58 is continuous ddxf isdiscontinuous at the the point

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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 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) ==...
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This document was uploaded on 02/02/2014.

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