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Integrales impropias
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Cristian Estevez
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Cálculo diferencial e integral
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Integrales impropias
1.
UNIVERSIDAD TÃCNICA DEL
NORTE FACULTAD DE INGENIERIAS EN CIENCIAS APLICADAS INGENIERÃA TEXTIL CÃLCULO DE UNA VARIABLE INTEGRALES IMPROPIAS NOMBRE: Cristian Estévez DOCENTE: MSc. Luis Chamorro FECHA: 27 de Enero del 2018 TEMA: Integrales Impropias Ibarra, 2017 â 2018
2.
En los ejercicios
15 a 32, determinar si la integral impropia es divergente o convergente. Evaluar la integral si es convergente. 15. â« ð ð ð â ð ð ð ðŠ = 1 ð¥2 lim ðââ â« ðð¥ 1 ð 1 x y lim ðââ [â 1 ð¥ ]b 1 1 1 lim ðââ [â 1 ð â (â 1 1 )] 2 0.25 ðŽ = [â 1 â + 1] 3 0.11 ðŽ = 1 ð¢2 Convergente 4 0.06
3.
16. â« ð ð ð â ð ð ð
ðŠ = ð ð ð lim ðââ 5â« ð ð ð ð ð ð x y lim ðââ [â 5 2ð¥2 ]b 1 1 5 lim ðââ [â 5 2ð2 â (â 5 2(12) )] 2 0,625 A = [â 5 2â2 + 2,5] 3 0,185 A = 2,5 ð¢2 Converge 4 0,078
4.
17. â« ð â ðð â ð ð ð
ðŠ = 3 â ð¥3 lim ðââ â« 3 â ð¥3 ð 1 ðð¥ x y lim ðââ 3 â« ð¥ â 1 3 ð 1 ðð¥ 1 9 lim ðââ [ 9 2 âð¥23 ]b 1 2 2,38 lim ðââ [ 9 âð¥23 2 â 9 âð¥23 2 ]b 1 3 2,08 lim ðââ [ 9 âð23 2 â 9 â123 2 ] 4 1,88 ðŽ = [ 9ââ23 2 â 4,5] ðŽ =[ ðŒðð· â 4,5] lim ðââ [ 9 âð23 2 â 9 â123 2 ] lim ðââ [ 18 6 â ð3 â 9 â123 2 ] lim ðââ [ 18 6 ââ3 â 9 â123 2 ] |ðŽ| = 4,5ð¢2 Converge
5.
18. â« ð â ðð â ð ð ð
ð = ð â ðð lim ðââ 4 â« ð¥ â 1 4 ð 1 ðð¥ x y lim ðââ [ 4ð¥ 3 4 3 ]b 1 1 4 lim ðââ [ 4ð 3 4 3 â 4(1) 3 4 3 ] 2 3,36 lim ðââ [ 4(â) 3 4 3 â 4(1) 3 4 3 ] 3 3,30 lim ðââ [ ðŒðð· â 4(1) 3 4 3 ] 4 2,82 lim ðââ [ 4ð 3 4 3 â 4(1) 3 4 3 ] lim ðââ [ 12 12ð 1 4 â 4(1) 3 4 3 ] ðŽ = [ 12 12(â) 1 4 â 4 â134 3 ] ðŽ = 1.33ð¢2 Converge
6.
19. â« ððâðð ð ð ð ââ lim ðâââ â«
ð¥ðâ2ð¥ ðð¥ 0 ð ð¢ = ð¥ ðð¢ = ðð¥ â« ðð£ = â« ðâ2ð¥ ðð¥ ð£ = â ðâ2ð¥ 2 lim ðâââ â ð¥ðâ2ð¥ 2 â â« â 0 ð ðâ2ð¥ 2 ðð¥ lim ðâââ [â ð¥ðâ2ð¥ 2 â ðâ2ð¥ 4 ]0 a lim ðâââ [â ðâ2ð2 2 â ðâ2ð 4 ] â [â (0)ð0 2 â ð0 4 ] ðŽ =[â + 1 4 ] ðŽ = â
7.
20. â« ð¥ð â ð¥ 2 â 0 ðð¥ lim ðââ â«
ð¥ð â ð¥ 2 ð 0 ðð¥ lim ðââ [â2ð â ð¥ 2( ð¥ + 2)]b 0 lim ðââ [â 2( ð + 2) ð ð 2 + 2(0 + 2) ð 0 2 ] ðŽ = [ â â + 4] lim ðââ [â 2( ð + 2) ð ð 2 + 4] lim ðââ [â 4 ð ð 2 + 4] ðŽ = [ 4 â + 4] ðŽ = 4 Converge
8.
21. â« ð¥2â 0 ðâð¥ ðð¥ lim ðââ â«
ð¥2 ð 0 ðâð¥ ðð¥ lim ðââ [â ðâð¥ ( ð¥2 +2ð¥+2)]b 0 lim ðââ [â ðâð ( ð 2 +2ð +2) + ðâ0 (0 2 +2(0)+2)] lim ðââ [â â2 +2â+2 ðâ +2] lim ðââ [â â â+2] lim ðââ [ðŒðð· +2] lim ðââ [ ðâð (2ð 2 +2ð+2) +2] lim ðââ [ ðâð ð 2 +2] lim ðââ [ðŒðð· +2] lim ðââ [ ðâð ð 2 +2] lim ðââ [â ðâð(ðâ2)+2] ðŽ = 2 Converge
9.
22. â« (
ð â ð) ðâðâ ð ð ð lim ðââ â« ( ð¥ â 1) ðâð¥ð 0 ðð¥ lim ðââ â« ð¥ðâð¥ð 0 ðð¥ â â« ðâð¥ð 0 ðð¥ lim ðââ [ ðð¥2 2 â ð¥2 2 â ðð¥]b 0 lim ðââ [ ðð2 2 â ð2 2 â ðð¥ â ð(0)2 2 + 02 2 + ð0] ðŽ = [ â 2 â â 2 â â] ðŽ = â 23. â« ðâðâ ð ððšð¬ ð ð ð lim ðââ â« ðâð¥ð 0 cos ð¥ ðð¥ lim ðââ [ 1 2 ðâð¥(sin ð¥ â cos ð¥)]b 0 lim ðââ [ 1 2 ðâð(sin ð â cos ð) â 1 2 ð0(sin 0 â cos0)] lim ðââ [ 1 2 ðââ(sinâ â cosâ) + 1 2 ] ðŽ = [ sin ââcosâ ðâ + 1 2 ] ðŽ = 1 2 Converge
10.
24. â« ðâððâ ð ð¬ð¢ð§
ð ð ð, ð > ð lim ðââ â« ðâðð¥ ð 0 sin ð¥ ðð¥ lim ðââ [â ðâðð¥(ðð ððð¥+ððð ð¥) ð2+1 ]b 0 lim ðââ [â ðâðð( ðð ððð+ððð ð) ð2+1 + ðâð0(ðð ðð0+ððð 0) ð2+1 ] lim ðââ [â ðâðð( ðð ððð+ððð ð) ð2+1 + 1 ð2+1 ] ðŽ = [â ( ðð ððâ+ððð â) (ð2+1)(ð ðâ) + 1 ð2+1 ] ðŽ = 1 ð2+1 25. â« ð ð(ð°ðð) ð â ð ð ð lim ðââ â« 1 ð¥(ðŒðð¥)3 ð 4 ðð¥ lim ðââ [â 1 2 ðð2 ð¥ ]b 4 lim ðââ [â 1 2 ðð2 ð + 1 2 ðð2(4) ] ðŽ = [â 1 2 ðð2â + 1 2 ðð2(4) ] ðŽ = 1 2 ðð2(4)
11.
ðŽ = 1 2 ðð2(4) ðŽ
= 7,68 ð¢2 Converge 26. â« ððð ð â ð ð ð lim ðââ â« ððð ð ð ð ð ð lim ðââ [ ðð2 ð¥ 2 ]b 1
12.
lim ðââ [ ðð2 ð 2 â ðð2 1 2 ] ðŽ = [ â 2 ] ðŽ
= â Divergente
13.
27.â« ð ð+ð ð â ââ ð ð lim ðâââ 2â« ð ð ð+ð ð ð ð
+ lim ðââ 2â« ð ð ð+ð ð ð ð lim ðâââ 2â« ð ð ð ð+ð ð ð ð + lim ðââ 2â« ð ð ð ð+ð ð ð ð lim ðâââ [ 1 ð ððð ð¡ð ð£ ð ]0 a +lim ðââ [ 1 ð ððð ð¡ð ð£ ð ]b 0 lim ðâââ [ 1 2 ððð ð¡ð ð¥ ð ]0 a + lim ðââ [ 1 2 ððð ð¡ð ð¥ ð ]b 0 lim ðâââ [ 1 2 ððð ð¡ð 0 ð â 1 2 ððð ð¡ð ð ð ] +lim ðââ [ 1 2 ððð ð¡ð ð ð â 1 2 ððð ð¡ð 0 ð ] ðŽ = [â 1 2 ððð ð¡ð ââ ð + 1 2 ððð ð¡ðâ] ðŽ = ð 4 ðŽ = ð 8 Converge
14.
28. â« ð ð (ð
ð+ð) ð â ð ð ð lim ðââ â« ð¥3 (ð¥2 + 1)2 ð 0 ðð¥ lim ðââ [ 1 2 ( 1 ð¥2+1 + ln(ð¥2 + 1))]b 0 lim ðââ [ 1 2 ( 1 ð2 + 1 + ln( ð2 + 1) â 1 2 ( 1 02 + 1 + ln(02 + 1)))] ðŽ = [ 1 2â2 + 1 + ln(â2 + 1) â 1 2 ] ðŽ = ln â â 1 2 ðŽ = ð 2 â 1 2 ðŽ = ð â 1 2 ðŽ = 1.07ð¢2 Converge
15.
29. â« ð ð ð+ðâð â ð ð ð lim ðââ â« 1 ð
ð¥+ðâð¥ ð 0 ðð¥ lim ðââ [ ððð ð¡ð (ð ð¥ )]b 0 lim ðââ [ððð ð¡ð (ð ð ) â ððð ð¡ð (ð0 )] ðŽ = [ ððð ð¡ð ( ðâ) â ððð ð¡ð (ð0 )] ðŽ = â â 1 ðŽ = ð 2 â 1 ðŽ = 0,57 ð¢2 Converge
16.
30. â« ð ð ð+ð
ð ð ð â ð lim ðââ â« ð ð¥ 1+ð ð¥ ðð¥â 0 lim ðââ [ln ( ð ð¥ +1)]b 0 lim ðââ [ln ( ð ð +1) â ln ( ð0 +1)] lim ðââ [ln ( ðâ +1) â ln(2)] ðŽ = â â ð 5 ðŽ = ð 2 â ð 5 ðŽ = 3ð 10 ðŽ = 0,94 ð¢2 ð¶ððð£ðððð 31. â« ððšð¬ ð ð ð ð â ð â« ð ð¥ 1 + ð ð¥ ðð¥ â 0 lim ðââ â« cos ðð¥ ðð¥ ð 0 lim ðââ [ 1 ð ð¥ ]b 0
17.
lim ðââ [ 1 ð ð â 1 ð ð ] ðŽ
= [â 1 â + 1] ðŽ =1
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