ALTERNATIVE TREATMENT:
AREA
AND INTEGRALS
1
Sigma Notation
2
Area
3
The Definite Integral
2
7
16
Reprinted with permission from Calculus: Early Transcendentals, Third Edition
by James Stewart ©1995 by Brooks/Cole Publishing Company,
A division of International Thomson Publishing Co.
2
❙❙❙❙
SECTION 1 SIGMA NOTATION
||||
SECTION 1 Sigma Notation
In finding areas and evaluating integrals we often encounter sums with many terms. A
convenient way of writing such sums uses the Greek letter (capital sigma, corresponding to our letter S) and is called sigma notation.
This tells us to
end with i=n.
This tells us
to add.
1 Definition If a m, a m1, . . . , a n are real numbers and m and n are integers such
that m n, then
n
µ ai
n
a
im
This tells us to
start with i=m.
i
a m a m1 a m2 a n1 a n
im
With function notation, Definition 1 can be written as
n
f i f m f m 1 f m 2 f n 1 f n
im
Thus, the symbol nim indicates a summation in which the letter i (called the index of
summation) takes on the values m, m 1, . . . , n. Other letters can also be used as the
index of summation.
EXAMPLE 1
4
i
(a)
2
12 2 2 3 2 42 30
i1
n
i 3 4 5 n 1 n
(b)
i3
5
2
(c)
j
2 0 2 1 2 2 2 3 2 4 2 5 63
j0
n
1
1
1
1
1 2
3
n
k1 k
3
i1
11
21
31
1
1
13
(e) 2
2
2
2
0 1 3
2 3
3 3
7
6
42
i1 i 3
(d)
4
(f)
222228
i1
EXAMPLE 2 Write the sum 2 3 3 3 n 3 in sigma notation.
SOLUTION There is no unique way of writing a sum in sigma notation. We could write
n
23 33 n 3 i
3
i2
n1
or
23 33 n 3 j 1
3
j1
n2
or
23 33 n 3 k 2
3
k0
The following theorem gives three simple rules for working with sigma notation.
SECTION 1 SIGMA NOTATION
2
ca
n
i
a
c
im
a
n
(b)
i
im
n
(c)
3
Theorem If c is any constant (that is, it does not depend on i ), then
n
(a)
❙❙❙❙
bi im
n
i
bi im
n
i
a
a
a
n
i
im
b
i
im
n
i
im
b
i
im
Proof To see why these rules are true, all we have to do is write both sides in expanded
form. Rule (a) is just the distributive property of real numbers:
ca m ca m1 ca n ca m a m1 a n Rule (b) follows from the associative and commutative properties:
a m bm a m1 bm1 a n bn am am1 an bm bm1 bn Rule (c) is proved similarly.
n
EXAMPLE 3 Find
1.
i1
n
1 1 1 1 n
SOLUTION
i1
n terms
|||| PRINCIPLE OF MATHEMATICAL INDUCTION
Let Sn be a statement involving the positive
integer n. Suppose that
1. S1 is true.
2. If Sk is true, then Sk1 is true.
Then Sn is true for all positive integers n.
EXAMPLE 4 Prove the formula for the sum of the first n positive integers:
n
i 1 2 3 n i1
nn 1
2
SOLUTION This formula can be proved by mathematical induction (see page 59) or by the
following method used by the German mathematician Karl Friedrich Gauss (1777–1855)
when he was ten years old.
Write the sum S twice, once in the usual order and once in reverse order:
S1
2
3
n 1 n
S n n 1 n 2 2
1
Adding all columns vertically, we get
2S n 1 n 1 n 1 n 1 n 1
On the right side there are n terms, each of which is n 1, so
2S nn 1
or
S
nn 1
2
EXAMPLE 5 Prove the formula for the sum of the squares of the first n positive
integers:
n
i
i1
2
12 2 2 3 2 n 2 nn 12n 1
6
4
❙❙❙❙
SECTION 1 SIGMA NOTATION
SOLUTION 1 Let S be the desired sum. We start with the telescoping sum (or collapsing
sum):
n
1 i
3
Most terms cancel in pairs.
i 3 2 3 13 3 3 2 3 4 3 3 3 n 13 n 3 i1
n 13 13 n 3 3n 2 3n
On the other hand, using Theorem 2 and Examples 3 and 4, we have
n
n
1 i 3
i 3 i1
3i
n
2
3i 1 3
i1
i
2
i1
3S 3
n
n
i1
i1
3
i1
nn 1
n 3S 32 n 2 52 n
2
Thus we have
n 3 3n 2 3n 3S 32 n 2 52 n
Solving this equation for S, we obtain
3S n 3 32 n 2 12 n
S
or
|||| See pages 59 and 61 for a more thorough
discussion of mathematical induction.
2n 3 3n 2 n
nn 12n 1
6
6
SOLUTION 2 Let Sn be the given formula.
1. S1 is true because
12 11 12 1 1
6
2. Assume that Sk is true; that is,
12 2 2 3 2 k 2 kk 12k 1
6
Then
12 2 2 3 2 k 12 12 2 2 3 2 k 2 k 12
kk 12k 1
k 12
6
k 1
k2k 1 6k 1
6
k 1
2k 2 7k 6
6
k 1k 22k 3
6
k 1k 1 12k 1 1
6
So Sk1 is true.
By the Principle of Mathematical Induction, Sn is true for all n.
We list the results of Examples 3, 4, and 5 together with a similar result for cubes and
fourth powers (see Exercises 37–40) as Theorem 3. These formulas are needed for finding
areas in the next section.
SECTION 1 SIGMA NOTATION
3
Theorem Let c be a constant and n a positive integer. Then
n
(a)
n
1n
c nc
(b)
i1
n
(c)
i
i1
nn 1
2
n
(e)
i
3
i1
i1
2
nn 12n 1
6
4
nn 12n 13n 2 3n 1
30
n
i
(d)
nn 1
2
i1
2
n
i
(f)
i1
n
EXAMPLE 6 Evaluate
i4i
2
3.
i1
SOLUTION Using Theorems 2 and 3, we have
n
i4i
n
2
3 i1
4i
i1
4
n
|||| The type of calculation in Example 7 arises
in the next section when we compute areas.
EXAMPLE 7 Find lim
n l i1
SOLUTION
n
lim
n l i1
3
n
3
n
nn 1
2
i
n
3
3
i1
2
3
i
i1
nn 1
2
nn 12nn 1 3
2
nn 12n 2 2n 3
2
i
n
2
i
n
n
3i 4
3
1 .
2
n
1 lim
n l i1
lim
nl
lim
nl
lim
nl
lim
nl
3 2
3
i n3
n
3
n3
n
i2 i1
3
n
n
1
i1
3 nn 12n 1
3
n
3
n
6
n
1 n
2 n
n1
n
1
1
1 1
2
n
12 1 1 2 3 4
2n 1
n
2
1
n
3
3
❙❙❙❙
5
6
❙❙❙❙
SECTION 1 SIGMA NOTATION
||||
1 Exercises
1–10
1.
3.
6
si
2.
i1
6
6
3
i
4.
4
k0
i
2k 1
2k 1
6.
10
8.
40. Prove formula (e) of Theorem 3 using the following method
published by Abu Bekr Mohammed ibn Alhusain Alkarchi in
about A.D. 1010. The figure shows a square ABCD in which
sides AB and AD have been divided into segments of lengths 1,
2, 3, . . . , n. Thus the side of the square has length nn 12
so the area is nn 12 2. But the area is also the sum of the
areas of the n “gnomons” G1 , G2 , . . . , Gn shown in the figure.
Show that the area of Gi is i 3 and conclude that formula (e) is
true.
x
k
n3
j
2
jn
1
n
j
10.
j0
■
11–20
3
k5
n1
■
of Example 5, Solution 1 [start with 1 i 4 i 4 .
8
i1
9.
i
39. Prove formula (e) of Theorem 3 using a method similar to that
i4
n
7.
induction.
1
i1
i1
i4
5.
38. Prove formula (e) of Theorem 3 using mathematical
Write the sum in expanded form.
||||
5
■
f x x
i
i
i1
■
■
■
■
D
■
■
■
■
n
Write the sum in sigma notation.
||||
11. 1 2 3 4 10
13.
1
2
23 34 45 19
20
5
14.
3
7
48 59 106 23
27
4
15. 2 4 6 8 2n
17. 1 2 4 8 16 32
21–36
■
■
i1
■
23.
25.
22.
6
8
j1
24.
k0
20
100
1
n
26.
2
i 2
28.
30.
n
n
2
3i 4
32.
34.
43–46
43. lim
1
n
n
44. lim
1
n
n
n l i1
2 3i
n l i1
45. lim
2
n
n
46. lim
3
n
n l i1
3 2i 2
n l i1
ii 1i 2
■
n
3
i 2
36.
i1
k
2
■
■
■
■
37. Prove formula (b) of Theorem 3.
5 i1 a
i
a i1 i1
■
■
■
n
a i
i1
■
■
2
i
n
i
n
3
2i
n
1
■
1
3
5
3i
n
■
2i
n
3
2 1
■
3i
n
■
■
■
■
■
47. Prove the formula for the sum of a finite geometric series with
k 2 k 1
first term a and common ratio r 1:
i1
■
i
Find each limit.
||||
n
i1
n
5
n
(d)
i1
n
i1
ai i1
i 1i 2
B
i1
n
2 5i i1
■
1
1
i
i1
n
2i
i
(b)
4
i1
i
(c)
i 14 ii 2
i2
n
■
■
4
4
i
i1
35.
■
i1
n
33.
■
cos k
j1
i0
31.
i3
3
n
42. Prove the generalized triangle inequality
i4
4
29.
■
6
3i 2
n1
27.
■
i3
8
21.
99
■
Find the value of the sum.
||||
. . .
5
100
i
(a)
20. 1 x x 2 x 3 1n x n
■
G¢
n
19. x x 2 x 3 x n
■
G∞
41. Evaluate each telescoping sum.
14 19 161 251 361
■
.
.
.
3
G£
2
G™
1
A1 2 3 4
16. 1 3 5 7 2n 1
1
1
Gn
..
.
12. s3 s4 s5 s6 s7
18.
C
■
■
■
n
ar
i1
i1
a ar ar 2 ar n1 ar n 1
r1
■
SECTION 2 AREA
n
48. Evaluate
2
.
i1
n
cos ix n
49. Evaluate
i1
2i 2 .
i
i1
m
50. Evaluate
n
j1
n
cos ix i1
n
51. Find the number n such that
i 78.
n
sin ix 52. (a) Use the product formula for sin x cos y (see 18a in
i1
Appendix D) to show that
1
1
1
sin 2 nx sin 2 n 1x
sin 12 x
where x is not an integer multiple of 2.
2 sin 2 x cos ix sin(i 2) x sin(i 2) x
1
sin 12 nx cos 12 n 1x
sin 12 x
53. Use the method of Exercise 52 to prove the formula
i1
||||
sin(n 12) x sin 12 x
2 sin 12 x
where x is not an integer multiple of 2. Deduce that
i j i1
7
(b) Use the identity in part (a) and telescoping sums to prove
the formula
3
i1
❙❙❙❙
1
SECTION 2 Area
We begin by attempting to solve the area problem: Find the area of the region S that lies
under the curve y f x from a to b. This means that S, illustrated in Figure 1, is bounded
by the graph of a continuous function f [where f x 0], the vertical lines x a and
x b, and the x-axis.
y
y=ƒ
x=a
S
FIGURE 1
0
S=s(x, y) | a¯x¯b, 0¯y¯ƒd
a
x=b
b
x
In trying to solve the area problem we have to ask ourselves: What is the meaning of
the word area ? This question is easy to answer for regions with straight sides. For a rectangle, the area is defined as the product of the length and the width. The area of a triangle
is half the base times the height. The area of a polygon is found by dividing it into
triangles (as in Figure 2) and adding the areas of the triangles.
A™
w
h
l
FIGURE 2
A=lw
A¡
A£
A¢
b
A= 21 bh
A=A¡+A™+A£+A¢
However, it isn’t so easy to find the area of a region with curved sides. We all have an
intuitive idea of what the area of a region is. But part of the area problem is to make this
intuitive idea precise by giving an exact definition of area.
Recall that in defining a tangent we first approximated the slope of the tangent line by
slopes of secant lines and then we took the limit of these approximations. We pursue a sim-
8
❙❙❙❙
SECTION 2 AREA
ilar idea for areas. We first approximate the region S by polygons and then we take the limit
of the areas of these polygons. The following example illustrates the procedure.
EXAMPLE 1 Let’s try to find the area under the parabola y x 2 from 0 to 1 (the parabolic
region S illustrated in Figure 3).
y
(1, 1)
y=≈
0
FIGURE 3
x
1
One method of approximating the desired area is to divide the interval 0, 1 into
subintervals of equal length and consider the rectangles whose bases are these
subintervals and whose heights are the values of the function at the right-hand endpoints
of these subintervals. Figure 4 shows the approximation of the parabolic region by four,
eight, and n rectangles
y
y
y
(1, 1)
0
FIGURE 4
1
4
1
2
3
4
1
(1, 1)
x
0
x
1
1
8
(1, 1)
0
1
1
n
(a)
(b)
(c)
Let S n be the sum of the areas of the n rectangles in Figure 4(c). Each rectangle
has width 1n and the heights are the values of the function f x x 2 at the points
1n, 2n, 3n, . . . , nn; that is, the heights are 1n2, 2n2, 3n2, . . . , nn2. Thus
Sn 1
n
1
n
2
1
n
2
n
2
1
n
3
n
2
1 1 2
1 2 2 3 2 n 2 n n2
1
n3
1
n
n
n
2
n
i
2
i1
Using the formula for the sum of the squares of the first n integers [Formula 1.3(d)],
we can write
Sn 1 nn 12n 1
n 12n 1
3
n
6
6n 2
For instance, the sum of the areas of the four shaded rectangles in Figure 4(a) is
S4 59
0.46875
616
x
SECTION 2 AREA
n
Sn
10
20
30
50
100
1000
0.385000
0.358750
0.350185
0.343400
0.338350
0.333834
❙❙❙❙
9
and the sum of the areas of the eight rectangles in FIgure 4(b) is
S8 917
0.3984375
664
The results of similar calculations are shown in the table in the margin.
It looks as if S n is becoming closer to 13 as n increases. In fact
lim S n lim
nl
nl
n 12n 1
6n 2
lim
1
6
lim
1
6
nl
nl
n1
n
1
2n 1
n
1
n
1
n
2
16 1 2 13
From Figure 4 it appears that, as n increases, S n becomes a better and better approximation to the area of the parabolic segment. Therefore we define the area A to be the limit
of the sums of the areas of the approximating rectangles, that is,
A lim S n 13
nl
In applying the idea of Example 1 to the more general region S of Figure 1, we have no
need to use rectangles of equal width. We start by subdividing the interval a, b into n
smaller subintervals by choosing partition points x 0 , x 1, x 2 , . . . , x n so that
a x 0 x 1 x2 x n1 x n b
Then the n subintervals are
x 0 , x 1 ,
x 1, x 2 ,
x 2 , x 3 , . . . ,
x n1, x n This subdivision is called the partition of a, b and we denote it by P. We use the notation x i for the length of the ith subinterval x i1, x i . Thus
x i x i x i 1
This length of the longest subinterval is denoted by P and is called the norm of P. Thus
P max x1, x 2 , . . . , x n
Figure 5 illustrates one possible partition of a, b.
Î⁄
FIGURE 5
0
a=x¸
Τ
⁄
΋
¤
Î xi
‹ . . . xi-1
Îxn
xi
. . .
xn-1 xn=b
x
By drawing the lines x a, x x1, x x 2 , . . . , x b, we use the partition P to divide
the region S into strips S1, S2, . . . , S n as in Figure 6. Next we approximate these strips Si by
rectangles R i . To do this we choose a number x *i in each subinterval x i1, x i and construct
a rectangle R i with base x i and height fx *
i as in Figure 7.
10
❙❙❙❙
SECTION 2 AREA
y
y
Îxi
y=ƒ
S¡
0
a
S™
⁄
S£
¤
Si
‹
. . . xi-1
Sn
xi . . . xn-1
R¡
b
0
x
a
R™
⁄
x*¡
FIGURE 6
R£
¤
x *™
f(x*i )
Ri
‹
xi
xi-1
x£*
Rn
xn-1
x*i
b
x
x n*
FIGURE 7
Each point x *i can be anywhere in its subinterval—at the right endpoint (as in Example 1)
or at the left endpoint or somewhere between the endpoints. The area of the ith rectangle
R i is
A i f x*i x i
The n rectangles R1, . . . , R n form a polygonal approximation to the region S. What we
think of intuitively as the area of S is approximated by the sum of the areas of these rectangles, which is
n
A
1
n
i
i1
f x * x
i
i
f x *1 x1 f x *n x n
i1
Figure 8 shows this approximation for partitions with n 2, 4, 8, and 12.
0
a
x¡*
⁄
(a) n=2
FIGURE 8
x™*
b x
0
y
y
y
y
a
⁄
¤
‹
b
0
x
a
(b) n=4
b
0
x
(c) n=8
a
b
x
(d) n=12
Notice that this approximation appears to become better and better as the strips become
thinner and thinner, that is, as P l 0 . Therefore we define the area A of the region S as
the limiting value (if it exists) of the areas of the approximating polygons, that is, the limit
of the sum (1) of the areas of the approximating rectangles. In symbols:
n
A lim
2
f x * x
Pl 0 i1
i
i
The preceding discussion and the diagrams in Figures 7 and 8 show that the definition
of area in (2) corresponds to our intuitive feeling of what area ought to be.
The limit in (2) may or may not exist. It can be shown that if f is continuous, then this
limit does exist; that is, the region has an area. [The precise meaning of the limit in
Definition 2 is that for every 0 there is a corresponding number 0 such that
n
A
f x * x
i
i1
i
whenever
P In other words, the area can be approximated by a sum of areas of rectangles to within an
arbitrary degree of accuracy () by taking the norm of the partition sufficiently small.
SECTION 2 AREA
❙❙❙❙
11
EXAMPLE 2
(a) If the interval 0, 3 is divided into subintervals by the partition P and the set of
partition points is 0, 0.6, 1.2, 1.6, 2.5, 3, find P .
(b) If f x x 2 4x 5 and x *i is chosen to be the left endpoint of the ith subinterval, find the sum of the areas of the approximating rectangles.
(c) Sketch the approximating rectangles.
SOLUTION
(a) We are given that x0 0 , x1 6 , x 2 1.2 , x 3 1.6 , x 4 2 , x 5 2.5 , and
x 6 3 , so
0.6
0
0.6
0.6
0.4 0.4 0.5
1.2 1.6
2
0.5
2.5
3
x
x1 0.6 0 0.6
x 2 1.2 0.6 0.6
x 3 1.6 1.2 0.4
x 4 2 1.6 0.4
x 5 2.5 2 0.5
x 6 3 2.5 0.5
(See Figure 9.) Therefore
FIGURE 9
P max 0.6, 0.6, 0.4, 0.4, 0.5, 0.5 6
(b) Since x *i x i1, the sum of the areas of the approximating rectangles is, by (1),
y
6
f x* x
i
6
i
i1
y=≈-4x+5
f x
i1
xi
i1
f 0 x1 f 0.6 x 2 f 1.2 x 3 f 1.6 x 4 f 2 x 5
f 2.5 x 6
50.6 2.960.6 1.640.4 1.160.4 10.5 1.250.5
0
0.6 1.2 2
3
1.6 2.5
FIGURE 10
x
7.021
(c) The graph of f and the approximating rectangles are sketched in Figure 10.
EXAMPLE 3 Find the area under the parabola y x 2 1 from 0 to 2.
SOLUTION Since f x x 2 1 is continuous, the limit (2) that defines the area must exist
for all possible partitions P of the interval 0, 2 as long as P l 0 . To simplify things
let us take the partition P that divides 0, 2 into n subintervals of equal length. (This is
called a regular partition.) Then the partition points are
x0 0, x1 and
2
4
2i
2n
, x2 , . . . , xi , . . . , xn 2
n
n
n
n
x1 x2 xi xn 2
n
so the norm of P is
P maxx i 2
n
The point x *i can be chosen to be anywhere in the ith subinterval. For the sake of definiteness, let us choose it to be the right-hand endpoint:
x *i x i 2i
n
12
❙❙❙❙
SECTION 2 AREA
Since P 2n , the condition P l 0 is equivalent to n l . So the definition of
area (2) becomes
n
n
fx *x
A lim
i
P l 0 i 1
n
lim
n l i1
lim
nl
lim
nl
lim
nl
i
2i
n
lim
2
1
n l i 1
f
2i
n
n
2
lim
n l i1
n
2
n
8i 2
2
n3
n
8 n 2
2 n
i 1
3 n i1
n i1
(by Theorem 1.2)
8 nn 12n 1
2
n
3
n
6
n
(by Theorem 1.3)
4
1
1 1
3
n
1
n
2
2
43 1 1 2 2 143
The sum in this calculation is represented by the areas of the shaded rectangles in
Figure 11. Notice that in this case, with our choice of x *i as the right-hand endpoint and
y
y
n=10 R¡¸=5.08
0
1
y
n=50 R∞¸=4.7472
n=30 R£¸Å4.8015
2
x
0
1
x
2
0
1
2
x
FIGURE 11 Right sums
y
y
n=10 L¡¸=4.28
0
1
FIGURE 12 Left sums
y
n=50 L∞¸=4.5872
n=30 L£¸Å4.5348
2
x
0
1
2
x
0
1
2
x
SECTION 2 AREA
❙❙❙❙
13
since f is increasing, f x *i is the maximum value of f on x i1, x i, so the sum Rn of the
areas of the approximating rectangles is always greater than the exact area A 143.
We could just as well have chosen x *i to be the left-hand endpoint, that is,
x *i xi1 2i 1n. Then f x *i is the minimum value of f on xi1, xi , so the sum
L n of the areas of the approximating rectangles in Figure 12 is always less than A.
The calculation with this choice is as follows:
n
fx *x
A lim
i
P l 0 i 1
n
lim
f
n l i1
n
lim
n l i1
n
lim
n l i1
lim
nl
lim
nl
lim
nl
i
2i 1
n
2
n
2i 1
n
2
2
n
1
8 2
2
i 2i 1 n3
n
8 n 2
16 n
8 n
2 n
i 3 i 3 1
1
3
n i1
n i1
n i1
n i1
8 nn 12n 1
16 nn 1
8
2
3
3n n
3
n
6
n
2
n
n
4
3
1
n
1 1
2
1
n
8
1
1
n
n
8
2
n2
43 1 1 2 0 1 0 2 143
Notice that we have obtained the same answer with the different choice of x *i . In fact,
we would obtain the same answer if x *i was chosen to be the midpoint of xi1, xi
(see Exercise 11) or indeed any other point of this interval.
EXAMPLE 4 Find the area under the cosine curve from 0 to b, where 0 b 2.
SOLUTION As in the first part of Example 3, we choose a regular partition P so that
P x1 x2 xn b
n
and we choose x *i to be the right-hand endpoint of the ith subinterval:
x *i xi ib
n
Since P bn l 0 as n l , the area under the cosine curve from 0 to b is
n
n
3
A lim
P l 0 i 1
lim
nl
b
n
fx *i xi lim
n
cos
i1
i
cos
n l i 1
i
b
n
b
n
b
n
To evaluate this limit we use the formula of Exercise 52 in Section 1:
n
cos ix i1
sin 12 nx cos 12n 1x
sin 12 x
14
❙❙❙❙
SECTION 2 AREA
with x bn. Then Equation 3 becomes
A lim
4
nl
Now
cos
n 1b
2n
b
n
n 1b
2n
b
sin
2n
sin 12 b cos
1
n
cos 1 b
b
l cos
2
2
as n l since cosine is continuous. Letting t bn and using Theorem 3.5.2, we have
lim
nl
b
n
1
t
lim
b
sin
2n
tl0
sin
t
2
lim 2 t
2
tl0
sin
2
t
2
Putting these limits in Equation 4, we obtain
A 2 sin
b
b
cos sin b
2
2
In particular, taking b 2, we have proved that the area under the cosine curve from
0 to 2 is sin2 1 (see Figure 13).
y
y=cos x
1
area=1
0
x
π
2
FIGURE 13
NOTE The area calculations in Example 3 and 4 are not easy. We will see in Section
5.3, however, that the Fundamental Theorem of Calculus gives a much easier method for
computing these areas.
■
||||
2 Exercises
4. f x 2x 1, 0, 4, 0, 0.5, 1, 2, 4, x *i left endpoint
1–8
|||| You are given a function f , an interval, partition points, and
a description of x *i within the ith subinterval.
(a) Find P .
(b) Find the sum of the areas of the approximating rectangles, as
given in (1).
(c) Sketch the graph of f and the approximating rectangles.
5. f x x 3 2, 1, 2, 1, 0.5, 0, 0.5, 1.0, 1.5, 2,
x *i right endpoint
6. f x 1x 1, 0, 2, 0, 0.5, 1.0, 1.5, 2,
x *1 0.25, x *2 1, x *3 1.25, x *4 2
7. f x 2 sin x, 0, , 0, 4, 2, 34, ,
1. f x 16 x 2, 0, 4, 0, 1, 2, 3, 4, x *i left endpoint
x *1 6, x *2 3, x *3 23, x *4 56
2. f x 16 x 2, 0, 4, 0, 1, 2, 3, 4, x *i right endpoint
8. f x 4 cos x, 0, 2, 0, 6, 4, 3, 2,
x *i left endpoint
3. f x 16 x 2, 0, 4, 0, 1, 2, 3, 4, x *i midpoint
■
■
■
■
■
■
■
■
■
■
■
■
❙❙❙❙
SECTION 2 AREA
9. (a) Sketch a graph of the region that lies under the parabola
15
19–20
|||| If you have a programmable calculator (or a computer),
it is possible to evaluate the expression (1) for the sum of areas of
approximating rectangles, even for large values of n, using looping.
(On a TI use the Is command, on a Casio use Isz, on an HP or in
BASIC use a FOR-NEXT loop.) Compute the sum of the areas of
approximating rectangles using equal subintervals and right endpoints for n 10, 30, and 50. Then guess the value of the exact area.
y x 2 2x 2 from x 0 to x 3 and use it to make a
rough visual estimate of the area of the region.
(b) Find an expression for R n , the sum of the areas of the n
approximating rectangles, taking x*i in (1) to be the right
endpoint and using subintervals of equal length.
(c) Find the numerical values of the approximating areas R n
for n 6, 12, and 24.
(d) Find the exact area of the region.
19. The region under y sin x from 0 to 20. The region under y 1x 2 from 1 to 2
; 10. (a) Use a graphing device to sketch a graph of the region that
lies under the curve y 4x x 3 from x 0 to x 2 and
use it to make a rough visual estimate of the area of the
region.
(b) Find an expression for R n , the sum of the areas of the n
approximating rectangles, taking x*i in (1) to be the right
endpoint and using subintervals of equal length.
(c) Find the numerical values of the approximating areas R n
for n 10, 20, and 30.
(d) Find the exact area of the region.
11. Find the area from Example 3 taking x*i to be the midpoint
■
CAS
■
■
■
■
■
■
■
■
■
■
21. Some computer algebra systems have commands that will draw
approximating rectangles and evaluate the sums of their areas,
at least if x*i is a left or right endpoint. (For instance, in Maple
use leftbox, rightbox, leftsum, and rightsum.)
(a) If f x sx, 1 x 4, find the left and right sums for
n 10, 30, and 50.
(b) Illustrate by graphing the rectangles in part (a).
(c) Show that the exact area under f lies between 4.6 and 4.7.
CAS
of x i1, x i. Illustrate the approximating rectangles with a
■
22. (a) If f x sinsin x, 0 x 2, use the commands
discussed in Exercise 21 to find the left and right sums for
n 10, 30, and 50.
(b) Illustrate by graphing the rectangles in part (a).
(c) Show that the exact area under f lies between 0.87
and 0.91.
sketch.
12. Find the area under the curve y x 3 from 0 to 1 using sub-
intervals of equal length and taking x*i in (2) to be the (a) left
endpoint, (b) right endpoint, and (c) midpoint of the ith subinterval. In each case, sketch the approximating rectangles.
23–24
13 –18
Use (2) to find the area under the given curve from a
to b. Use equal subintervals and take x*i to be the right endpoint of
the ith subinterval. Sketch the region.
||||
13. y 2x 1 ,
a 0, b 5
14. y x 2 3x 2 ,
15. y 2x 4x 5 ,
16. y x 3 2x ,
18. y x 4 3x 2 ,
■
■
■
■
■
■
n l i1
■
■
■
■
3
n
1
■
■
3i
n
■
in a circle with radius r. By dividing the polygon into n
congruent triangles with central angle 2n, show that
A n 12 nr 2 sin2 n.
a 0, b 3
■
■
n
24. lim
■
26. (a) Let A n be the area of a polygon with n equal sides inscribed
a 0, b 1
■
n l i1
i
tan
4n
4n
[Hint: Use equal subintervals and right endpoints, and use
Exercise 53 in Section 1.]
a 0, b 2
2
25. Find the area under the curve y sin x from 0 to .
a 3, b 2
17. y x 2x x ,
3
n
23. lim
■
a 1, b 4
2
|||| Determine a region whose area is equal to the given limit.
Do not evaluate the limit.
■
■
■
■
■
■
(b) Show that lim n l A n r 2. [Hint: Use Equation 3.5.2.]
16
❙❙❙❙
SECTION 3 THE DEFINITE INTEGRAL
||||
SECTION 3 The Definite Integral
We saw in the preceeding section that a limit of the form
n
A lim
1
fx *x
i
P l 0 i 1
i
arises when we compute an area. It turns out that this same type of limit occurs in a wide
variety of situations even when f is not necessarily a positive function. In Chapters 5 and
8 we will see that limits of the form (1) also arise in finding lengths of curves, volumes of
solids, areas of surfaces, centers of mass, fluid pressure, and work, as well as other quantities. We therefore give this type of limit a special name and notation.
2 Definition of a Definite Integral If f is a function defined on a closed interval
a, b , let P be a partition of a, b with partition points x 0 , x1 , . . . , x n, where
a x0 x1 x2 xn b
Choose points x *i in x i1, x i and let x i x i x i1 and P maxx i . Then
the definite integral of f from a to b is
y
b
a
n
f x dx lim
fx *x
i
P l 0 i 1
i
if this limit exists. If the limit does exist, then f is called integrable on the
interval a, b.
NOTE 1 The symbol x was introduced by Leibniz and is called an integral sign. It is
an elongated S and was chosen because an integral is a limit of sums. In the notation
xab f x dx, f x is called the integrand and a and b are called the limits of integration;
a is the lower limit and b is the upper limit. The symbol dx has no meaning
by itself; xab f x dx is all one symbol. The procedure of calculating an integral is called
integration.
■
The definite integral xab f x dx is a number; it does not depend on x. In fact, we
could use any letter in place of x without changing the value of the integral:
NOTE 2
■
y
b
a
NOTE 3
■
b
b
f x dx y f t dt y f r dr
a
a
The sum
n
3
f x* x
i
i
i1
y
0 a
FIGURE 1
b
x
that occurs in Definition 2 is called a Riemann sum after the German mathematician
Bernhard Riemann (1826–1866). The definite integral is sometimes called the Riemann
integral. If f happens to be positive, then the Riemann sum can be interpreted as a sum of
areas of approximating rectangles [Compare (3) with (2.1).] If f takes on both positive and
negative values, as in Figure 1, then the Riemann sum is the sum of the areas of the rectangles that lie above the x-axis and the negatives of the areas of the rectangles that lie
below the x-axis (the areas of the gold rectangles minus the areas of the blue rectangles).
EXAMPLE 1 Let f x 1 5x and consider the partition P of the interval 2, 1 by
means of the set of partition points 2, 1.5, 1, 0.3, 0.2, 1. In this example
SECTION 3 THE DEFINITE INTEGRAL
❙❙❙❙
17
a 2, b 1, n 5, and x0 2, x1 1.5, x2 1, x3 0.3, x4 0.2, and
x5 1. the lengths of the subintervals are
x1 1.5 2 0.5
x 2 1 1.5 0.5
x 3 0.3 1 0.7
x 4 0.2 0.3 0.5
x 5 1 0.2 0.8
Thus the norm of the partition P is
P max0.5, 0.5, 0.7, 0.5, 0.8 0.8
Suppose we choose x*1 1.8, x*2 1.2, x*3 0.3, x *4 0, and x*5 0.7. Then
the corresponding Riemann sum is
5
f x* x
i
i
f 1.8x1 f 1.2x2 f 0.3x3 f 0x4 f 0.7x5
i1
80.5 50.5 0.50.7 10.5 4.50.8
2.75
Notice that, in this example, f is not a positive function and so the Riemann sum does
not represent a sum of areas of rectangles. But it does represent the sum of the areas of
the gold rectangles (above the x-axis) minus the sum of the areas of the blue rectangles
(below the axis) in Figure 2.
FIGURE 2
|
NOTE 4
An integral need not represent an area. But for positive functions, an integral
can be interpreted as an area. In fact, comparing Definition 2 with the definition of area
(2.2), we see the following:
■
For the special case where f x 0,
y
b
a
f x dx the area under the graph of f from a to b
In general, a definite integral can be interpreted as a difference of areas:
y
y
b
a
+
0 a
FIGURE 3
+
_
b
x
f x dx A1 A2
where A1 is the area of the region above the x-axis and below the graph of f and A2 is the
area of the region below the x-axis and above the graph of f . (This seems reasonable from
a comparison of Figures 1 and 3.)
NOTE 5
■
The precise meaning of the limit that defines the integral in Definition 2 is as
follows:
xab f x dx I means that for every 0 there is a corresponding number 0 such that
n
I
f x* x
i
i1
i
for all partitions P of a, b with P and for all possible choices of x *i in x i1, x i.
18
❙❙❙❙
SECTION 3 THE DEFINITE INTEGRAL
|||| Bernhard Riemann received his Ph.D. under
the direction of the legendary Gauss at the University of Göttingen and remained there to teach.
Gauss, who was not in the habit of praising
other mathematicians, spoke of Riemann’s “creative, active, truly mathematical mind and gloriously fertile originality.” The definition (2)
of an integral that we use is due to Riemann. He
also made major contributions to the theory of
functions of a complex variable, mathematical
physics, number theory, and the foundations of
geometry. Riemann’s broad concept of space and
geometry turned out to be the right setting, 50
years later, for Einstein’s general relativity theory.
Riemann’s health was poor throughout his life,
and he died of tuberculosis at the age of 39.
This means that a definite integral can be approximated to within any desired degree of
accuracy by a Riemann sum.
In Definition 2 we are dealing with a function f defined on an interval a, b,
so we are implicitly assuming that a b. But for some purposes it is useful to extend the
definition of xab f x dx to the case where a b or a b as follows:
NOTE 6
■
b
a
If a b, then y f x dx y f x dx.
a
b
a
If a b, then y f x dx 0.
a
EXAMPLE 2 Express
n
lim
x *
i
P l 0 i 1
3
x *i sin x *
i xi
as an integral on the interval 0, .
SOLUTION Comparing the given limit with the limit in Definition 2, we see that they will
be identical if we choose
f x x 3 x sin x
We are given that a 0 and b . Therefore, by Definition 2, we have
n
lim
x *
P l 0 i 1
i
3
3
x *i sin x *
i xi y x x sin x dx
0
EXAMPLE 3 Evaluate the following integrals by interpreting each in terms of areas.
(a)
1
0
s1 x 2 dx
(b)
3
0
x 1 dx
(a) Since f x s1 x 2 0, we can interpret this integral as the area under the curve
y s1 x 2 from 0 to 1. But, since y 2 1 x 2, we get x 2 y 2 1, which shows
that the graph of f is the quarter-circle with radius 1 in Figure 4. Therefore
y= œ„„„„„
1-≈
or
≈+¥=1
1
y s1 x
2
0
0
y
SOLUTION
y
1
y
1
x
FIGURE 4
dx 14 12 4
(In Section 8.3 we will be able to prove that the area of a circle of radius r is r 2 by
evaluating the integral x0r sr 2 x 2 dx using the techniques of Chapter 8.)
(b) The graph of y x 1 is the line with slope 1 shown in Figure 5. We compute the
integral as the difference of the areas of the two triangles:
y
3
0
x 1 dx A 1 A 2 12 2 2 12 1 1 1.5
y
(3, 2)
y=x-1
A¡
0 A™
FIGURE 5
_1
1
3
x
SECTION 3 THE DEFINITE INTEGRAL
❙❙❙❙
19
The integrals in Example 3 were simple to evaluate because we were able to express
them in terms of areas of simple regions, but not all integrals are that easy. In fact, the integrals of some functions don’t even exist. So the question arises: Which functions are integrable? A partial answer is given by the following theorem, which is proved in courses on
advanced calculus.
y
4 Theorem If f is either continuous or monotonic on a, b, then f is integrable
on a, b; that is, the definite integral xab f x dx exists.
0
a
b
x
FIGURE 6
Discontinuous integrable function
y
0
If f is discontinuous at some points in a, b, then xab f x dx might exist or it might not
exist (see Exercises 70 and 71). If f has only a finite number of discontinuities and these
are all jump discontinuities, then f is called piecewise continuous and it turns out that f
is integrable. (See Figure 6.)
It can be shown that if f is integrable on a, b, then f must be a bounded function
on a, b; that is, there exists a number M such that f x M for all x in a, b. Geometrically, this means that the graph of f lies between the horizontal lines y M and
y M. In particular, if f has an infinite discontinuity at some point in a, b, then f is not
bounded and is therefore not integrable. (See Exercise 70 and Figure 7.)
If f is integrable on a, b, then the Riemann sums (3) must approach xab f x dx
as P l 0 no matter how the partitions P are chosen and no matter how the points x *i
are chosen in x i1, x i. Therefore, if it is known beforehand that f is integrable on a, b
(for instance, if it is known that f is continuous or monotonic), then in calculating the value
of an integral we are free to choose partitions P and points x *i in any way we like as long
as P l 0 . For purposes of calculation, it is often convenient to take P to be a regular
partition; that is, all the subintervals have the same length x. Then
x
FIGURE 7
x x1 x2 xn Nonintegrable function
ba
n
x0 a, x1 a x, x2 a 2x, . . . , x i a i x
and
If we choose x *i to be the right endpoint of the ith subinterval, then
x *i x i a i x a i
ba
n
Since P x b an , we have P l 0 as n l , so Definition 2 gives
y
b
a
n
f x dx lim
fx * x
i
P l 0 i 1
lim
n
n l i 1
f ai
ba
n
ba
n
Since b an does not depend on i, Theorem 1.2 allows us to take it in front of the
sigma sign, and we have the following formula for calculating integrals.
5
Theorem If f is integrable on a, b, then
y
b
a
f x dx lim
nl
ba
n
n
f
i1
ai
ba
n
20
❙❙❙❙
SECTION 3 THE DEFINITE INTEGRAL
EXAMPLE 4 Evaluate
y
3
0
x 3 5x dx.
SOLUTION Here we have f x x 3 5x, a 0, and b 3. Since f is continuous, we
know it is integrable and so Theorem 5 gives
y
3
0
x 3 5x dx lim
nl
lim
nl
lim
nl
lim
nl
3
n
n
f
i1
3i
n
lim
i3 45
n2
3
n
nl
n
i1
3i
n
3
5
n
81
n4
i1
81
n4
nn 1
2
81
4
1
1
n
3i
n
n
i
i1
2
2
45 nn 1
n2
2
45
1
1
2
n
814 452 94 2.25
This integral cannot be interpreted as an area because f takes on both positive and negative values. But it can be interpreted as the difference of areas A 1 A 2 , where A 1 and A 2
are shown in Figure 8.
Figure 9 illustrates the calculation by showing the positive and negative terms in the
right Riemann sum R n for n 40. The values in the table show the Riemann sums
approaching the exact value of the integral, 2.25, as n l .
FIGURE 8
n
Rn
40
100
500
1000
5000
1.7873
2.0680
2.2139
2.2320
2.2464
FIGURE 9
A much simpler method for evaluating the integral in Example 4 will be given in
Section 5.4 after we have proved the Fundamental Theorem of Calculus.
The Midpoint Rule
We often choose the sample point x *i to be the right endpoint of the i th subinterval because
it is convenient for computing the limit. But if the purpose is to find an approximation to
an integral, it is usually better to choose x *i to be the midpoint of the interval, which we
denote by x i . Any Riemann sum is an approximation to an integral, but if we use midpoints
and a regular partition we get the following approximation:
Midpoint Rule
y
b
a
where
and
n
f x dx f x x x f x f x i
1
i1
x ba
n
x i 12 x i1 x i midpoint of x i1, x i n
SECTION 3 THE DEFINITE INTEGRAL
EXAMPLE 5 Use the Midpoint Rule with n 5 to approximate
y
2
1
❙❙❙❙
21
1
dx.
x
SOLUTION The partition points are 1, 1.2, 1.4, 1.6, 1.8, and 2.0, so the midpoints of the five
1
intervals are 1.1, 1.3, 1.5, 1.7, and 1.9. The width of the intervals is x 2 15 5 ,
so the Midpoint Rule gives
y
1
y= x
y
2
1
1
dx x f 1.1 f 1.3 f 1.5 f 1.7 f 1.9
x
1
5
1
1
1
1
1
1.1
1.3
1.5
1.7
1.9
0.691908
0
1
2
x
FIGURE 10
Since f x 1x 0 for 1 x 2, the integral represents an area and the approximation given by the Midpoint Rule is the sum of the areas of the rectangles shown in
Figure 10.
At the moment we don’t know how accurate the approximation in Example 5 is, but in
Section 8.7 we will learn a method for estimating the error involved in using the Midpoint
Rule. At that time we will discuss other methods for approximating definite integrals.
If we apply the Midpoint Rule to the integral in Example 4, we get the picture in Figure 11. The approximation M40 2.2563 is much closer to the true value 2.25 than the
right endpoint approximation, R 40 1.7873 shown in Figure 9.
Properties of the Definite Integral
We now develop some basic properties of integrals that will help us to evaluate integrals
in a simple manner.
FIGURE 11
M40 2.2563
Properties of the Integral Suppose that all of the following integrals exist. Then
1.
y
b
2.
y
b
3.
y
b
4.
y
b
5.
y
b
a
a
a
a
a
c dx cb a, where c is any constant
b
b
f x tx dx y f x dx y tx dx
a
a
b
cf x dx c y f x dx, where c is any constant
a
b
b
f x tx dx y f x dx y tx dx
a
b
a
b
f x dx y f x dx y f x dx
a
a
The proof of Property 1 is requested in Exercise 15. This property says that the integral
of a constant function f x c is the constant times the length of the interval. If c 0 and
a b, this is to be expected because cb a is the area of the shaded rectangle in
Figure 12.
y
y=c
c
area=c(b-a)
FIG URE 12
j
b
a
c dx=c(b-a)
0
a
b
x
22
❙❙❙❙
SECTION 3 THE DEFINITE INTEGRAL
Proof of Property 2 Since
xab f x tx dx exists, we can compute it using a regular parti-
tion and choosing x *i to be the right endpoint of the ith subinterval, that is, x *i x i. Using
the fact that the limit of a sum is the sum of the limits, we have
y
b
a
n
f x tx x
f x tx dx lim
i
n l i1
i
n
lim
nl
n
f x i x i1
tx x
i
i1
n
(by Theorem 1.2)
n
f x x lim tx x
lim
i
n l i1
i
n l i1
b
b
y f x dx y tx dx
a
y
y=ƒ
0
a
FIG URE 13
c
b
x
a
Property 2 says that the integral of a sum is the sum of the integrals. Property 3 can be
proved in a similar manner (see Exercise 65) and says that the integral of a constant times
a function is the constant times the integral of the function. In other words, a constant (but
only a constant) can be taken in front of an integral sign. Property 4 is proved by writing
f t f t and using Properties 2 and 3 with c 1.
Property 5 is somewhat more complicated and is proved at the end of this section, but
for the case where f x 0 and a c b, it can be seen from the geometric interpretation in Figure 13. For positive functions f , xab f x dx is the total area under y f x from
a to b, which is the sum of xac f x dx (the area from a to c) and xcb f x dx (the area from c
to b).
EXAMPLE 6 Use the properties of integrals and the results
y
b
a
x dx b2 a2
2
y
2
0
cos x dx 1
(from Exercise 21 in this section and Example 4 in Section 2) to evaluate the following
integrals.
(a)
y
2
0
x 3 cos x dx
(b)
y x dx
5
4
SOLUTION
(a) Usiing Properties 2 and 3 of integrals, we get
y
2
0
x 3 cos x dx y
2
0
(b) Since
x
x dx 3 y
2
0
cos x dx 2
3
8
x if x 0
x if x 0
we use Property 5 to split the integral at 0:
y x dx y x dx y x dx
5
0
4
4
5
0
0
5
0
5
y x dx y x dx y x dx y x dx
4
4
0
0
0 2 42 5 2 0 2 20.5
1
2
1
2
Notice that Properties 1–5 are true whether a b, a b, or a b. The following
properties, however, are true only if a b.
SECTION 3 THE DEFINITE INTEGRAL
❙❙❙❙
23
Order Properties of the Integral Suppose the following integrals exist and a b.
6. If f x 0 for a x b, then
y
b
f x dx 0.
a
7. If f x tx for a x b, then
y
b
a
b
f x dx y tx dx.
a
8. If m f x M for a x b, then
b
mb a y f x dx Mb a
a
9.
y
M
y=ƒ
m
0
a
b x
FIGURE 14
y
b
a
f x dx y
f x dx
b
a
If f x 0, then xab f x dx represents the area under the graph of f , so the geometric
interpretation of Property 6 is simply that areas are positive. But the property can be
proved from the definition of an integral (Exercise 66). Property 7 says that a bigger function has a bigger integral. It follows from Properties 6 and 4 because f t 0.
Property 8 is illustrated by Figure 14 for the case where f x 0. If f is continuous we
could take m and M to be the absolute minimum and maximum values of f on the interval a, b. In this case Property 8 says that the area under the graph of f is greater than the
area of the rectangle with height m and less than the area of the rectangle with height M .
Proof of Property 8 Since m f x M , Property 7 gives
y
b
a
b
b
m dx y f x dx y M dx
a
a
Using Property 1 to evaluate the integrals on the left- and right-hand sides, we obtain
b
mb a y f x dx Mb a
a
The proof of Property 9 is left as Exercise 67.
EXAMPLE 7 Use Property 8 to estimate the value of
y
4
1
sx dx.
SOLUTION Since f x sx is an increasing function, its absolute minimum on 1, 4 is
m f 1 1 and its absolute maximum on 1, 4 is M f 4 s4 2. Thus Property 8 gives
y
y=œ„x
2
4
14 1 y sx dx 24 1
1
1
4
3 y sx dx 6
or
0
FIGURE 15
1
4
x
1
The result of Example 7 is illustrated in Figure 15. The area under y sx from 1 to 4
is greater than the area of the lower rectangle and less than the area of the large rectangle.
EXAMPLE 8 Show that
y
4
1
s1 x 2 dx 7.5.
SOLUTION The minimum value of f x s1 x 2 on 1, 4 is m f 1 s2, since f is
increasing. Thus Property 8 gives
y
4
1
s1 x 2 dx s24 1 3s2 4.24
24
❙❙❙❙
SECTION 3 THE DEFINITE INTEGRAL
This result is not good enough, so instead we use Property 7. Notice that
1 x 2 x 2 ? s1 x 2 sx 2 x
Since x x for x 0, we have s1 x 2 x for 1 x 4. Thus, by Property 7,
y
4
1
4
1
s1 x 2 dx y x dx 2 4 2 12 7.5
1
[Here we have used the fact that xab x dx b 2 a 2 2 from Exercise 21.]
Proof of Property 5 We first assume that a c b. Since we are assuming that
xab f x dx
exists, we can compute it as a limit of Riemann sums using only partitions P that include
c as one of the partition points. If P is such a partition, let P1 be the corresponding partition of a, c determined by those partition points of P that lie in a, c. Similarly, P2 will
denote the corresponding partition of c, b. Note that P1 P and P2 P . Thus,
if P l 0, it follows that P1 l 0 and P2 l 0. If xi 1 i n is the set of partition
points for P and n k m, where k is the number of subintervals in a, c and m is the
number of subintervals in c, b, then xi 1 i k is the set of partition points for P1. If
we write tj xkj for the partition points to the right of c, then ti 1 j m is the set of
partition points for P2. Thus we have
a x0 x1 xk xk1 xn b
c t1 tm b
Choosing x *i xi and letting tj tj tj1, we compute xab f x dx as follows:
y
b
a
n
f x lim
f x x
i
P l 0 i1
i
k
lim
P l 0
n
i1
P l 0
m
f xixi i1
f t t
j
f x x
P1 l 0 i1
i
i
c
j
j1
k
lim
f xixi
ik1
k
lim
f xixi m
f t t
lim
P2 l 0 j1
j
j
b
y f x dx y f t dt
a
c
Now suppose that c a b. By what we have already proved, we have
y
b
b
f x y f x dx y f x dx
a
Therefore
y
a
a
b
f x y f x dx y f x dx
c
a
a
b
c
c
a
b
y f x dx y f x dx
a
c
(See Note 6.) The proofs are similar for the remaining four orderings of a, b, and c.
❙❙❙❙
SECTION 3 THE DEFINITE INTEGRAL
||||
3 Exercises
Riemann sums for the function f x s1 x 2 on the interval
1, 2 with n 100. Explain why these estimates show that
1–6
|||| You are given a function f , an interval, partition points
that define a partition P, and points x *i in the ith subinterval.
(a) Find P . (b) Find the Riemann sum (3).
2
1.805 y s1 x 2 dx 1.815
1. f x 7 2x, 1, 5, 1, 1.6, 2.2, 3.0, 4.2, 5, x *i midpoint
1
2. f x 3x 1, 2, 2, 2, 1.2, 0.6, 0, 0.8, 1.6, 2,
Deduce that the approximation using the Midpoint Rule with
n 10 in Exercise 11 is accurate to two decimal places.
x *i midpoint
3. f x 2 x 2, 2, 2, 2, 1.4, 1, 0, 0.8, 1.4, 2,
15–20
4. f x x x 2, 2, 0, 2, 1.5, 1, 0.7, 0.4, 0,
15.
y
b
5. f x x 3, 1, 1, 1, 0.5, 0, 0.5, 1,
17.
y
4
6. f x sin x, 2, , 2, 1, 0, 1, 2, ,
19.
y
b
x *i right endpoint
x *i left endpoint
x *1 1, x *2 0.4, x *3 0.2, x *4 1
x *1 1.5, x *2 0.5, x *3 0.5, x *4 1.5, x *5 3
■
■
■
■
■
■
■
■
■
■
Use Theorem 5 to evaluate the integral.
||||
a
1
0
■
■
c dx
16.
y
7
x 2 2 dx
18.
y
5
x 3 4x dx
20.
y
1
■
■
■
■
■
a
b
22. Prove that y x 2 dx y
a
f
23–28
1
8. The table gives the values of a function obtained from an
experiment. Use them to estimate x06 f x dx using three equal
subintervals with (a) right endpoints, (b) left endpoints, and
(c) midpoints. If the function is known to be a decreasing function, can you say whether your estimates are less than or
greater than the exact value of the integral?
0
f x
9–12
1
9.3
2
9.0
8.3
3
4
6.5
5
10.5
11.
■
5
x 3 dx,
y
2
n5
s1 x 2 dx,
1
■
■
■
10.
n 10
■
12.
■
■
y
3
1
y
1
dx,
2x 7
tan x dx,
0
■
■
26.
y
3
y (1 x ) dx
28.
y 3x 5 dx
27.
1 2x dx
0
2
3
2
2
■
■
29–32
■
■
■
n4
2x *
30. lim
sx * x ,
31. lim
cos x x ,
32. lim
i
2 x dx
3
0
■
■
■
■
■
■
0, 1
5x *i xi,
i
i
i
Pl 0 i1
■
0, tan xi
xi,
xi
■
||||
i
1, 4
■
2, 4
■
■
■
■
■
■
■
■
■
■
■
■
Express the limit as a definite integral.
■
13. If you have a CAS that evaluates midpoint approximations
and graphs the corresponding rectangles (use middlesum and
middlebox commands in Maple), check the answer to Exercise 11 and illustrate with a graph. Then repeat with n 20
and n 30.
tions for Exercise 19 in Section 2), compute the left and right
2
n
33–35
14. With a programmable calculator or computer (see the instruc-
■
1
s4 x 2 dx
n
29. lim
Pl 0 i1
■
■
2
Express the limit as a definite integral on the given inter-
||||
n
CAS
■
val.
n
n4
■
■
b3 a3
.
3
y (1 s9 x ) dx
■
4
■
2
25.
3
1
Pl 0 i1
Use the Midpoint Rule with the given value of n to
approximate the integral. Round the answer to four decimal places.
0
■
n
||||
y
■
y
y
6
7.6
2.3
x 3 5x 4 dx
Evaluate the integral by interpreting it in terms of areas.
||||
Pl 0 i1
x
0
2 3x x 2 dx
24.
23.
x
1
1
6 2x dx
b2 a2
.
2
b
21. Prove that y x dx four equal subintervals with (a) right endpoints, (b) left
endpoints, and (c) midpoints.
0
2
■
7. The graph of a function f is given. Estimate x08 f x dx using
9.
25
33. lim
i4
n5
34. lim
1
n
n l i1
nl
n
35. lim
n l i1
■
■
n
i1
[Hint: Consider f x x 4.]
1
1 in2
3 1
■
■
5
2i
n
6
■
■
2
n
■
■
26
❙❙❙❙
SECTION 3 THE DEFINITE INTEGRAL
1
36. Evaluate y x 2 cos x dx.
54.
1
9
37. Given that y sx dx 38
3
4
4
, what is y st dt ?
■
a Riemann sum with right endpoints and n 8.
(b) Draw a diagram like Figure 1 to illustrate the approximation
in part (a).
(c) Evaluate x04 x 2 3x dx.
(d) Interpret the integral in part (c) as a difference of areas and
illustrate with a diagram like Figure 3.
39–44 |||| Use the properties of integrals to evaluate each integral.
You may assume from Section 2 that
b
0
41.
43.
44.
y
1
y
4
y
1
1
y
4
f x dx
42.
y
6
y
1
3
0
45– 48
y
46.
y
8
■
■
■
■
6
■
■
■
■
■
47.
y
10
48.
y
5
■
6
f x dx y
0
6
5
■
■
sin 3 x dx y
4
y
50.
y
2
51.
y
6
52.
2
y sin x dx 6
6
3
0
■
sin 2 x dx
2
s5 x dx y sx 1 dx
1
1
6
1
dx y
dx
4
x
8x
53. 2 y
1
1
0
59.
y
1
56.
y
x 2 2x dx
58.
y
s1 x 4 dx
60.
y
3
1
■
■
■
■
■
2
0
■
sx 3 1 dx
3
■
4
34
4
cos x dx
sin2x dx
■
■
■
■
61.
y
3
■
■
s1 x 2 dx 2 s2
■
62.
y
5
63.
y
64.
y
sx 4 1 dx 263
1
sx 2 1 dx 10.5
2
2
0
0
■
x sin x dx 28
x 2 cos x dx 33
■
■
■
■
■
■
■
■
■
65. Prove Property 3 of integrals.
[Hint: Use the Extreme
(b) f x sec x
f x dx y f x dx
3
49.
4
■
interval 0, 2?
(a) f x x 2 sin x
■
■
■
49–54 |||| Use the properties of integrals to verify the inequality
without evaluating the integrals.
1
■
69. Which of the following functions are integrable on the
2
4
y
1
dx
x
in a, b. Prove that xab f x dx 0 .
Value Theorem and Property 8.]
7
0
57.
1
f x dx y f x dx
■
■
68. Suppose that f is continuous on a, b and f x 0 for all x
0
■
■
[Hint: f x f x f x .]
f x dx y f x dx
3
■
Use Property 8 to estimate the value of the integral.
67. Prove Property 9 of integrals.
5
2
■
66. Prove Property 6 of integrals.
12
3
5
■
4
f x dx y f x dx y f x dx
1
2
■
||||
3
y
5 cos x 4x dx
Write the given sum or difference as a single integral in
the form xab f x dx .
45.
||||
55.
1
1
■
55–60
4 7x dx
2x if 1 x 0
3x 2 if 0 x 1
where f x 3
■
■
f x dx
|||| Use properties of integrals, together with Exercises 21
and 22, to prove the inequality.
f x dx y f x dx y f x dx
3
■
40.
2x 2 3x 1 dx
1
■
2
0
61–64
cos x dx sin b
s3 dx
4
f x sin 2x dx y
■
■
and you may use the results of Exercises 21 and 22.
39.
2
0
9
38. (a) Find an approximation to the integral x04 x 2 3x dx using
y
y
■
(c) f x (d) f x 70. Let
x 1 if 0 x 1
2 x if 1 x 2
x 12 if x 1
1
if x 1
1
f t x
0
if 0 x 1
if x 0
(a) Show that f is not continuous on 0, 1.
(b) Show that f is unbounded on 0, 1.
(c) Show that x01 f x dx does not exist, that is, f is not integrable on 0, 1. [Hint: Show that the first term in the
Riemann sum, fx *1 x 1, can be made arbitrarily large.]
■
SECTION 3 THE DEFINITE INTEGRAL
71. Let
0 if x is rational
f x 1 if x is irrational
Show that f is bounded but not integrable on a, b.
[Hint: Show that, no matter how small P is, some Riemann
sums are 0 whereas others are equal to b a.]
72. Evaluate x12 x 3 dx using a partition of 1, 2 by points of a
geometric progression: x0 1, x1 21n, x2 2 2n, . . . ,
xi 2 in, . . . , xn 2 nn 2. Take x*i xi and use the
formula in Exercise 47 in Section 1 for the sum of a
geometric series.
73. Find x12 x 2 dx. Hint: Use a regular partition but choose x*i to
be the geometric mean of x i1 and x i (x*i sx i1 x i ) and use
the identity
1
1
1
mm 1
m
m1
❙❙❙❙
A27
2
; 74. (a) Draw the graph of the function f x cosx in the view-
ing rectangle 0, 2 by 1, 1.
(b) If we define a new function t by tx x0x cos t 2 dt, then
tx is the area under the graph of f from 0 to x [until f x
becomes negative, at which point tx becomes a difference
of areas.] Use the graph of f from part (a) to estimate the
value of tx when x 0, 0.2, 0.4, 0.6, . . . up to x 2.
At what value of x does tx start to decrease?
(c) Use the information from part (a) to sketch a rough graph
of t.
(d) Sketch a more accurate graph of t by using your calculator or computer to estimate t0.2, t0.4, . . . . (Use the
integration command, if available, or the Midpoint
Rule.)
(e) Use your graph of t from part (d) to sketch the graph of t
using the interpretation of tx as the slope of a tangent
line. How does the graph of t compare with the graph
of f ?