5.5-5.6 - 350 93. Chapter 5 Integration Example CAS...

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Unformatted text preview: 350 93. Chapter 5 Integration Example CAS commands: Maple: f := `f`; q1 := Diff( Int( f(t), t=a..u(x) ), x ); d1 := value( q1 ); Example CAS commands: Maple: f := `f`; q2 := Diff( Int( f(t), t=a..u(x) ), x,x ); value( q2 ); 94. 85-94. Example CAS commands: Mathematica: (assigned function and values for a, and b may vary) For transcendental functions the FindRoot is needed instead of the Solve command. The Map command executes FindRoot over a set of initial guesses Initial guesses will vary as the functions vary. Clear[x, f, F] {a, b}= {0, 21}; f[x_] = Sin[2x] Cos[x/3] F[x_] = Integrate[f[t], {t, a, x}] Plot[{f[x], F[x]},{x, a, b}] x/.Map[FindRoot[F'[x]==0, {x, #}] &,{2, 3, 5, 6}] x/.Map[FindRoot[f'[x]==0, {x, #}] &,{1, 2, 4, 5, 6}] Slightly alter above commands for 75 - 80. Clear[x, f, F, u] a=0; f[x_] = x2 c 2x c 3 u[x_] = 1 c x2 F[x_] = Integrate[f[t], {t, a, u(x)}] x/.Map[FindRoot[F'[x]==0,{x, #}] &,{1, 2, 3, 4}] x/.Map[FindRoot[F''[x]==0,{x,#}] &,{1, 2, 3, 4}] After determining an appropriate value for b, the following can be entered b = 4; Plot[{F[x], {x, a, b}] 5.5 INDEFINTE INTEGRALS AND THE SUBSTITUTION RULE 1. Let u oe 3x du oe 3 dx " 3 ' ' sin 3x dx oe ' du oe dx " 3 " " sin u du oe c 3 cos u b C oe c 3 cos 3x b C " 4 2. Let u oe 2x# du oe 4x dx x sin a2x# b dx oe ' " 4 du oe x dx " " sin u du oe c 4 cos u b C oe c 4 cos 2x# b C " # 3. Let u oe 2t du oe 2 dt ' ' sec 2t tan 2t dt oe ' du oe dt " # " # sec u tan u du oe " # sec u b C oe t 2 2 3 " # sec 2t b C t 4. Let u oe 1 c cos 2 du oe sin ^1 c cos t # # ^sin t # dt oe ' 2u# du oe t # dt 2 du oe sin 2 3 dt u$ b C oe t ^1 c cos # $ b C Section 5.5 Indefinite Integrals and the Substitution Rule 5. Let u oe 7x c 2 du oe 7 dx 351 ' 28(7x c 2)c& dx oe ' " 7 (28)uc& du oe ' 4uc& du oe cuc% b C oe c(7x c 2)c% b C " 4 " 7 du oe dx 6. Let u oe x% c " du oe 4x$ dx $ ' x$ ax% c 1b dx oe ' # du oe x$ dx " 1# " 4 u# du oe u 1# bCoe ax% c 1b b C $ 7. Let u oe 1 c r$ du oe c3r# dr c3 du oe 9r# dr $ # # 9r ' dr 1cr oe ' c3uc"# du oe c3(2)u"# b C oe c6 a1 c r$ b "# bC 8. Let u oe y% b 4y# b 1 du oe a4y$ b 8yb dy 3 du oe 12 ay$ b 2yb dy ' ' ' 12 ay% b 4y# b 1b ay$ b 2yb dy oe ' 3u# du oe u$ b C oe ay% b 4y# b 1b b C # $ 3 # 9. Let u oe x$# c 1 du oe x sin# ^x$# c 1 dx oe ' " x x"# dx 2 3 2 3 du oe x dx 2 3 u ^# c " 4 sin# u du oe sin 2u b C oe " 3 ^x$# c 1 c " 6 sin ^2x$# c 2 b C oe 11. (a) Let u oe cot 2) du oe c2 csc# 2) d) c " du oe csc# 2) d) # (b) Let u oe csc 2) du oe c2 csc 2) cot 2) d) c " du oe csc 2) cot 2) d) # # # # # 12. (a) Let u oe 5x b 8 du oe 5 dx (b) Let u oe 5x b 8 du oe 13. Let u oe 3 c 2s du oe c2 ds c " du oe ds # ' ' ' ' 3 c 2s ds oe ' u ^c " du oe c " ' u"# du oe ^c " ^ 2 u$# b C oe c " (3 c 2s)$# b C 3 3 # # # " # " # 14. Let u oe 2x b 1 du oe 2 dx (2x b 1)$ dx oe ' u$ ^ " du oe # 15. Let u oe 5s b 4 du oe 5 ds " 5s b 4 ds oe ' " u " ^5 du oe " 5 ' " 5 du oe ds " uc"# du oe ^ 5 ^2u"# b C oe 2 5 16. Let u oe 2 c x du oe cdx cdu oe dx # 3 (2 c x) "c dx oe ' 3(cdu) u oe c3 ' uc# du oe c3 S u 1 < b C oe c % # " x " cos ^ x dx oe " c 2x c " sin ^ 2 4 x # ' cos# acub du oe bC ' ' ' ' csc# 2) cot 2) d) oe c ' " csc# 2) cot 2) d) oe ' c " u du oe c " S u < b C oe c u b C oe c 4 csc# 2) b C 4 # # # " 5 dx 5x b 8 oe' oe' " 5 " S u < dx 5x b 8 2 5 du oe # 10. Let u oe c " du oe x dx ' cos# aub du oe ^ u b # " 4 " sin 2u b C oe c 2x b " 4 2 sin ^c x b C " # " u du oe c " S u < b C oe c u b C oe c 4 cot# 2) b C # # 4 du oe " # " 5 ' du oe dx " 5 uc"# du oe ^2u"# b C oe 2 5 2 5 u"# b C oe 2 5 5x b 8 b C (5x b 8)c"# (5) dx 2 5 du oe dx 5xb8 2 5 ubCoe 5x b 8 b C ' du oe dx u$ du oe ^ " S u < b C oe # 4 " 8 (2x b 1)% b C 5s b 4 b C 3 2cx bC 352 Chapter 5 Integration &% ) 1 c )# d) oe ' % u ^c " du oe c " ' u"% du oe ^c " ^ 4 u&% b C oe c 2 a1 c )# b b C # # # 5 5 % 17. Let u oe 1 c )# du oe c2) d) c " du oe ) d) # ' 18. Let u oe 2y# b 1 du oe 4y dy # ' ' ' ' ' ' ' ' 4y dy 2y b 1 oe' " u du oe ' uc"# du oe 2u"# b C oe 22y# b 1 b C " 2 x 19. Let u oe 1 b x du oe # # dx 2 du oe " x dx " x ^" b x dx oe ' 2 du u oec2 bCoe u " 2 x c2 1 b x bC dx ^1 b x% b C 20. Let u oe 1 b x du oe $ & dx 2 du oe " x " # ^1 b x x " dx oe ' u$ (2 du) oe 2 ^ 4 u% b C oe " 3 21. Let u oe 3z b 4 du oe 3 dz cos (3z b 4) dz oe ' (cos u) ^ " 3 du oe dz " 3 du oe " " ' cos u du oe 3 sin u b C oe 3 sin (3z b 4) b C 22. Let u oe tan x du oe sec# x dx tan# x sec# x dx oe ' u# du oe " 3 u$ b C oe " 3 tan$ x b C 23. Let u oe cos x du oe csin x dx cdu oe sin x dx tan x dx oe ' sin x cos x dx oe c' " # 1 u du oe cln lul b C oe cln lcos xl b C oe ln lcos xlc1 b C oe ln lsec xl b C 24. Let u oe tan ^ x du oe # " tan( ^ x sec# ^ x dx oe ' u( (2 du) oe 2 ^ 8 u) b C oe # # # sec# ^ x dx 2 du oe sec# ^ x dx # # " 4 tan) ^ x b C # r r r# S 18 c 1< dr oe ' u& (6 du) oe 6 ' u& du oe 6 S u < b C oe S 18 c 1< b C 6 $ ' 27. Let u oe x$# b 1 du oe ' ' ' x"# sin ^x$# b 1 dx oe ' (sin u) ^ 2 du oe 3 4 3 3 # x"# dx 2 3 du oe x"# dx 2 3 ' ' sin u du oe 2 3 (ccos u) b C oe c 2 cos ^x$# b 1 b C 3 28. Let u oe x%$ c 8 du oe x"$ sin ^x%$ c 8 dx oe ' (sin u) ^ 3 du oe 4 x"$ dx 3 4 du oe x"$ dx 3 4 sin u du oe 3 4 (ccos u) b C oe c 3 cos ^x%$ c 8 b C 4 29. Let u oe cos (2t b 1) du oe c2 sin (2t b 1) dt c " du oe sin (2t b 1) dt # # sin (2t b 1) cos (2t b 1) # " dt oe ' c # du u oe " #u bCoe " # cos (2t b 1) bC & % ' r % S7 c r 10 < & 26. Let u oe 7 c $ 25. Let u oe $ r 18 c 1 du oe & r 6 dr 6 du oe r# dr ' r 10 $ dr oe ' u$ (c2 du) oe c2 ' u$ du oe c2 S u < b C oe c " S7 c # 4 du oe c " r% dr c2 du oe r% dr # r 10 < % bC Section 5.5 Indefinite Integrals and the Substitution Rule 30. Let u oe 2 b sin t du oe cos t dt $ $ 353 6 cos t (2 b sin t) 6 u 32. Let u oe sec z du oe sec z tan z dz ' sec z tan z sec z " t 34. Let u oe csc ) du oe Sccsc ) cot )< S # ' cos ) ) sin ) 35. Let u oe s$ b 2s# c 5s b 5 du oe a3s# b 4s c 5b ds # u # 36. Let u oe 1 b t% du oe 4t$ dt ' t$ a1 b t% b dt oe ' u$ ^ " du oe 4 $ # " 4 " 4 du oe t$ dt " 16 " ^ 4 u% b C oe a 1 b t% b b C % 37. Let u oe 1 c & 38. Let u oe x# b ". Then du oe #xdx and " du oe xdx and x# oe u c ". Thus ' x$ x# b " dx oe ' au c "b " u du # # oe " # ' au$# c u"# bdu oe " ' # u&# c # u$# " b C oe " u&# c " u$# b C oe " ax# b "b&# c " ax# b "b$# b C # & $ & $ & $ 39. Let u oe sin x du oe cos x dx ' acos xb esin x dx oe ' eu du oe eu b C oe esin x b C ' asin2 )b esin ) d) oe ' eu du oe eu b C oe esin ) b C 2 2 40. Let u oe sin2 ) du oe 2 sin )cos ) d) oe sin2 ) d) 42. Let u oe 1 b e x du oe e x 1 1 ' x1 e 2 1 x c ' 1 xe x sec2 ^ex b 1dx oe 2' sec2 u du oe 2 tan u b C oe 2 tan^ex b 1 b C c1 x2 dx 1 sec S1 b e x <tan S1 b e x < dx oe c' sec u tan u du oe csec u b C oe csecS1 b e x < b C 1 1 cdu oe e x 1 1 x2 dx c 41. Let u oe ex b 1 du oe 1 ex dx 2 x 2 du oe # # c ' x x 1 dx oe ' " x du oe " x x c 1 dx oe ' x " x dx " x 1 c " x dx oe ' u du oe ' u"# du oe 1 x x e dx oe 1 x e x dx # $ u du oe bCoe # ' as$ b 2s# c 5s b 5b a3s# b 4s c 5b ds oe ' # ' " t cos ^ " t c 1 dt oe ' (cos u)(cdu) oe c ' cos u du oe csin u b C oe csin ^ " c 1 b C t " < # ) d) c2 du oe d) oe ' " ) cot ) csc ) d) oe ' c2 du oe c2u b C oe c2 csc ) b C oe c # 33. Let u oe # ' " ) sin " ) cos " ) " " d) oe ' cu du oe c # u# b C oe c # sin# dz oe ' " u du oe ' uc"# du oe 2u"# b C oe 2sec z b C " t c 1 oe tc" c 1 du oe ctc# dt cdu oe # # 31. Let u oe sin " ) du oe ^cos " ^c )" d) cdu oe ) #c ' dt oe ' du oe 6 ' uc$ du oe 6 S u < b C oe c3(2 b sin t)c# b C c# " ) cos " ) " ) d) bC dt " ) cot ) csc ) d) 2 sin ) bC as b 2s c 5s b 5b # bC 2 3 u$# b C oe 2 3 ^1 c " $# b C x 354 Chapter 5 Integration 43. Let u oe ln x du oe 1 dx x 1 ' x ln x dx oe ' 1 du oe ln lul b C oe ln lln xl b C u 44. Let u oe ln t du oe 1 dt t ' ln t dt oe ' ln tt t dz ' 1be 1 2 dt oe " # ' lnt t dt oe " ' u du oe " u2 # # e b 1 dz z 2 b C oe 1 aln tb2 b C 4 45. Let u oe ecz b 1 du oe cecz dz cdu oe ecz dz c z oe caln a1 b e b c ln ez b b C oe z c ln a1 b ez b b C 46. Let u oe x2 du oe 2x dx " du oe x dx # 1 1 b ez z e e z e z ' dx x x4 c 1 oe' x dx x 2 ax 2 b 2 c 1 oe " # ' 47. Let u oe 2 r du oe 2 dr 3 du oe dr 3 3 2 ' 9 b54r dr oe 5 ' 9 2 1 2 dr 1 b ^ 2 r 3 oe 5 9 5 5 2 ' 1 b u du oe 6 tanc1 u b C oe 6 tanc1 ^ 3 r b C 3 2 2 48. Let u oe e) du oe e) d) ) ) ) ) ' e ' 1 d) e2 c 1 oe' e d) 2 e ae b c 1 oe' oe 2 3 u$# b C oe 2 3 atanc" xb < $# bCoe " u 2 3 atanc" xb$ b C oe ln kuk b C oe ln ktan yk b C ; oe ln kuk b C oe ln ksinc" yk b C # "c # "c 54. asin " yb 1 b y dy oe ' 1 # c " ' : sin y y dy oe ' " u du, where u oe sinc" y and du oe dy 1 c y # "c # "c 53. " y b a1 b y b dy oe 1 # b " ' atan ' Stan y y c" dy oe ' du, where u oe tanc" y and du oe # # 52. dx ' 1tanx x dx oe ' u"# du, where u oe tanc" x and du oe 1 b x b "c $ oe u 3 bCoe 3 $ "c asin xb bC # # 51. # "c sin ' a xb 1cx dx oe ' u# du, where u oe sinc" x and du oe "c oe ceu b C oe cecos x bC dx 1 c x # # 50. cos 1cx "c e ' x dx oe c ' eu du, where u oe cosc" x and du oe "c oe eu b C oe esin x bC c dx 1 c x # # 49. sin x 1cx dx oe ' eu du, where u oe sinc" x and du oe c c c oe' z dz oe ' 1 oe c' u du oe cln lul b C oe cln aecz b 1b b C oe cln ^ 1 bz e b C e z du u u2 c 1 oe " secc1 u b C oe " secc1 ax2 b b C # # du u u2 c 1 oe secc1 u b C oe secc1 ^e) b C "c dx 1 c x dy 1by Section 5.5 Indefinite Integrals and the Substitution Rule 55. (a) Let u oe tan x du oe sec# x dx; v oe u$ dv oe 3u# du 6 dv oe 18u# du; w oe 2 b v dw oe dv bCoe dx oe ' dx oe ' # # # $ # # # $ # 355 ' ' ' ' (b) Let u oe tan x du oe 3 tan x sec x dx 6 du oe 18 tan# x sec# x dx; v oe 2 b u dv oe du (c) Let u oe 2 b tan$ x du oe 3 tan# x sec# x dx 6 du oe 18 tan# x sec# x dx $ # $ # # 18 tan x sec x a2 b tan xb # # $ # 6 du (2 b u) oe' 18 tan x sec x a2 b tan xb # # $ # 6 du u 6 6 oe c u b C oe c 2 b tan 56. (a) Let u oe x c 1 du oe dx; v oe sin u dv oe cos u du; w oe 1 b v# dw oe 2v dv oe' " # " 3 1 b sin# (x c 1) sin (x c 1) cos (x c 1) dx oe ' 1 b sin# u sin u cos u du oe ' v1 b v# dv w dw oe w$# b C oe " 3 (b) Let u oe sin (x c 1) du oe cos (x c 1) dx; v oe 1 b u dv oe 2u du ' ' 1 b sin# (x c 1) sin (x c 1) cos (x c 1) dx oe ' u 1 b u# du oe ' oe ^ " ^ 2 v$# b C oe # 3 # " 3 v$# b C oe (c) Let u oe 1 b sin (x c 1) du oe 2 sin (x c 1) cos (x c 1) dx 1 b sin# (x c 1) sin (x c 1) cos (x c 1) dx oe ' " 3 " # oe a1 b sin# (x c 1)b $# bC " 1# 57. Let u oe 3(2r c 1)# b 6 du oe 6(2r c 1)(2) dr # # ' (2r c 1) cos 3(2r c 1) b 6 3(2r c 1) b 6 dr oe ' S cos u ^" u < 1# oe " 6 sin 3(2r c 1)# b 6 b C " < # ) 58. Let u oe cos ) du oe Scsin )< S oe 4 cos ) bC 59. Let u oe 3t# c 1 du oe 6t dt 2 du oe 12t dt $ s oe ' 12t a3t# c 1b dt oe ' u$ (2 du) oe 2 ^ " u% b C oe 4 s oe 3 when t oe 1 3 oe " # (3 c 1)% b C 3 oe 8 b C C oe c5 s oe 60. Let u oe x# b 8 du oe 2x dx 2 du oe 4x dx y oe ' 4x ax# b 8b c"$ dx oe ' uc"$ (2 du) oe 2 ^ 3 u#$ b C oe 3u#$ b C oe 3 ax# b 8b # #$ y oe 0 when x oe 0 0 oe 3(8) 61. Let u oe t b 1 1# b C C oe c12 y oe 3 ax b 8b s oe ' 8 sin# ^t b 1 dt oe ' 8 sin# u du oe 8 ^ u c " sin 2u b C oe 4 ^t b 1# c 2 sin ^2t b 1 b C; 4 6 # 1 s oe 8 when t oe 0 8 oe 4 ^ 1# c 2 sin ^ 1 b C C oe 8 c 1 b 1 oe 9 c 1 6 3 3 1 s oe 4^t b 1# c 2 sin ^2t b 1 b 9 c 1 oe 4t c 2 sin ^2t b 1 b 9 6 3 6 1 1# du oe dt #$ $ $ ' sin ) ) cos ) d) oe ' sin ) ) cos ) $ $ 18 tan x sec x a2 b tan xb 6 oe c 2bu $ dx oe ' 18u du oe a2 b u b 6 c 2 b tan x b C # # ' 6 dv (2 b v) oe' 6 dw w 6 oe 6 ' wc# dw oe c6wc" b C oe c # b v b C 6 dv v 6 6 6 oe c v b C oe c 2 b u b C oe c # b tan x bC x bC " # dw oe v dv $# a 1 b v# b $# bCoe # " 3 a1 b sin# ub $# bCoe " # " 3 a1 b sin# (x c 1)b " # bC " # v dv oe ' $# dv oe u du v"# dv bC " 3 a1 b u # b $# bCoe " 3 a1 b sin# (x c 1)b " # " # u du oe ' du oe sin (x c 1) cos (x c 1) dx u"# du oe " # ^ 2 u$# b C 3 du oe (2r c 1) dr; v oe u dv oe " 6 " # u du " 6 dv oe " 1 # u du " du oe ' (cos v) ^ 6 dv oe sin v b C oe " 6 sin u b C d) c2 du oe sin ) ) d) 4 u d) oe ' c2 du u oe c2 ' uc$# du oe c2 ^c2uc"# b C oe bC " # u% b C oe " # a3t# c 1b b C; " # % a3t# c 1b c 5 % #$ b C; # #$ c 12 356 Chapter 5 Integration 1 4 62. Let u oe roe 1 8 r oe ' 3 cos# ^ 1 c ) d) oe c ' 3 cos# u du oe c3 ^ u b 4 # 1 3 when ) oe 0 1 oe c 38 c 4 sin 1 8 # 3 3 1 1 3 r oe # ) c 4 sin ^ # c 2) b 8 b 4 r oe 1 # 3 2 c ) cdu oe d) 3 sin 2u b C oe c 3 ^ 1 c ) c 4 sin ^ 1 c 2) b C; # 4 # 1 3 3 ^1 3 c 4 sin ^ 1 c 2) b 1 b bC Coe # b 4 roec# 4 c) # # " 4 3 4 )c 3 4 cos 2) b 1 8 b 3 4 63. Let u oe 2t c ds dt oe ' c4 sin ^2t c 1 dt oe ' (sin u)(c2 du) oe 2 cos u b C" oe 2 cos ^2t c 1 b C" ; # # ds dt du oe 2 dt c2 du oe c4 dt at t oe 0 and s oe ' ^2 cos ^2t c 1 b 100 dt oe ' (cos u b 50) du oe sin u b 50u b C# oe sin ^2t c 1 b 50 ^2t c 1 b C# ; # # # at t oe 0 and s oe 0 we have 0 oe sin ^c 1 b 50 ^c 1 b C# C# oe 1 b 251 # # s oe sin ^2t c 1 b 100t c 251 b (1 b 251) s oe sin ^2t c 1 b 100t b 1 # # oe 100 we have 100 oe 2 cos ^c 1 b C" C" oe 100 # ds dt oe 2 cos ^2t c 1 b 100 # 64. Let u oe tan 2x du oe 2 sec# 2x dx 2 du oe 4 sec# 2x dx; v oe 2x dv oe 2 dx dy dx oe ' 4 sec# 2x tan 2x dx oe ' u(2 du) oe u# b C" oe tan# 2x b C" ; dy dx " # dv oe dx at x oe 0 and y oe ' asec# 2x b 3b dx oe ' asec# v b 3b ^ " dv oe # " # oe 4 we have 4 oe 0 b C" C" oe 4 " # dy dx oe tan# 2x b 4 oe asec# 2x c 1b b 4 oe sec# 2x b 3 " # " # tan v b 3 v b C# oe # tan 2x b 3x b C# ; at x oe 0 and y oe c1 we have c1 oe (0) b 0 b C# C# oe c1 y oe tan 2x b 3x c 1 65. Let u oe 2t du oe 2 dt 3 du oe 6 dt s oe ' 6 sin 2t dt oe ' (sin u)(3 du) oe c3 cos u b C oe c3 cos 2t b C; at t oe 0 and s oe 0 we have 0 oe c3 cos 0 b C C oe 3 s oe 3 c 3 cos 2t s ^ 1 oe 3 c 3 cos (1) oe 6 m # 66. Let u oe 1t du oe 1 dt 1 du oe 1# dt v oe ' 1# cos 1t dt oe ' (cos u)(1 du) oe 1 sin u b C" oe 1 sin (1t) b C" ; at t oe 0 and v oe 8 we have 8 oe 1(0) b C" C" oe 8 v oe s oe 8t c cos (1t) b 1 s(1) oe 8 c cos 1 b 1 oe 10 m ds dt oe 1 sin (1t) b 8 s oe ' (1 sin (1t) b 8) dt oe ' sin u du b 8t b C# oe ccos (1t) b 8t b C# ; at t oe 0 and s oe 0 we have 0 oe c1 b C# C# oe 1 67. All three integrations are correct. In each case, the derivative of the function on the right is the integrand on the left, and each formula has an arbitrary constant for generating the remaining antiderivatives. Moreover, sin# x b C" oe 1 c cos# x b C" C# oe 1 b C" ; also ccos# x b C# oe c cos 2x c " b C# C$ oe C# c " oe C" b " . # # # # 68. Both integrations are correct. In each case, the derivative of the function on the right is the integrand on the left, and each formula has an arbitrary constant for generating the remaining antiderivatives. Moreover, tan x sec xc1 ^C c " b C oe sec x b # bCoe # # # # # # a constant 69. (a) S " < 60 c 0 " 0 oe c Vmax [1 c 1] oe 0 #1 (b) Vmax oe 2 Vrms oe 2 (240) 339 volts ' 1 60 " Vmax sin 1201t dt oe 60 <cVmax ^ 1201 cos (1201t)` ! "'! oe c Vmax [cos 21 c cos 0] #1 Section 5.6 Substitution and Area Between Curves (c) # 357 0 aVmax b# sin# 1201t dt oe aVmax b# # 0 ^ 1 c cos 2401t dt oe # # aVmax b # 0 oe 5.6 SUBSTITUTION AND AREA BETWEEN CURVES 1. (a) Let u oe y b 1 du oe dy; y oe 0 u oe 1, y oe 3 u oe 4 ' ' ' ' ' ' 3 0 y b 1 dy oe y b 1 dy oe ' 4 1 u"# du oe < 2 u$# ` " oe ^ 2 (4)$# c ^ 2 (1)$# oe ^ 2 (8) c ^ 2 (1) oe 3 3 3 3 3 u"# du oe < 2 u$# ` ! oe ^ 2 (1)$# c 0 oe 3 3 " 2 3 % 14 3 (b) Use the same substitution for u as in part (a); y oe c1 u oe 0, y oe 0 u oe 1 0 1 ' 1 0 2. (a) Let u oe 1 c r# du oe c2r dr c " du oe r dr; r oe 0 u oe 1, r oe 1 u oe 0 # 1 0 r 1 c r# dr oe r 1 c r# dr oe ' 0 1 " c " u du oe <c " u$# ` " oe 0 c ^c 3 (1)$# oe # 3 ! " 3 (b) Use the same substitution for u as in part (a); r oe c1 u oe 0, r oe 1 u oe 0 1 1 ' 0 0 c " u du oe 0 # 1 4 3. (a) Let u oe tan x du oe sec# x dx; x oe 0 u oe 0, x oe 4 0 0 uoe1 (b) Use the same substitution as in part (a); x oe c 1 u oe c1, x oe 0 u oe 0 4 0 4 1 # ! 4. (a) Let u oe cos x du oe csin x dx cdu oe sin x dx; x oe 0 u oe 1, x oe 1 u oe c1 0 (b) Use the same substitution as in part (a); x oe 21 u oe 1, x oe 31 u oe c1 1 2 5. (a) u oe 1 b t% du oe 4t$ dt 0 1 (b) Use the same substitution as in part (a); t oe c1 u oe 2, t oe 1 u oe 2 c " ' ' ' 1 1 t$ a1 b t% b dt oe $ ' 2 2 " 4 u$ du oe 0 " # 6. (a) Let u oe t# b 1 du oe 2t dt 7 0 du oe t dt; t oe 0 u oe 1, t oe 7 u oe 8 ) 45 8 t at# b 1b "$ dt oe ' 8 1 " # u"$ du oe <^ " ^ 3 u%$ ` " oe ^ 3 (8)%$ c ^ 3 (1)%$ oe 4 8 8 # u"$ du oe c " # (b) Use the same substitution as in part (a); t oe c7 u oe 8, t oe 0 u oe 1 0 7 t at# b 1b "$ dt oe ' " 8 # 1 ' 8 1 " # u"$ du oe c 45 8 7. (a) Let u oe 4 b r# du oe 2r dr 1 # # 5r 'c a4 b r b 1 dr oe 5 ' du oe r dr; r oe c1 u oe 5, r oe 1 u oe 5 5 5 " # uc# du oe 0 % % % ' 1 c 1 ' 1 ' 3 cos# x sin x dx oe ' c c3u# du oe ccu$ d c" oe c(c1)$ c ac(1)$ b oe 2 " 1 1 3 3 cos# x sin x dx oe 'c 1 t$ a1 b t% b dt oe $ ' 2 " 4 c tan x sec# x dx oe ' 0 u du oe ' u " # ! c" oe0c 1 c3u# du oe 2 " 4 du oe t$ dt; t oe 0 u oe 1, t oe 1 u oe 2 # 2 16 u u$ du oe ' 16 " oe # # tan x sec# x dx oe ' 1 u du oe ' u " oe # " 1 # c0oe " # " # oec" # c 1 16 oe 15 16 # aVmax b # " <t c ^ 2401 sin 2401t` "'! oe ! aVmax b # " " " <^ 60 c ^ 2401 sin (41) c ^0 c ^ #401 sin (0)` oe 1c 1 ' 1 60 ' 1 60 ' 1 60 (1 c cos 2401t) dt aVmax b 1#0 c c 358 Chapter 5 Integration (b) Use the same substitution as in part (a); r oe 0 u oe 4, r oe 1 u oe 5 # # ' ' ' ' 1 5r 0 a4 b r b dr oe 5 ' 5 4 " # uc# du oe 5 <c " uc" ` % oe 5 ^c " (5)c" c 5 ^c " (4)c" oe # # # 3 # & " 8 8. (a) Let u oe 1 b v$# du oe 1 10v # #$ 0 1 (b) Use the same substitution as in part (a); v oe 1 u oe 2, v oe 4 u oe 1 b 4$# oe 9 4 10v # #$ 1 2 9. (a) Let u oe x# b 1 du oe 2x dx 2 du oe 4x dx; x oe 0 u oe 1, x oe 3 u oe 4 3 0 4x x b 1 # dx oe ' 4 2 1 u (b) Use the same substitution as in part (a); x oe c3 u oe 4, x oe 3 u oe 4 # 10. (a) Let u oe x% b 9 du oe 4x$ dx $ % ' ' ' ' x 0 x b 9 0 (b) Use the same substitution as in part (a); x oe c1 u oe 10, x oe 0 u oe 9 x 1 x b 9 $ % 11. (a) Let u oe 1 c cos 3t du oe 3 sin 3t dt 6 0 0 (b) Use the same substitution as in part (a); t oe 3 6 1 12. (a) Let u oe 2 b tan 1c ' ' ' t # du oe t # " # sec# 2 1 0 2 t ^2 b tan # sec# dt oe ' t # dt 2 du oe sec# # u (2 du) oe cu# d " oe 2# c 1# oe 3 c1 # $ (b) Use the same substitution as in part (a); t oe 2 2 u oe 1, t oe t ^2 b tan # sec# t # dt oe 2 ' 3 1 u du oe cu# d " oe 3# c 1# oe 8 " 3 13. (a) Let u oe 4 b 3 sin z du oe 3 cos z dz 0 (b) Use the same substitution as in part (a); z oe c1 u oe 4 b 3 sin (c1) oe 4, z oe 1 u oe 4 1c 1 ' ' 14. (a) Let u oe 3 b 2 cos w du oe c2 sin w dw c " du oe sin w dw; w oe c 1 u oe 3, w oe 0 u oe 5 # # 0 # 1c 1 2 cos z 4 b 3 sin z dz oe dz oe ' du oe cos z dz; z oe 0 u oe 4, z oe 21 u oe 4 4 4 " u " ^ 3 du oe 0 cos z 4 b 3 sin z ' 4 4 " u " ^ 3 du oe 0 sin w 2 (3 b 2 cos w) dw oe ' 5 3 " uc# ^c # du oe " # cuc" d $ oe # (1 c cos 3t) sin 3t dt oe ' 1 6 u oe 1, t oe # " " 6 2 " 3 " u du oe ' 3 S u < " oe # # c ' 3 4x 3 x b 1 dx oe ' 10 4 4 2 u 1 dx oe ' 9 " 4 " uc"# du oe < 4 (2)u"# ` * oe dx oe ' " 10 4 9 (1 c cos 3t) sin 3t dt oe # a1 b v b dv oe ' 9 " u # a1 b v b dv oe ' v"# dv 20 3 2 " u ^ 20 du oe 3 ' 20 3 2 du oe 10v dv; v oe 0 u oe 1, v oe 1 u oe 2 10 3 1 " # " 1 uc# du oe c 20 < u ` " oe c 20 < # c 1 ` oe 3 3 1 7 " " ^ 20 du oe c 20 < u ` * oe c 20 ^ 9 c 2 oe c 20 ^c 18 oe 3 3 3 3 # 70 #7 du oe ' 1c 1 1 1 4 1 2uc"# du oe <4u"# ` " oe 4(4)"# c 4(1)"# oe 4 % du oe 0 " 4 du oe x$ dx; x oe 0 u oe 9, x oe 1 u oe 10 "! " # " (10)"# c # (9)"# oe 10 c 3 # uc"# du oe c ' 10 9 " 4 uc"# du oe " 3 3 c 10 # 1 6 1 c du oe sin 3t dt; t oe 0 u oe 0, t oe " ! " 6 " (1)# c 6 (0)# oe 1 3 " 6 u oe 1 c cos ' 1 # oe1 1 " 3 " u du oe ' 3 S u < " oe # u oe 1 c cos 1 oe 2 " 2 " (2)# c 6 (1)# oe c1 # t # dt; t oe u oe 2 b tan ^ c1 oe 1, t oe 0 u oe 2 4 1 # uoe3 & " # " ^ " c " oe c 15 5 3 Section 5.6 Substitution and Area Between Curves (b) Use the same substitution as in part (a); w oe 0 u oe 5, w oe # 359 15. Let u oe t& b 2t du oe a5t% b 2b dt; t oe 0 u oe 0, t oe 1 u oe 3 ' 1 0 t& b 2t a5t% b 2b dt oe 16. Let u oe 1 b y du oe 1 2 # # ' ' 4 dy 2 y ^1 b y 17. Let u oe cos 2) du oe c2 sin 2) d) c " du oe sin 2) d); ) oe 0 u oe 1, ) oe # 1 1 #c ) 18. Let u oe tan ^ 6 du oe " 6 ) ) sec# ^ 6 d) 6 du oe sec# ^ 6 d); ) oe 1 u oe tan ^ 1 oe 6 u oe tan 1 1 4 oe1 " 19. Let u oe 5 c 4 cos t du oe 4 sin t dt u oe 5 c 4 cos 1 oe 9 1 " 4 du oe sin t dt; t oe 0 u oe 5 c 4 cos 0 oe 1, t oe 1 5 4 ' 5 (5 c 4 cos t)"% sin t dt oe ' 9 1 5u"% ^ " du oe 4 ' 9 1 u"% du oe < 5 ^ 4 u&% ` " oe 9&% c 1 oe $&# c " 4 5 1 4 * 20. Let u oe 1 c sin 2t du oe c2 cos 2t dt c " du oe cos 2t dt; t oe 0 u oe 1, t oe # 1 uoe0 " 5 ' 4 (1 c sin 2t)$# cos 2t dt oe ' 0 1 c " u$# du oe <c " ^ 2 u&# ` " oe ^c 1 (0)&# c ^c 1 (1)&# oe 2 5 5 5 # ! 21. Let u oe 4y c y# b 4y$ b 1 du oe a4 c 2y b 12y# b dy; y oe 0 u oe 1, y oe 1 u oe 4(1) c (1)# b 4(1)$ b 1 oe 8 ' 1 a4y c y# b 4y$ b 1b c#$ a12y# c 2y b 4b dy oe ' 8 1 uc#$ du oe <3u"$ ` " oe 3(8)"$ c 3(1)"$ oe 3 " 3 ) 22. Let u oe y$ b 6y# c 12y b 9 du oe a3y# b 12y c 12b dy du oe ay# b 4y c 4b dy; y oe 0 u oe 9, y oe 1 % 2 3 2 (4)"# c 3 (9)"# oe 2 3 ' uoe4 1 ay$ b 6y# c 12y b 9b 2 3 3 # c"# ay# b 4y c 4b dy oe ' 4 9 " 3 " uc"# du oe < 3 ^2u"# ` * oe oec 23. Let u oe )$# du oe #1 )"# d) 1 2 3 3 du oe ) d); ) oe 0 u oe 0, ) oe 1# u oe 1 24. Let u oe 1 b c c oe " # c " 4 sin 2 ! 1 tc# sin# ^1 b " dt oe t c ' " t " du oe ctc# dt; t oe c1 u oe 0, t oe c # u oe c1 1 2 ' ! ' ! 3 ) cos# ^)$# d) oe ' u cos# u ^ 2 du oe < 2 ^ # b 3 3 " 4 sin 2u` ! oe 1 2 3 ^1 b # " 4 2 sin 21 c 3 (0) oe 1 3 1 csin# u du oe <c ^ u c 2 " 4 c" " sin 2u` ! oe c '^c # c " 4 0 sin (c2) c ^ # c 1 3 3 < % % % %c ! 1 ' 3 2 ) ) cot& ^ 6 sec# ^ 6 d) oe ' 1 uc& (6 du) oe '6 S u 4 <" c " "$ 3 " 3 oe <c 2u ` "$ oe c 2(1) c :c 3 #S # " " 3 ,)oe # # cosc$ 2) sin 2) d) oe uc$ ^c " du oe c " # # 6 1 ' ! 2 sin w (3 b 2 cos w) dw oe ' 1 # uoe3 3 5 " uc# ^c # du oe " # ' uc# du oe 3 5 " 15 ' 3 0 u"# du oe < 2 u$# ` ! oe 3 $ 2 3 (3)$# c 2 (0)$# oe 23 3 dy 2 y ; y oe 1 u oe 2, y oe 4 u oe 3 3 2 oe ' 3 " u du oe ' uc# du oe ccuc" d # oe ^c 1 c ^c 1 oe 3 2 $ " 6 ' 1 2 ' 1 6 u oe cos 2 ^ 1 oe 6 c " 4(1) " # 1 2 " uc$ du oe 'c 2 S u <" c# "# 1 oe " 4 ^1 oe 3 4 ! ! ! ! 31 # ; oe 12 (2 c 3) " 4 sin 0" 360 Chapter 5 Integration 1 4 25. Let u oe tan ) du oe sec# ) d); ) oe 0 u oe 0, ) oe u oe 1; 1 4 '0 1 4 ^1 b etan ) sec# ) d) oe ' 14 0 sec# ) d) b '0 eu du oe ctan )d 0 1 b ceu d " oe <tan ^ 1 c tan (0)` b ae" c e! b ! 4 oe (1 c 0) b (e c 1) oe e 26. Let u oe cot ) du oe c csc# ) d); ) oe 1 4 u oe 1, ) oe 0 1 2 u oe 0; 12 '14 ^1 b ecot ) csc# ) d) oe '14 1 2 12 csc# ) d) c '1 eu du oe cc cot )d 14 c ceu d ! oe <c cot ^ 1 b cot ^ 1 ` c ae! c e" b " 2 4 oe (0 b 1) c (1 c e) oe e 27. 1 '0 '0 3 28. 4 sin ) 1 c 4 cos ) d) oe cln k1 c 4 cos )kd ! 1 3 1$ 29. Let u oe ln x du oe '1 '2 '2 '2 2 2 ln x x dx oe '0 " x dx; x oe 1 u oe 0 and x oe 2 u oe ln 2; ln 2 ln 2 2u du oe cu# d 0 oe (ln 2)# " x 30. Let u oe ln x du oe 4 dx; x oe 2 u oe ln 2 and x oe 4 u oe ln 4; # dx x ln x oe 'ln 2 ln 4 " u ln du oe cln ud ln 4 oe ln (ln 4) c ln (ln 2) oe ln ^ ln 4 oe ln S ln 2 < oe ln ^ 2ln 22 oe ln 2 ln 2 ln 2 ln 2 " x 31. Let u oe ln x du oe 4 # dx; x oe 2 u oe ln 2 and x oe 4 u oe ln 4; ln 4 # dx x(ln x) oe 'ln 2 uc# du oe <c " ` ln 2 oe c ln"4 b u ln 4 " ln # oe c ln"# b " ln 2 " oe c 2 ln # b 32. Let u oe ln x du oe 16 dx 2xln x oe x # " # 'ln 2 " x dx; x oe 2 u oe ln 2 and x oe 16 u oe ln 16; ln 16 ln 16 uc"# du oe <u"# ` ln 2 oe ln 16 c ln 2 oe 4 ln 2 c ln 2 oe 2ln 2 c ln 2 oe ln 2 33. Let u oe cos 1 du oe c " sin # 2 # x # dx c2 du oe sin 1 x # dx; x oe 0 u oe 1 and x oe 2 34. Let u oe sin t du oe cos t dt; t oe 1 4 4 35. Let u oe sin 1 ) 3 du oe 1 " 3 ) cos ) ) 3 36. Let u oe cos 3x du oe c3 sin 3x dx c2 du oe 6 sin 3x dx; x oe 0 u oe 1 and x oe 6 sin 3x cos 3x du u oe c2 cln kukd 1 1 1 12 12 1 1 '0 6 tan 3x dx oe '0 dx oe c2 '1 2 2 2 2 cot du u 3 oe 6 cln kukd 1 2 2 oe 6 Sln ) 3 2 cos 3 sin 3 1 1 ' d) oe ' d) oe 6 '1 2 1 1 1 ' 2 cot t dt oe ' 2 cos t sin t # tan x # du u oe cc2 ln kukd 1 1 " 2 1 4 du u uoe and t oe dt oe '1 1 2 oe cln kukd " # oe c ln " ) 3 d) 6 du oe 2 cos 3 2 d) ; ) oe 1 2 sin x cos x '0 dx oe '0 dx oe c2 '1 2 oe c2 ln " 2 1 # u oe 1; oe ln 2 1 # " 2 uoe 3 # c ln " < oe 6 ln 3 oe ln 27 # 1 1# 2 oe c2 ln c with ) oe 0 u oe c3 and ) oe 4 sin ) 1 c 4 cos ) c 1 u oe c1 '0 1 u oe 1 and t oe 1 u oe 3 '0 1 sin t 2 c cos t 1 dt oe cln k2 c cos tkd ! oe ln 3 c ln 1 oe ln 3; or let u oe 2 c cos t du oe sin t dt with t oe 0 sin t 2 c cos t dt oe '1 3 " u du oe cln kukd $ oe ln 3 c ln 1 oe ln 3 " oe ln k1 c 2k oe c ln 3 oe ln " ; or let u oe 1 c 4 cos ) du oe 4 sin ) d) 3 3 d) oe ' 1 3 " u du oe cln kukd c" oe c ln 3 oe ln c$ " 3 " ln # oe " # ln 2 oe " ln 4 1 # uoe " 2 ; oe 2 ln 2 oe ln 2 " # and ) oe 1 u oe 3 # ; uoe " 2 ; " 2 c ln 1 oe 2 ln 2 oe ln 2 Section 5.6 Substitution and Area Between Curves c 2 1 # # 361 6 3 oe cc tanc" ud $ oe c tanc" 1 b tanc" 3 oe c 1 b 4 3 " oe ctanc" ud " 40. % 1 $ oe tanc" 3 c tanc" 1 oe 4 oe c4 41. oe 4 tanc" 1 4 4 oe ctan ud 14 oe tan 1 1$ 1 3 c tan 1 4 oe 3 c 1 oe csin ud 1' oe sin 2 2 1$ 1 3 c sin 1 6 oe 3 c " # oe csecc" kukd c# 2 3 c # oe secc" c2 c secc" kc2k oe oe csecc" kukd c# c # oe secc" c2 c secc" kc2k oe 47. Let u oe 4 c x# du oe c2x dx c " du oe x dx; x oe c2 u oe 0, x oe 0 u oe 4, x oe 2 u oe 0 # oe < 2 u$# ` ! oe 3 % 2 3 (4)$# c 2 (0)$# oe 3 16 3 48. Let u oe 1 c cos x du oe sin x dx; x oe 0 u oe 0, x oe 1 u oe 2 ! # # # ! 1 ' (1 c cos x) sin x dx oe ' 2 u du oe ' u " oe 2 # 2 # c 0 # oe2 ! ! ! ! c Aoec ' 0 2 x4 c x# dx b ' # c # c 46. 2 3 dy y9y c 1 c c ' oe' 2 2 du u u c 1 # c # c 45. 1 dy y4y c 1 c c ' oe' 2 2 du u u c 1 , where u oe 2y and du oe 2 dy; y oe c1 u oe c2, y oe c 1 4 c 1 3 1 oe c 1# , where u oe 3y and du oe 3 dy; y oe c 2 u oe c2, y oe c 3 1 4 c 1 3 1 oe c 1# 2 x4 c x# dx oe c ' 4 c " u"# du b # ' # 1 # 44. 3 "c '22 cos asec xb x x c 1 dx oe ' 3 6 cos u du, where u oe secc" x and du oe dx x x c 1 # 1 # 43. 1 "c # ' 22 sec asec xb x x c 1 dx oe ' 3 oe < " sinc" u ` 0 3 # 2 2 oe " # Ssinc" 3 sec# u du, where u oe secc" x and du oe # # 42. ds 9 c 4s oe " # 2 4 '03 # '01 4 ds 4cs s " oe <4 sinc" # ` ! oe 4 ^sinc" " # c sinc" 0 oe 4 ^ 1 c 0 oe 6 21 3 '03 2 4 du 9 c u , where u oe 2s and du oe 2 ds; s oe 0 u oe 0, s oe 2 # c sinc" 0< oe " # ^ 1 c 0 oe 4 1 8 dx x x c 1 ; x oe 2 u oe ;xoe 2 3 0 4 c " u"# du oe 2 # 1 # 4 dt t a1 b ln tb # du 1 b u , where u oe ln t and 1% tanc" ud ! oe 4 ^tanc" 1 c tanc" 0 4 1 '1e oe 4'0 # 39. ex dx 1 b e2x '0ln oe '1 3 du 1bu , where u oe ex and du oe ex dx; x oe 0 u oe 1, x oe ln 3 u oe 3 1 3 # # 38. # 1 1 ' 1c ! 37. 1 ' 2 2 cos ) d) 1 b (sin )) oe2' 1 du 1bu , where u oe sin ) and du oe cos ) d); ) oe c 1 u oe c", ) oe # 1 # uoe" oe c2 " tanc" ud c" oe 2 atanc" 1 c tanc" (c1)b oe 2 < 1 c ^c 1 ` oe 1 4 4 du 1bu 4 csc x dx 1 b (cot x) oe c' 1 , where u oe cot x and du oe c csc# x dx; x oe 1 3 1 6 u oe 3 , x oe 1 4 uoe1 oe 1 1# c 1 4 oe " t 1 1# du oe dt; t oe 1 u oe 0, t oe e 4 uoe 1 4 3 2 4 uoe 3 2 # 1 4 ,xoe2 uoe 1 3 uoe 1 6 ,xoe2 uoe 1 3 2 # u oe c 2 2 3 u oe c 2 ' 4 " # u"# du oe ' 4 u"# du 362 Chapter 5 Integration 49. Let u oe 1 b cos x du oe csin x dx cdu oe sin x dx; x oe c1 u oe 1 b cos (c1) oe 0, x oe 0 u oe 1 b cos 0 oe 2 ! ! 3 3 50. Let u oe 1 b 1 sin x du oe 1 cos x dx Because of symmetry about x oe c 1 , A oe 2 # ! oe 51. For the sketch given, a oe 0, b oe 1; f(x) c g(x) oe 1 c cos# x oe sin# x oe ! ! Aoe (" c cos 2x) # dx oe " # 52. For the sketch given, a oe c 1 , b oe 1 ; f(t) c g(t) oe 3 3 53. For the sketch given, a oe c2, b oe 2; f(x) c g(x) oe 2x# c ax% c 2x# b oe 4x# c x% ; 2 & $ 54. For the sketch given, c oe 0, d oe 1; f(y) c g(y) oe y# c y$ ; ! % $ 55. For the sketch given, c oe 0, d oe 1; f(y) c g(y) oe a12y# c 12y$ b c a2y# c 2yb oe 10y# c 12y$ b 2y; oe ^ 10 c 0 c (3 c 0) b (1 c 0) oe 3 4 3 ! ! ! ! Aoe ' 1 a10y# c 12y$ b 2yb dy oe ' 1 10y# dy c 56. For the sketch given, a oe c1, b oe 1; f(x) c g(x) oe x# c ac2x% b oe x# b 2x% ; 1 & $ oe ^2 c 8 1# c " # oe2c 2 3 c " # oe 5 6 58. We want the area between the x-axis and the curve y oe x# , 0 Y x Y 1 :6?= the area of a triangle (formed by x oe 1, $ 59. AREA oe A1 b A2 A1: For the sketch given, a oe c3 and we find b by solving the equations y oe x# c 4 and y oe cx# c 2x simultaneously for x: x# c 4 oe cx# c 2x 2x# b 2x c 4 oe 0 2(x b 2)(x c 1) x oe c2 or x oe 1 so c b oe c2: f(x) c g(x) oe ax# c 4b c acx# c 2xb oe 2x# b 2x c 4 A1 oe c ! x b y oe 2, and the x-axis) with base 1 and height 1. Thus, A oe ' 1 x# dx b " (1)(1) oe ' x " b # 3 ! " " # oe " 3 b ' 2 3 a2x# b 2x c 4b dx $ # ! (formed by y oe x and y oe 1) with base 1 and height 1. Thus, A oe ' 2 S1 c # 57. We want the area between the line y oe 1, 0 Y x Y 2, and the curve y oe c Aoe ' 1 ax# b 2x% b dx oe ' x b 3 2x 5 " " c" ! ! ! Aoe ' c Aoe ' 2 a4x# c x% b dx oe ' 4x c 3 1 ay# c y$ b dy oe 1c 1c oe " # 3 sec# t dt b 2 ' 1 ' 3 ' 3 3 (1 c cos 2t) dt oe " # [tan t]c1$ b 2[t c 1$ 1$ sin 2t # ]c1$ oe 3 b 4 1 3 c 3 oe x 5 " # c# oe ^ 32 c 3 32 5 c <c 32 c ^c 32 ` oe 3 5 64 3 c 64 5 oe 320c192 15 1 y# dy c ' 1 y$ dy oe ' y " c ' y " oe 3 4 ! " " (" c 0) 3 c (" c 0) 4 oe " 3 c " 4 oe ' 1 12y$ dy b ' 1 2y dy oe < 10 y$ ` ! c < 12 y% ` ! b < 2 y# ` ! 3 4 # " " 2 oe ^ " b 2 c <c 3 b ^c 5 ` oe 3 5 2 3 b 4 5 oe 10 b 12 15 oe 22 15 x 4 , 738?= the area of a triangle x 4 < dx c " (1)(1) oe 'x c # 1c 1c 1c 1c 1 1c Aoe 3 ^ " sec# t b 4 sin# t dt oe # " # sec# t dt b 4 3 sin# t dt oe " # 3 sec# t dt b 4 1 1 1 1 1 ' 3 1 1 ' 1 ' sin u du oe [ccos u]1 oe (ccos 1) c (ccos 0) oe 2 ! 1 c cos 2x ; # 1 # ' (1 c cos 2x) dx oe " # <x c " # sin 2x ` 1 # ! oe " # [(1 c 0) c (0 c 0)] oe " # 3 sec# t c ac4 sin# tb oe sec# t b 4 sin# t; ' ' 3 ' ! 2 1 # (cos x) (sin (1 b 1 sin x)) dx oe 2 1 1c 1c Aoec ' 0 3 (sin x) 1 b cos x dx oe c ' 2 3u"# (cdu) oe 3 " 1 0 ' 2 u"# du oe <2u$# ` ! oe 2(2)$# c 2(0)$# oe 2&# # ' du oe cos x dx; x oe c 1 u oe 1 b 1 sin ^c 1 oe 0, x oe 0 u oe 1 # # ' 1 # " (sin u) ^ 1 du ' 3 3 (" c cos 2t) # 41 3 dt oe 128 15 " 1# " " # x 1# " ! c " # " # oe 5 6 Section 5.6 Substitution and Area Between Curves # 363 oe ' 2x b 3 $ 2x # c 4x" c# c$ oe ^c 16 b 4 b 8 c (c18 b 9 b 12) oe 9 c 3 " c# 16 3 oe 11 3 ; A2: For the sketch given, a oe c2 and b oe 1: f(x) c g(x) oe acx# c 2xb c ax# c 4b oe c2x# c 2x b 4 2 $ oec2 c1b4c 3 Therefore, AREA oe A1 b A2 oe 60. AREA oe A1 b A2 A1: For the sketch given, a oe c2 and b oe 0: f(x) c g(x) oe a2x$ c x# c 5xb c acx# b 3xb oe 2x$ c 8x 2 A2: For the sketch given, a oe 0 and b oe 2: f(x) c g(x) oe acx# b 3xb c a2x$ c x# c 5xb oe 8x c 2x$ 0 % # A2 oe ' 2 a8x c 2x$ b dx oe ' 8x c 2 2x 4 Therefore, AREA oe A1 b A2 oe 16 61. AREA oe A1 b A2 b A3 A1: For the sketch given, a oe c2 and b oe c1: f(x) c g(x) oe (cx b 2) c a4 c x# b oe x# c x c 2 2 A2: For the sketch given, a oe c1 and b oe 2: f(x) c g(x) oe a4 c x# b c (cx b 2) oe c ax# c x c 2b 1 A3: For the sketch given, a oe 2 and b oe 3: f(x) c g(x) oe (cx b 2) c a4 c x# b oe x# c x c 2 2 # $ A3 oe ' 3 ax# c x c 2b dx oe ' x c 3 11 6 x # 9 # c 2x" oe ^ 27 c 3 # 9 # Therefore, AREA oe A1 b A2 b A3 oe 62. AREA oe A1 b A2 b A3 b b ^9 c 2 for x: 0 0 oe " 3 (8 c 4) oe 4 ; 3 x 3 A3: For the sketch given, a oe 2 and b oe 3: f(x) c g(x) oe S x c x< c 3 2 % oe " 3 ax$ c 4xb " 3 A3 oe " 3 ' 3 ax$ c 4xb dx oe " 3 ' x c 2x# " oe 4 # 4 3 $ " 3 <^ 81 c 2 9 c ^ 16 c 8` oe 4 4 oe 19 4 ^ 81 c 14 oe 4 25 12 ; Therefore, AREA oe A1 b A2 b A3 oe b 4 3 b 25 12 oe 32b25 1# % $ $ f(x) c g(x) oe x 3 " " c S x c x< oe c 3 ax$ c 4xb A2 oe c 3 3 $ $ x 3 cxoe x 3 x 3 c xoe0 4 3 x 3 (x c 2)(x b 2) oe 0 x oe c2, x oe 0, or x oe 2 so b oe 2: ' 2 ax$ c 4xb dx oe $ A2: For the sketch given, a oe 0 and we find b by solving the equations y oe % c A1 oe " 3 ' 0 ax$ c 4xb dx oe " 3 ' x c 2x# " 4 ! c# " oe 0 c 3 (4 c 8) oe 4 ; 3 x 3 c x and y oe " 3 $ A1: For the sketch given, a oe c2 and b oe 0: f(x) c g(x) oe S x c x< c 3 $ # $ c A2 oe c ' 2 ax# c x c 2b dx oe c ' x c 3 # $ c A1 oe c ' 1 ax# c x c 2b dx oe ' x c 3 x # # % c A1 oe ' 0 a2x$ c 8xb dx oe ' 2x c 4 c A2 oe c ' 1 a2x# b 2x c 4b dx oe c ' 2x b x# c 4x" 3 16 3 oe c ^ 2 b 1 c 4 b ^c 16 b 4 b 8 3 3 b 4 b 8 oe 9; 11 3 b9oe 38 3 8x # " ! c# # ! oe 0 c (8 c 16) oe 8; " oe (16 c 8) oe 8; c 2x" c" c# oe ^c " c 3 # c" " # b 2 c ^c 8 c 3 4 # 4 2 b 4 oe 7 3 c " # oe 14c3 6 oe " # 1" 6 ; x # c 2x" $ oe c ^8 c 3 9 # c 4 b ^c 1 c 3 4 2 1 2 b 2 oe c3 b 8 c oe 9; # 8 c 6 c ^ 3 c 5 6 c 4 oe 9 c 9 # 8 c 3; c 8 oe 9 c 3 oe 49 6 x 3 oe x 3 c4xoe 3 " 3 ax$ c 4xb x 3 simultaneously " 3 # ! ' 2 a4x c x$ b oe '2x# c x 4 " 364 Chapter 5 Integration 63. a oe c2, b oe 2; f(x) c g(x) oe 2 c ax# c 2b oe 4 c x# 2 $ 8 oe 2 ^ 24 c 3 oe 3 64. a oe c1, b oe 3; f(x) c g(x) oe a2x c x# b c (c3) oe 2x c x# b 3 1 oe ^9 c 27 3 b 9 c ^1 b 1 3 c 3 oe 11 c " 3 65. a oe 0, b oe 2; f(x) c g(x) oe 8x c x% A oe & ' 2 0 a8x c x% b dx oe 48 5 oe ' 8x c 2 # x 5 " oe 16 c ! # 32 5 oe 80 c 32 5 66. Limits of integration: x# c 2x oe x x# oe 3x x(x c 3) oe 0 a oe 0 and b oe 3; f(x) c g(x) oe x c ax# c 2xb oe 3x c x# 0 oe 27 # c9oe 27 c 18 # oe 9 # 67. Limits of integration: x# oe cx# b 4x 2x# c 4x oe 0 2x(x c 2) oe 0 a oe 0 and b oe 2; f(x) c g(x) oe acx# b 4xb c x# oe c2x# b 4x 0 oe c 16 b 3 16 # oe c32 b 48 6 oe 8 3 68. Limits of integration: 7 c 2x# oe x# b 4 3x# c 3 oe 0 3(x c 1)(x b 1) oe 0 a oe c1 and b oe 1; f(x) c g(x) oe a7 c 2x# b c ax# b 4b oe 3 c 3x# 1 " oe 3 <^1 c " c ^c1 b 3 ` oe 6 ^ 2 oe 4 3 3 $ c Aoe ' 1 a3 c 3x# b dx oe 3 'x c x 3 " " c" # $ Aoe ' 2 ac2x# b 4xb dx oe ' c2x b 3 $ # Aoe ' 3 a3x c x# b dx oe ' 3x c 2 x 3 " $ ! $ c Aoe ' c Aoe ' 2 a4 c x# bdx oe '4x c 32 3 x 3 " # c# oe ^8 c 8 c ^c8 b 8 3 3 3 a2x c x# b 3b dx oe 'x# c x 3 b 3x" oe 32 3 $ c" 4x 2 " # ! Section 5.6 Substitution and Area Between Curves 69. Limits of integration: x% c 4x# b 4 oe x# x% c 5x# b 4 oe 0 ax# c 4b ax# c 1b oe 0 (x b 2)(x c 2)(x b 1)(x c 1) oe ! x oe c2, c1, 1, 2; f(x) c g(x) oe ax% c 4x# b 4b c x# oe x% c 5x# b 4 and g(x) c f(x) oe x# c ax% c 4x# b 4b oe cx% b 5x# c 4 365 b ' 2 1 acx% b 5x# c 4bdx c# $ & $ oe ^" c 5 5 3 b 4 c ^ 32 c 5 60 3 40 3 " b 8 b ^ 5 c 5 3 " b 4 c ^c 5 b c" oe c 60 b 5 oe 300 c 180 15 oe8 70. Limits of integration: xa# c x# oe 0 x oe 0 or a# c x# oe 0 x oe 0 or a# c x# oe 0 x oe ca, 0, a; c Aoe oe oe " # " 3 ' 0 a cxa# c x# dx b $# ! ' a 0 xa# c x# dx $# a ' 2 aa# c x# b 3 aa# b $# " ca c " ' 2 aa# c x# b # 3 $ " ! c 'c " aa# b 3 $# "oe 2a 3 71. Limits of integration: y oe kxk oe J cx, x Y 0 and x, x 0 5y oe x b 6 or y oe x b 6 ; for x Y 0: cx oe x b 6 5 5 5 5 5cx oe x b 6 25(cx) oe x# b 12x b 36 x# b 37x b 36 oe 0 (x b 1)(x b 36) oe 0 x oe c1, c36 (but x oe c36 is not a solution); for x 0: 5x oe x b 6 25x oe x# b 12x b 36 x# c 13x b 36 oe 0 (x c 4)(x c 9) oe 0 x oe 4, 9; there are three intersection points and Aoe oe ^ 36 10 c 25 10 c 2 3 b c" 2 ^ 100 c 3 10 4 $# ! c 36 10 b 0 b 2 ^3 9 $# c 225 10 c 72. Limits of integration: x# c 4, x Y c2 or x 2 y oe kx# c 4k oe oe 4 c x# , c2 Y x Y 2 2x c 8 oe x b 8 x oe 16 x oe ,, 4; x# oe 0 x oe 0; by symmetry of the graph, # # # # for c2 Y x Y 2: 4 c x# oe x # b 4 8 c 2x# oe x# b 8 # for x Y c2 and x 2: x# c 4 oe x 2 b4 # # oe ' (x b 6) b 2 (cx)$# " 10 3 # c ' 0 1 ^ x b 6 c cxdx b 5 ! ' 4 0 ^ x b 6 c xdx b 5 % ' 9 4 ^ x c b ' (x b 6) c 2 x$# " b ' 2 x$# c 10 3 3 $ & oe 'c x b 5 & 5x 3 c 4x" c" b 'x c 5 5x 3 b 4x" c c Aoe c 2 ' 1 acx% b 5x# c 4bdx b ' 1 1 ax% c 5x# b 4bdx " x b ' c5 b 5x 3 c 4x" 5 3 # " 40 3 " c 8 c ^c 5 b 5 3 c 4 b ^c 32 b 5 c 4 xb6 5 dx * % (x b 6) 10 " 2 3 4$# b 100 10 oe c 50 b 10 20 3 oe 5 3 366 Chapter 5 Integration ! $ $ # # Aoe2 ' 2 0 'S x b 4< c a4 c x# b"dx b 2 2 64 6 ' 4 2 'S x b 4< c ax# c 4b"dx oe 2 ' x " b 2 '8x c 2 2 56 3 # x 6 " % # oe 2 ^ 8 c 0 b 2 ^32 c # 8 c 16 b 6 oe 40 c oe 64 3 73. Limits of integration: c oe 0 and d oe 3; f(y) c g(y) oe 2y# c 0 oe 2y# 0 $ Aoe ' 3 2y# dy oe ' 2y " oe 2 9 oe 18 3 ! $ 74. Limits of integration: y# oe y b 2 (y b 1)(y c 2) oe 0 c oe c1 and d oe 2; f(y) c g(y) oe (y b 2) c y# 1 " oe ^4 b 4 c 8 c ^" c 2 b 3 oe 6 c # 3 # 8 3 c " # b2c 75. Limits of integration: 4x oe y# c 4 and 4x oe 16 b y y# c 4 oe 16 b y y# c y c 20 oe 0 (y c 5)(y b 4) oe 0 c oe c4 and d oe 5; oe oe " 4 " 4 ^c 125 b 3 ^c 189 b 3 76. Limits of integration: x oe y# and x oe 3 c 2y# y# oe 3 c 2y# 3y# oe 3 3(y c 1)(y b 1) oe 0 c oe c1 and d oe 1; f(y) c g(y) oe a3 c 2y# b c y# " c" " 3 oe 3 2 ^1 c 77. Limits of integration: x oe cy# and x oe 2 c 3y# cy# oe 2 c 3y# 2y# c 2 oe 0 2(y c 1)(y b 1) oe 0 c oe c1 and d oe 1; f(y) c g(y) oe a2 c 3y# b c acy# b oe 2 c 2y# oe 2 a1 c y# b 1 " oe 2 ^ 1 c " c 2 ^c 1 b 3 oe 4 ^ 2 oe 3 3 $ c Aoe2 ' $ oe 3 'y c y 3 " " oe 3 ^1 c " c 3 ^c1 b 3 3 oe4 1 a1 c y# b dy oe 2 'y c y 3 c oe 3 c 3y# oe 3 a1 c y# b A oe 3 # $ oe " 4 'c y b 3 c Aoe " 4 ' 5 4 acy# b y b 20b dy c% 25 " b 100 c 4 ^ 64 2 3 9 b 180 oe 243 2 8 y # b 20y" & b 16 # ' # # b f(y) c g(y) oe ^ 16 4 y c S y c4 4 < oe cy b y b 20 4 c 80 1 1 a1 c y# b dy " 8 3 " c" $ # c Aoe ' 2 ay b 2 c y# b dy oe ' y b 2y c # y 3 " # c" " 3 oe 9 # Section 5.6 Substitution and Area Between Curves 78. Limits of integration: x oe y#$ and x oe 2 c y% y#$ oe 2 c y% c oe c1 and d oe 1; f(y) c g(y) oe a2 c y% b c y#$ c 367 Aoe oe '2y c ' 1 1 ^2 c y% c y#$ dy " c" " 5 oe ^2 c " c 3 c ^c2 b 5 5 oe 2 ^2 c " c 3 oe 12 5 5 5 79. Limits of integration: x oe y# c 1 and x oe kyk 1 c y# y# c 1 oe kyk 1 c y# y% c 2y# b 1 oe y# a1 c y# b y% c 2y# b 1 oe y# c y% 2y% c 3y# b 1 oe 0 a2y# c 1b ay# c 1b oe 0 2y# c 1 oe 0 or y# c 1 oe 0 y# oe " # are not solutions y oe ,, 1; for c1 Y y Y 0, f(x) c g(x) oe cy1 c y# c ay# c 1b Substitution shows that oe 1 c y# c y a1 c y# b 0 1 0 80. AREA oe A1 b A2 Limits of integration: x oe 2y and x oe y$ c y# y$ c y# oe 2y y ay# c y c 2b oe y(y b 1)(y c 2) oe 0 y oe c1, 0, 2: for c1 Y y Y 0, f(y) c g(y) oe y$ c y# c 2y 1 for 0 Y y Y 2, f(y) c g(y) oe 2y c y$ b y# A2 oe ^4 c 2 oe 0 c ^" b 4 " 3 c 1 oe 5 12 ; 16 4 8 8 b 3 c 0 oe 3; 5 12 Therefore, A1 b A2 oe b 8 3 oe 37 12 81. Limits of integration: y oe c4x# b 4 and y oe x% c 1 x% c 1 oe c4x# b 4 x% b 4x# c 5 oe 0 ax# b 5b (x c 1)(x b 1) oe 0 a oe c1 and b oe 1; f(x) c g(x) oe c4x# b 4 c x% b 1 oe c4x# c x% b 5 1 oe ^c 4 c 3 " 5 b 5 c ^ 4 b 3 " 5 c 5 oe 2 ^c 4 c 3 " 5 & $ c Aoe ' 1 ac4x# c x% b 5b dx oe 'c 4x c 3 x 5 $ % '! a2y c y$ b y# b dy oe 'y# c y4 $ % c A1 oe ' $ oe 2 'y c y 3 " c" 0 ay$ c y# c 2yb dy oe ' y c 4 #$ # ! b 2 ^ " " 2 a1 c3y b # c 1 c Aoe2 ' '1 c y# c y a1 c y# b"# " dy "# oe 2' a1 c y# b dy c 2 ' y a1 c y# b dy 0 1 & y 5 c 3 y&$ " 5 b 3 5 or y# oe 1 y oe ,, ,, 2 # 2 # or y oe ,, 1. "# , and by symmetry of the graph, c ! c" " oe 2 <(! c 0) c ^c1 b 3 ` b ^ 2 c 0 oe 2 3 y 3 c y# " ! c" b y 3 " # ! b 5x" " c" 104 15 b 5 oe 368 Chapter 5 Integration 82. Limits of integration: y oe x$ and y oe 3x# c 4 x$ c 3x# b 4 oe 0 ax# c x c 2b (x c 2) oe 0 (x b 1)(x c 2)# oe 0 a oe c1 and b oe 2; f(x) c g(x) oe x$ c a3x# c 4b oe x$ c 3x# b 4 1 $ % oe ^ 16 c 4 24 3 83. Limits of integration: x oe 4 c 4y# and x oe 1 c y% 4 c 4y# oe 1 c y% y% c 4y# b 3 oe 0 Sy c 3< Sy b 3< (y c 1)(y b 1) oe 0 c oe c1 and d oe 1 since x 0; f(y) c g(y) oe a4 c 4y# b c a1 c y% b " c" c oe c2 and d oe 2; f(y) c g(y) oe a3 c y# b c S cy < 4 2 $ # oe 3 <^2 c 8 12 c ^c 2 b 8 ` 12 85. a oe 0, b oe 1; f(x) c g(x) oe 2 sin x c sin 2x Aoe 0 oe <c2(c1) b " ` c ^c2 1 b " oe 4 # # 86. a oe c 1 , b oe 1 ; f(x) c g(x) oe 8 cos x c sec# x 3 3 3 # 1c Aoe oe S8 c 3< c Sc8 87. a oe c1, b oe 1; f(x) c g(x) oe a1 c x# b c cos ^ 1x # 1 oe ^1 c " 3 2 c 1 c ^c1 b " 3 2 2 b 1 oe 2 ^2 c 1 oe 3 $ c Aoe ' 1 ' 1 ' (2 sin x c sin 2x) dx oe <c2 cos x b 3 3 a8 cos x c sec# xb dx oe [8 sin x c tan x] c1$ 3 # 1 <1 c x# c cos ^ 1#x ` dx oe 'x c c # oe 3 S1 c y 4 < Aoe3 ' 2 S1 c y 4 < dy oe 3 'y c 16 12 # y 1# " c# oe 3 ^4 c oe 12 c 4 oe 8 cos 2x ` 1 2 ! b 3< oe 63 x 3 c 2 1 4 3 sin ^ 1#x " c 4 1 # # 3 c y# oe c y 4 # 3y 4 c3oe0 3 4 (y c 2)(y b 2) oe 0 # 84. Limits of integration: x oe 3 c y# and x oe c y 4 & $ oe '3y c 4y 3 b y 5 " oe 2^3 c c oe 3 c 4y# b y% A oe c Aoe ' 2 ax$ c 3x# b 4b dx oe ' x c 4 27 4 3x 3 b 4x" # c" 1 b 8 c ^ 4 b " c 4 oe ' 1 1 a3 c 4y# b y% b dy 4 3 " b 5 oe 56 15 1$ " c" Section 5.6 Substitution and Area Between Curves 88. A oe A1 b A2 a" oe c1, b" oe 0 and a# oe 0, b# oe 1; f" (x) c g" (x) oe x c sin ^ 1#x and f# (x) c g# (x) oe sin ^ 1#x c x by symmetry about the origin, A" b A# oe 2A" A oe 2 # 369 ' 1 0 <sin ^ 1x c x` dx # 2 oe 2 'c 1 cos ^ 1#x c x # 2 2 " oe 2 <^c 1 0 c " c ^c 1 1 c 0` # ! 4 c1 1 " 2 c oe 2 ^ 1 c " oe 2 ^ 4211 oe # 89. a oe c 1 , b oe 1 ; f(x) c g(x) oe sec# x c tan# x 4 4 1% oe ' 1 dx oe [x]c1% oe 4 90. c oe c 1 , d oe 1 ; f(y) c g(y) oe tan# y c ac tan# yb oe 2 tan# y 4 4 1% oe 2[tan y c y]c1% oe 2 <^1 c 1 c ^c1 b 1 ` 4 4 1c oe 2 asec# y c 1b A oe oe 4 ^1 c 1 oe 4 c 1 4 91. c oe 0, d oe 1 ; f(y) c g(y) oe 3 sin ycos y c 0 oe 3 sin ycos y # Aoe3 0 oe c2(0 c 1) oe 2 92. a oe c1, b oe 1; f(x) c g(x) oe sec# ^ 1x c x"$ 3 c Aoe ' 1 1 3 3 <sec# ^ 13x c x"$ ` dx oe < 1 tan ^ 13x c 4 x%$ ` " c" 6 3 1 3 3 oe S 1 3 c 3 < c ' 1 Sc3< c 3 " oe 4 4 93. A oe A" b A# Limits of integration: x oe y$ and x oe y y oe y$ y$ c y oe 0 y(y c 1)(y b 1) oe 0 c" oe c1, d" oe 0 and c# oe 0, d# oe 1; f" (y) c g" (y) oe y$ c y and f# (y) c g# (y) oe y c y$ by symmetry about the origin, 0 " oe 2 ^" c 4 oe # " # % # A" b A# oe 2A# A oe 2 1 ' 2 sin ycos y dy oe c3 < 2 (cos y)$# ` ! 3 ' 1 1 1c oe 4 1 ' 4 4 csec# x c asec# x c 1bd dx 1 4 1c Aoe 1 ' 4 4 asec# x c tan# xb dx 1 c ^c 1 oe 4 1 # 1c ' 4 4 2 asec# y c 1b dy 1# ay c y$ b dy oe 2 ' y c # y 4 " " ! ...
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This note was uploaded on 04/17/2008 for the course MA 113 taught by Professor Massman during the Spring '08 term at Rose-Hulman.

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