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Unformatted text preview: nn sum number 28 28lim numberf252252xk = black ∆x
max ∆x →0
max
k 252 k →0 k=1 C ∆x k =
k=1 Chapter 5 b =
lim
C (b − a) = C (b − a). By Deﬁnition 5.5.1,
f (x) dx = C (b − a).
Chapter
max ∆xk →0
252 252
Chapter 5 5
a
44. Choose any large positive integer N and any partition of [0, a]. Then choose x∗ in the ﬁrst interval
1
so small that f (x∗ ) of 1 > N . For [a, b] contains x∗ rational . Then with this partition and
∆x
1
1
43. Each subinterval ∆xa partition ofexample chooseboth<of [01 /Nand irrationalxnumbers. If all x∗ are chosen to be
∗
44.44. Choose anynlarge positive integer andand any partition[0, a]., a]. Then choose in1the the ﬁrst interval
Choose any large positive integer Nn N any partition of
Then choose x∗ n in ﬁrst intervalk
n
n
1
∗ that f (x∗ )∆x > N .∗ For example choose x∗ < ∆x /N . Then with this∗partition and
∗ x∗ )∆xk > f (x )∆x1 > N . This shows that the sum is dependent on partition
∗
so of x
choice small , f f1 f ( 1 )∆x . For (1)∆xk =
1
so rational 1then(xn()∆xx∗>1 Nk = 1 example choose ∆xk<1 ∆x−/N . Then with thisfpartition = b − a. If all x∗ are
small that
x1 = b 1 a1so
k
lim
(xk )∆xk and
k
k
max ∆xk →0
kn k=1
=1
k=1
k=1
∗ (x∗ ) 5.5.1 ∗ ( not x1 > N k This shows that the sum is dependent on partition
∗ x∗ , Deﬁnitionxk > fisx∗ )∆satisﬁed.=1
choice of
.
and/or points,
1
choice of x1 , 1 sof (xf )∆xk >n (x1 )∆x1 > N . This shows that the sum is dependent on partition
f
kk
irrational then =1 lim
f (x∗ )∆xk = 0. Thus f is not integrable on [a, b] because the preceding limits are not
k=1k
k
so ∆x −1 1]
45. and/or points, so max Deﬁnition so is notnot satisﬁed. by Theorem 5.5.2
(a)and/orcontinuous on k[→0 k=1 5.5.1is integrable there
f is points, Deﬁnition ,5.5.1 f is satisﬁed.
equal. ≤ 1 so f is bounded on
(b) f (x)is continuous on [−1, 1] [−1,is integrable there point of discontinuity, so by Part (a) of
1], and f has one by Theorem 5.5.2
45.45. (a) isfcontinuous f is [integrablesoisfintegrable there by Theorem 5.5.2
(a) f
Theorem 5.5.8 on −1, 1] so f on [−1, 1]
45. (a)f (xf(x)1≤ 1 f isfbounded on [on1[,−1, andand f has by Theorem discontinuity, by Part (a) (a) of
is continuous on [−1,
so is bounded f integrable one point of of 5.5.2.
so
(b)(b) f  ) ≤ bounded on [1,1]1] so − is 1], 1], (x) =there, onefpointdiscontinuity,on [0,1]by Part of
(c)  is not so
f
because lim f f has ∞ so is not integrable so
+
Theorem 5.5.8isfintegrable on [−1[,−1, 1]
is integrable on →1]
x0
Theorem 5.5.8 f
(b) f is bounded is on [1,1] on [−1 1], and= has, one so 1 isofintegrable on [0,1]by
≤ bounded
f = so point
discontinuity, so
(c) (c) (isf(x)not 1 so f on boundedpoint x ,=limxf (x)+∞lim ,sinisfnotnotnot exist. fon [0,1]part (a) of Theorem 5.5.8
(d) f x)not discontinuous[1,1] because limxf (0because +∞ f
is
at the because → )
0
does integrable is continuous
x→0
x→0
f is integrable on [−1, 1].
x
1 1 By
elsewhere. −1 ≤ f (x) ≤ 1 for x in [−1, 1] so f is bounded there. doesPart (a), Theorem 5.5.8,
is continuous
(d)(d)(xf (is)discontinuous at the the point= 0= 0 because xsin sindoes notnot exist. isf continuous
f ) x is discontinuous at point x x because lim lim
exist. f
→0 x x
is is not bounded−1 1].
integrable on [
(c)f f elsewhere. −1 ≤ on, x) ≤ 1because lim , 1]xso= +x→0so f is there. By Part (a),[0,1].
[1,1] for x in [−1 f ( ) f is∞,
not integrable on Theorem 5.5.8,
bounded
x 1]
elsewhere. −1 ≤ f (xf ( 1 for x in [−1,→0 so f is bounded there. By Part (a), Theorem 5.5.8,
)≤
is integrable on 1].
f isfintegrable on [−1[,−1, 1].
1
(d)
(x) is discontinuous at the point x = 0 because lim sin does not exist. f is continuous elsewhere.
EXERCISEfSET 5.6
x→0
x
−1 2f (x) ≤ 1 for x in [−1, 1] f is bounded there. By part (a), Theorem 5.5.8, f is integrable on [−1, 1].
≤
so
2
EXERCISE SET
EXERCISE (2 − x)5.65.6 x − x2 /2) = 4 − 4/2 = 2
SET dx = (2
1. (a)
02 2
0 2
Exercise(2Set)dx)=1 (2x(2xx−/x2 /2 = 4= 44−24=22= 2
(2 x
dx
1. (a) 1 − x− 5.6 = − 2 2) 2)
−/ /
1. (a)
0
0
0
(b) 0 2 dx = 2x
2
= 2(1) − 2(−1) = 4
2
1 11
−1 1
−
1
1. (a) 3 (2 2dx )= 2x (2x= 2(1)2) 2(= 1) − 4/2 = 2.
− x dx =
− x2 / − − 4 = 4
3 −0 = 4
(b)(b) 0 2dx = 2x
= 2(1) − 2( 1)
2
( −1
= 9/2 + 3 − (1/2 + 1) = 6
(c)
1
−1 x + 1)dx = (x−/2 + x)
−1
1
13 3
3 3
1
1
( 1) 1)dx = (x2 / x) x
= /2 −
(1/ 1) 1)
(c) (c) (x +x += 2x5 2 /=+2 +−)2(−1)9= 3+ 3 − 2 +2 += 6= 6
9
9
(b) 5 1 2dxdx = (x 2 2(1) 1 =19/2 + 4. (1/
1
−1
2
−= 25/2
1
2. (a)
xdx = x /2
(b)
5dx = 5x = 5(9) − 5(3) = 30
5 5
9 9
05 5
39 9
0
3
2 3 xdx = x2 /2 = 25/23
2
2
2. (a) xdx = x /2 =225/2
(b) 5dx5= 5= 5x 5(9) − 5(3) = 30 30
dx x = = 5(9) − 5(3) =
2. (a)
(b)
(c) 0 (0(x + 1)dx = xx/22 + x) ==...
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This document was uploaded on 02/02/2014.
 Spring '14

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