Answer Key
University
California State UniversityCourse
MATH 150B | Calculus IIPages
1
Academic year
2023
anon
Views
36
11 can pm x dr lim i: 1/1/1 du . x dx 2 =1/2 [Dim In/x'+1); lim J b-o.oo : 12/day ln(b' +1) + lim In(citi)] 1/2 In 1x's'll 1/2 [ 00 . 0 ] + Diverges (b) paid ( (x+2) dx = lim braw I 1/2 It 2-x dx Sxoz X dx x(2+2) x(x)2) X(x42) = him b-as 1/ 2 12 S - 1 2/12 dx 1[lnx - ln(x12)] 1/ 2 [ lam base 0+2 - Inc/1) ] + +2 1 2 ln(f) (c) S 1/1/1 1 dx lim pot dx S 11/1/1 dx sin' 1-x2 bag /1-x2 (x)+c = line sint(s) sint(b) 17/2 bon (d) so (x-3) 2 dx = 2 lim be3 pb (x-3)2 17-13 dx + him a-3 P' (x-3)2 dx] S 1/2 du =2 [ lan bo3 x-3 1/3/10 lim b-03+ u -1 = + -2 [ lim ba3 b-3 - I 1/1 + lim ba3+ I/ + a:12 G-3 x-3 I u = x-3 2 [ 00-1-1-00 Diverges du . dx
13.cas lim n nso j 4n'+1 2 /n°(4+1) (b) lim now (.9) n =0 Geometric Sequence 181.9
(b) as SECTION k+l-k = k (k+1) k(k.1) k(k+1) ker KIKID 2 2 (c) k=o 3 1-13 (d) = )=2 1/3 1/1/10 2/3 (e) k=o E 8 E(2) = (33) 1-3/8 (f) 5 4.( 1/3) (35) 34/3 1-1/3 1-3/5 7/3 15 ca) first term Ans r==1/2 1rl=1/2
16.00) DO E f(x) = 1 I k=2 k(lnk) 2 x(lnx) x f'(x) = [(lnx) 2 lnx] < 0 for x >2 x'(lnx)4 (i) f IS continuous cur x (ln x)2 >0 for x 22 (ii) " positive cuz x(lnx)2 >0 for x 22 (iii) " decreasing cuz f'ix) >0 a E 1/4 Diverges ( Harmenic Series) kez 8 (c) E k k+1 e24 lim 442 kel = lim kr = 0 for Lal 4 Th 1+k Diverges by Comperison Test 8 E Diverges, Harmonic Series without 1st term k=1
(e) & L.C.T W k-1 E k=1 k3+5 k k3 I k Converges, P-series, P= 2 k=1 (leading terms) k-1 him k3+5 k2(k-1) k 3 - k = = 8 2'1 - 1 =L = k3+5 3 k 5 k3 + k2 The Series Converges by L.C.T. (f) kk lim x k! = 8 Growth Rate kal x x! 1 Diverger by Divergence Test 8 & 17.ca) iN (-1)k 1 k E (-1) 1 8 = k= k Ink kn k Ink W 1/4 klnk k21 Use integral Test Diverges by Integral Test. 8 E (-1) + bn = f(x) = kzl klnk n In n x lnx p(xx = - (lxx +1) I i) lim 0 x (ln x)2 n ln n ii) f(x) is decreasing cus Pix
(b) EN (.1) see (1) iN (-1)" sec( E sec(t) bn =see (1) has k=1 kol him n-g sec (i) lim n-a cos = = cas O Diverges by Divergence Test - (c) all (-1)" k a EN k (-1) 2 - 2 w = key E t2 ka kn 2k (k+1)2 lim ( leading terms) 2 kel = lim 24 kga &2 . lim know = 2.2 1/2 = k k-soo 2 is 1/2 2k Converges absolutely by Ratio Test (d) 8 EN (-1)" k kol k'+1 (leading terms) lim (-) k = k-9 00 lim k ki+1 know = 1 The Diverges by Divergence Test 8 (e) W (--1)' 00 kn k key " N C-14 k34 . km EN DO 1/1/ THE < 1/1/2 = 1 3/2 for k> 1 L3 E Converges, P-series, P kal The series converges absolutely by Comparison Test.
(f) oo E-(1)" k in & L.C.T k., Vk'+1 iN : (-1) k iN k'+1 kn 'VTe 1/15 kas 43+1 & (leading terms) lim 3/2 is Diverges P-series, pal/2 ka- = 14 9 k +1 k 3/2 k 1/2 as kg 133.1 W 1/1 I Diverger by L.C.T. as W C-19 k b n f(x) = x kol Vk'+1 Vx'st (leading terms) fix) = 2.x 2 i) lim n n 2(x' +1)( (vx+1) noo slet = or n 1/2 n /2 ii) bn is decreasing cuz f(12)
Math 150B Homework 2 Answers
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