Ch 5 solutions

# 58 is continuous ddxf isdiscontinuous at the the point

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 1￿1 ￿ −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 += 2x￿5 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 = 25/2￿3 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) ==...
View Full Document

## This document was uploaded on 02/02/2014.

Ask a homework question - tutors are online