
tities by using the indefinite integration (antidifferentiation) methods of Sections 5.1
and 5.2. In Example 5.3.3, we will illustrate this procedure by expressing the area A
in our real estate example as a definite integral and evaluating it using the funda-
mental theorem of calculus.