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Unformatted text preview: s, t, u
3r3s3t3u r1 s1 t1 u1
n
n
n
n n 1
4
3r3s3t3u
r1 s1 t1 u1
n
n 1
3r 6
r1 s1 2 n
r1 1
3r 4
r1
n 6
r1 13
2
4
6 n 1
2 19
8
19
8 11 81
160 n 1
3 2r n n 1
3s 1
3t t1 n 1
3 2s s1
n u1 1
3u n s1
n 1
3 2r 1
3t n 1
3t n 13
2 1
8
n 1
8
39
16 t1 1
3s 4 n 1
3s 2 1
3r 1
3r 3
r1 s1 n n r1 s1 t1
n n 2 1
3s n 1
2 r1 13
2 13 n
2
19 n
8
13 n
2 1
3 4r 3 1
2
1
8
39
160 n 1
2 13
2
3 n 13
2 n 1
2 1
2 1 81
80 n 1
80 and we conclude that Þ 0 x 4 d x
1 nÝ
lim Þ 0 s n d
1 39 .
160 Some Exercises on the RiemannStieltjes Integral 302 1. Prove that the integral Þ 1 3x 2 dx
4 exists and has the value 63. Solution: This exercise will become obsolete when we reach the fundamental theorem of calculus
later on in the chapter. The solution given here is a bare hands approach and repeats portions of the proof
that monotone functions are integrable.
We define f x 3x 2 for each number x 1, 4 . Given any positive integer n, if
P n x 0 , x 1 , , x n
is the regular npartition of the interval 1, 4 then we define two step functions s n and S n by making
sn x Sn x f x
whenever x is a point of the partition P n and, in each interval x j 1 , x j of the partition P n we make s n and
S n take the constant values f x j 1 and f x j respectively. Since
n Þ1 Sn
4 sn f xj f xj 4 1 j1 n 1 n 3
n 3f 4
as n f xj f xj 1 j1 f1
n 135
n 0 Ý we know that f is integrable and that
lim
Þ 1 S n n Ý Þ 1 s n Þ 1 f.
4 lim nÝ 4 4 Now since
3j
xj 1 n
for each j 0, , n, we have Þ1 Sn
4
as n n
j1 3
n 3 3j
1 n 9 14n 2 15n 3
2n 2 2 63 Ý. 0, 1 . Given a positive integer n, we shall take P n to be the
2. In this exercise we take f x x for x
partition of 0, 1 defined by the equation
2
2
2
2
P n 0 2 , 1 2 , 2 2 , , n 2 .
nnn
n
Prove that if we define a step function S n on 0, 1 by making
Sn x x
whenever x is a point of the partition P n and giving S n the constant value j/n in each interval
j 1 2 j2
,2
n2
n
of the partition P n , then
1
1
lim
Þ 0 x dx n Ý Þ 0 S n 2 .
3 Solution: For each positive integer n we define the function S n as described in the exercise and we
303 define a step function s n on 0, 1 by making
Sn x x
whenever x is a point of the partition P n and giving s n the constant value j
j 1 2 j2
,2
n2
n
We see at once that s n f S n for each n and that Þ0 Sn
1 lim
nÝ
as n n sn
j1 j j
n j2
n2 1
n j 1 1 /n in each interval 2 1
n n2 0 Ý. We conclude that the pair of sequences s n and S n squeezes f on the interval 0, 1 and that Þ 0 S n Þ 0 f.
1 lim nÝ 1 Therefore
lim
lim
Þ0 f n Ý Þ0 Sn n Ý
1 1 2
n Ý 4n 32n
lim
6n n
j1 j2
n2 j
n j 1 2 n2 1 2.
3 3. Prove that Þ0 1 x dx 3 .
4
This exercise is very similar to Exercise 2. This time we take
3
3
3
3
P n 0 3 , 1 3 , 3 3 , , n 3
nnn
n
for each n and, for each n, we define S n to be the step function that takes the value 3 x at each
point of P n and whose constant value in each interval
j 1 3 j3
,3
n3
n
is j/n. We o...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.
 Fall '08
 STAFF
 Math, Calculus

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