hw-1-sec-5.3-solns

hw-1-sec-5.3-solns - 388 CHAPTER 5 INTEGRALS $64 573 m , 1...

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Unformatted text preview: 388 CHAPTER 5 INTEGRALS $64 573 m , 1 . . 7. f(t) = t3 + 1 and g(x) = /1 ta + 1 dt, so by FTCl, g = f(:c) = $3 + 1. Note that the lower 11m1t, 1, could be any real number greater than — 1 and not affect this answer. 8. f(t) = (2 + t4)5 and 9(13) : flz(2 + t4)5 dt, so g’(ac) : f(;v) : (2 + $4)5. 9. f(t) : t2 sint and g(y) = f: t2 sintdt, so by FTCl, g'(y) 2 fly) : y2 sin y. 10. f(:c) : x/m2 +4andg(7‘) : for V352 +4dm, so by FTCl,g’(r) : f(r) = \/T2 +4. 11. F(x) z/ \/1 +sect dt = —/ V1 +sect dt => F'(:c) = —1 v1 +sect dt = —\/1 +secm 1 I :1: 12. G(a:):/ cosx/idt=—/ cosx/Edt :> G’(x)=—i cosx/Edtz—cos m a: 1 1 du 1 dh dh du 1 . : —, — : —-—l —_ = __ 3 Letu 3: Then dm x2 Also, dx du dm, so I d VI 4 d u 4 du 4 du —sin4(1/x) h dxl SIn td du 2 sm tdt dx sm udm x2 du dh dh du 1 . = 2, — : I -— : _._._ 4 Letu x Then dag 23 Also, dm du dip, so $2 U h’(x):%/ x/lfl—r3drzfi/ \/1+'r3dr-:ll—:=\/1+u3(2:r):2x\/1+(m2)3=2xx/1+:176. 0 0 15. Letu = tanx. Then 5% 2 see2 as. Also, 23—: = ((3—ng, so tans: u y’=i Vt—l—x/Edtzi/ \/t+\/Edt-—d—u=Vu+fifl=Vtanm+\/tanxsec2;r. d3: 0 du O dot dsv 16. Letu = cosx. Then 2—: : —sinm. Also, % = 2—13: Z—Z, so I d CO” 2 10 d u 2 10 du 2 10 du 2 10 . 3; dx 1 (14—11) (11) du 1(1-1—11) d1} dm (1+u) div ( +cos ac) smx _ dw _ dy ~ dy dw 17. Letw — 1 — 31’. Then dd: _ 3. Also, d3: — dw d$,so 1 3 1 3 w 3 3 3 1 _ y/Zi u duzi u du.d_w:_i u du.d_w:_ w RPM dz: l4751+le dw w 1+u2 dm dw 1 1+u2 d3: 1+w2 1+(1—3:1:)2 1 du 2 dy dy du 1 . : —. —— : ——. -~ I —— 8 Letu $2 Then dLr 333 Also, d3: du day, so d 0 d 0 du d “ du 2 25in3(1/$2) / - 3 . 3 . 3 . 3 =— tdtz— t -—-=— dt--—=— —— 2—. y dag 1M2 s1n du/u sm dt dm du 0 sm t div sm u( 363) $3 SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS D 389 ® ff(5—2t+3t2)dt= [5t~t2+t3]:=(2O—16+64)—(5—1+1)268—5263 @fol(1+%u4—§u9>du=[nan—#1015:(Haas—0% 1 23. folm4/5dm = [3.69/5] : g — 0 :3 O 8 24. ff wag; = f18m1/3 d1: = [gx4/3ll = as“ — 14/3) = §<24 — 1) = %(16 — 1) = %(15) = $75 2 3 2 -4 r3 2 3 1 2 1 7 t — 2— — :— —-—- :— @/1 t4 /1‘ 3i_3i1 _3 it3i1 1<8 1) 8 Q ffcosado = [sine]? =sin27T—sin7r =0—0=0 2 27. f02 $(2 +x5)d:c : f02(23: +336) dm = [x2 + %x7]0 = (4 + 1—?) — (0+0) = L? fol<3+mfi>dm= fel<3+ww>dw = [326+ = 1m 3 —01 = 9$—1 _ 9 CE 1 __ 9 1/2 —1/2 _ 2 3/2 1/29 29./1 fl (ix—fl (E—$>dx—/l (x —a: )dsc—[Eat -—2:1: L =(%-27-2-3)-(%—2)=12—(—%)=95? 30- f02(y~ 1)(2'y+ 1)dy= f02(2y2 —y — 1)dy = [%y3 - éy2 ~y]§ = (% — ~ 2) —0=§ @fJ/4sec2tdt = [tant]g/4 = tang —tan0 = 1 —0 = 1 32. few/4sec6 tan6d0 = [sec0]3/4 = secg —sec0 : fl— 1 33. ff(1+2y)2dy=ff(1+4y+4y2)dy= [y+2y2+§y3]f=(2+8+%)—(1+2+§)=%—%=3§3 2s4+1 2 2 2 13 12 8 1 1 7 1 17 34. d = — d = — —— = ~-- — —— =— —=— /1 s2 S [1(3 +8 ) 5 [35 sil (3 2) <3 1) 3+2 6 sinzr if 0§$ <7r/2 35. Iff(x) 2 then cossc if 7r/2gmg7r . 7r . . +[s1nxlw/2 = —COS g +cosO+s1n7r — Sln 1r/2 O 1 f0" f(ac)dx 2 far/2 sinmdm+f:/2 cosasdx = [— cosac] 2 :—0+1+0—1=0 , Note that f is integrable by Theorem 3 in Section 5.2. 36 Iffl) 2 if ~2SZL‘SO h . x: ten 4ww2 if0<m§2 2 [32 f(:r) dz: = ff? 2dx + [02(4 — 1:2)dx = [29432 + [4m — §x3]0 = [0 — (—4)] + (I; — 0) 2 % Note that f is integrable by Theorem 3 in Section 5.2. 37. f = 93—4 is not continuous on the interval [—2, 1], so 'FTCZ cannot be applied. In fact, f has an infinite discontinuity at l _ . a: = 0, so f_2 x 4 dz does not eXISt. ...
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hw-1-sec-5.3-solns - 388 CHAPTER 5 INTEGRALS $64 573 m , 1...

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