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chapter5_10p12 - CHAPTERS at the first part of the...

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Unformatted text preview: CHAPTERS REVIEW 13 403 at the first part of the integral. let 3: : 2 tan 9 =:» dx = 2 sec"2 9 d9. 2 if 1 Luz/25'“: ed92/eec9di9zlnise03-i-ten91, W 2sec6 2+4 I“ E .50 efigure.tan6 = §,andsec9 : _. 1%) do: = lim [in 3% +§‘ ’ Chllz+2ll {Hm From Lh ! 0 = lim t—um : lim [in(3”£jfgl) 4.11126] [In '5' +24“ _ Cln(t+2) — (1111 (01:12)] [HE'S : 1n(1im LEE) +1112“—l r—nza (“"21" . 1+t/‘rt3+4 2 — 11m _ C71 2 -———~———‘ C-l' ‘"°° CUE-+2) C'11m(t+2) £4.90 IfC < 1, L 2 9c and I diverges. It'C = L. L = 2 and 1 converges to in '3 + i112“ = 1112. IfC > 1., L = I) and I diverges to -=x. 52.1— m I — C (0 2722—1 333+l [11102 + 1}”2 —in(:5t+1}(:f3] r2 1 "J" r2 — i (. + )0” = 111 l1111i ’——-' (”:3 (n+1) use“) For C S U. the integral diverges. For C‘ > O. we have ri—— ./a?+1 H _ t/vt-’+l 1 . 1. l : lim ———«—f—. )iem—i L z 1' f—fiv : ___—..—# —' £31 (3H 11“” I25: C(rsr. + i}"-””'1 C Hoc- (:3: L i For 0/3 <1 <= C' < 3. L = eoandf diverges. FurC:3.L=:—§:mdl=in§, ForC > 3. L z I) and I diverges to — oo. 5 Review CONCEPT CHECK ‘l. (a) EL; fix?) Art is an expression for a Riemann sum ofa function f. 2:: is a point in the i111 subintervnl [2:14 , I1] and Ar. is the length of the subintervals. (b) See Figure l in Section 5.2. (c) In Section 5.2. see Figure 3 and the paragraph beside it. 2. (a) See Definition 5.2.2. (in) See Figure 2 in Section 5.2. (c) In Section 5.2. see Figure 4 and the paragraph above it. ...
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