5.5 Sustitucion - Section 5.5 Indefinite Integrals and the...

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Section 5.5 Indefinite Integrals and the Substitution Rule 323 value( q2 ); 71-80. Example CAS commands: : (assigned function and values for a, and b may vary) Mathematica 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, 2 }; f[x_] = Sin[2x] Cos[x/3] 1 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_] = x 2x 3 2 ±± u[x_] = 1 x ± 2 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; P l o t [ { F [ x ] , { x , a , b } ] 5.5 INDEFINTE INTEGRALS AND THE SUBSTITUTION RULE 1. Let u 3x du 3 dx du dx oÊo Ê o " 3 sin 3x dx sin u du cos u C cos 3x C '' oo ± ² o ± ² "" " 33 3 2. Let u 2x du 4x dx du x dx o Ê o # " 4 x sin 2x dx sin u du cos u C cos 2x C ab ## " ± ² o ± ² 44 4 3. Let u 2t du 2 dt du dt o Ê o " # sec 2t tan 2t dt sec u tan u du sec u C sec 2t C ² o ² " # 4. Let u 1 cos du sin dt 2 du sin dt o ± Ê o Ê o tt t 22 " 1 cos sin dt 2u du u C 1 cos C ˆ‰ ˆ ± o o ² o ± ² 2 2 t # #$ 5. Let u 7x 2 du 7 dx du dx o ± Ê o Ê o " 7 28(7x 2) dx (28)u du 4u du u C (7x 2) C ' ± o o o ± ² o ± ± ² ± & ± & ± & ± % ± % " 7 6. Let u x du 4x dx du x dx o ± "Ê o %$$ " 4 x x 1 dx u du C x 1 C $% # % ± o o ² o ± ² 411 u
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324 Chapter 5 Integration 7. Let u 1 r du 3r dr 3 du 9r dr o ± Ê o ± Ê ± o $# # 3u du 3(2)u C 6 1 r C ' 1r È ± ± "Î# "Î# $ "Î# o ± o ± ² o ± ± ² ' ab 8. Let u y 4y 1 du 4y 8y dy 3 du 12 y 2y dy o ² ² ² ² %# $ $ a b 12 y 4y 1 y 3u du u C y 4y 1 C '' a b $ # $ #$ ²² ² o o ² o ²²² 9. Let u x 1 du x dx du x dx o ± Ê o Ê o $Î# "Î# # 32 3 È x sin x 1 dx sin u du sin 2u C x 1 sin 2x 2 C È ˆ‰ ˆ ˆ‰ˆ Î # # $ Î # $ Î # # "" " ± o o ± ² o ± ± ± ² 22 u 33 4 3 6 10. Let u du dx o ± xx cos dx cos u du cos sin 2u C sin C ' " " " ## # # 4 2 x 4x u2 ˆ ˆ ‰ a b o ± o o ² ² o ± ² ± ² sin C o ² 2x 4 x 2 11. (a) Let u cot 2 du 2 csc 2 d du csc 2 d o ± Ê ± o )) ) ) ) " # csc 2 cot 2 d u du C C cot 2 C # # " # ) ) o ± o ± ² o ± ² o ± ² Š uu 44 (b) Let u csc 2 du 2 csc 2 cot 2 d du o ± Ê ± o ) ) ) ) ) " # u du C C csc 2 C # # " # ) ) o ± o ± ² o ±² o ± ² Š 12. (a) Let u 5x 8 du 5 dx du dx o ² Ê o Ê o " 5 du u du 2u C u C 5x 8 C ' dx 5x 8 55 5 5 5 u ÈÈ ² " " ± "Î# "Î# "Î# oo o ² o ² o Š È (b) Let u 5x 8 du (5x 8) (5) dx du o ² Ê o ² Êo È " # ± "Î# ² 2d x 5 5x 8 È du u C 5x 8 C dx 2 2 2 5x 8 5 È ² ² o ² ² È 13. Let u 3 2s du 2 ds du ds o ± ± Ê ± o " # 3 2s ds u du u du u C (3 2s) C ' È È ˆ ˆ ± o ± o ± o ± ² o ± ± ² " " # "Î# $Î# $Î# 2 14. Let u 2x 1 du 2 dx du dx o ² " # (2x 1) dx u du C (2x 1) C ' ² o o o ² o ² ² $$ $ % " " # ˆ Š u 48 15. Let u 5s 4 du 5 ds du ds o ² " 5 ds du u du 2u C 5s 4 C ' " " " ² ± "Î# "Î# 5s 4 u 5 5 2 ooo ² o ² ² ˆ ˆ È 16. Let u 2 x du dx du dx o ± ± Ê ± o dx 3 u du 3 C C ''' 3u 3 (2 x) u 1 2 x 3( du) ±± ± ± ± # ± o ± ² o ² Š 17. Let u 1 du 2 d du d o ± ± Ê ± o ) # " # 1 d u du u
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This note was uploaded on 03/19/2012 for the course ELECTRIC 311 taught by Professor Rene during the Spring '12 term at Uni San Francisco de Quito.

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5.5 Sustitucion - Section 5.5 Indefinite Integrals and the...

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