![1/ Hallar el Γ‘rea de la regiΓ³n encerradas por los grΓ‘ficos
a) ν(ν₯) = ν₯2 β 4, ν(ν₯) = ν₯ β 4
y
y= x2 - 4
1 x
Y= x- 4
νΌνν‘ννν νννννν: ν₯2 β 4 = ν β 4 β ν2 β ν = 0
ν(ν β 1) = 0
ν = 0 ; ν = 1
1
ν΄ = β« (ν₯ β 4) β ( ν₯2
0
1
β 4) νν₯ = β« ν β 4 β ν₯2 + 4 νν₯
0
1
ν΄ = β« ν β ν2
0
νν₯ =
ν₯2
2
β
ν3
3
β νν ν ν 1
ν΄ =
12
2
β
13
3
β [
ν2
2
β
ν3
3
]
ν΄ =
1
6
ν2](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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- 1. RepΓΊblica bolivariana de Venezuela Ministerio del poder popular para educaciΓ³n superior Instituto universitario de tecnologΓa Antonio JosΓ© de sucre Barquisimeto Ejercicios unidad 5 Alumno Wilder gallardo CI:24.668.002
- 2. 1/ Hallar el Γ‘rea de la regiΓ³n encerradas por los grΓ‘ficos a) ν(ν₯) = ν₯2 β 4, ν(ν₯) = ν₯ β 4 y y= x2 - 4 1 x Y= x- 4 νΌνν‘ννν νννννν: ν₯2 β 4 = ν β 4 β ν2 β ν = 0 ν(ν β 1) = 0 ν = 0 ; ν = 1 1 ν΄ = β« (ν₯ β 4) β ( ν₯2 0 1 β 4) νν₯ = β« ν β 4 β ν₯2 + 4 νν₯ 0 1 ν΄ = β« ν β ν2 0 νν₯ = ν₯2 2 β ν3 3 β νν ν ν 1 ν΄ = 12 2 β 13 3 β [ ν2 2 β ν3 3 ] ν΄ = 1 6 ν2
- 3. b) ν¦ = ν₯ 3, ν¦ = 4ν₯ ν3 = 4ν ν3 β 4ν = 0 ν (ν2 β 4) = 0 ν (ν + 2)(ν β 2) = 0 ν = 0 ; ν = 2 ; ν = β2 Y=x3 Y=4x -2 2 Y= SoluciΓ³n: ν΄ ν = 2 ν΄1 ν΄1 β 0 β€ ν β€ 2; ν3 β€ ν β€ 4ν 2 ν΄1 = β« 4ν β ν3 νν₯ 0 Integrando ν΄1 = 4 ν2 2 β ν4 4 β de 0 a 2 ν΄1 = 2 (2)2 β 24 4 = 4 ν’2
- 4. ν΄ ν = 2 β 4 ν’2 ; ν΄ ν = 8 ν’2 c) ν₯ = 12 ν¦ , ν₯ = 0, ν¦ = 1, ν¦ = ν2 ν = 12 ν ; ν = 0 ; ν = 0 ; ν = ν2 Y X=12/y y=e2 y=1 x Tipo II ν΄ = 1 β€ ν β€ ν2 ; 0 β€ ν β€ 12 ν ν΄ = β« 12 ν¦ ν2 1 νν¦ = 12 lnβ©νβͺ β νν 1 ν ν2 ν΄ = 12 [νν ν2 β ln(1)] ν΄ = 12 β 2 ν΄ = 24 ν’2
- 5. d) ν(ν₯) = tan ν₯ 2 , νν ννν ν₯ ν¦ ννν νννν‘νν ν₯ = 0, ν₯ = 1 2 ν A: Tipo I 0 β€ ν β€ ν 2 ; 0 β€ ν β€ ν‘νν ν 2 y y=tan(x/2) x=Ο/2 x ν΄ = β« ν‘νν ν 2 ν 2 0 νν₯ ; νΆννν β« ννν νΎν₯νν₯ = 1 νΎ ln ννν νΎ ν + νΆ ν΄ = 1 1 2 νΏν |ννν 1 2 ν₯| β νν 0 ν ν 2 1 cos ν΄ = 2 [νΏν | ν 4 | β Ln | 1 cos(0) |] ν΄ = 2 νΏν | 1 β2 2 2 = νΏν (2) ν’2 | = 2 νΏν 2 = νΏν β2
- 6. 2/ hallar el volumen del solido de revoluciΓ³n generado por la regiΓ³n encerrada por las curvas dadas (utilice el mΓ©todo del disco, arandelas y cortezas cilΓndricas) a) Un arco de y=cos2x, alrededor del eje x SoluciΓ³n MΓ©todo Disco 0 β€ ν β€ ν 4 ; 0 β€ ν β€ cos 2ν y y=cos2x x Ο/4 ν 4 ν = ν β« (cos 2ν)2 0 νν₯ = ν β« [ 1 + cos 4ν 2 ] ν 4 0 νν₯ ν = ν 2 ν 4 β« 0 νν₯ + ν 2 ν 4 β« cos 4ν 0 νν₯ = ν 2 ν + ν 2 ν νν 4ν 4 β νν 0 ν ν 4 ν 2 ν = [ ν 4 ( ) + ν 8 ν νν 4 ν 4 ] β [0 + ν 8 ν νν 0] = ν2 8
- 7. b) ν₯ = 4ν¦, ν₯ = 3βν¦, ννννννννν νν νν νννν‘ν ν₯ = 8 MΓ©todo Disco y 1/8 -1/8 x=8 4ν = βν 3 (4ν)3 = βν 3 3 64 ν3 = ν 64 ν3 β ν = 0 ν (64 ν2 β 1) = 0 β (ν = 0) ; ( ν = β 1 8 ) ; ( ν = 1 8 ) ν1 β β 1 8 β€ ν β€ 0 ; 4ν β€ ν β€ βν 3
- 8. 0 ν1 = ν β« (βν 3 β 8)2 1 8 β β (4ν β 8)2 νν¦ 2 3 0 ν1 = ν β« ν 1 8 β 1 3 + 64 β 16 ν2 + 64 ν β 64 νν¦ β 16 ν ν1 = ν [ 5 3 5 3 ν β 16 4 3 4 3 ν β 16 ν3 3 + 64 ν2 2 ] β νν β 1 8 ν 0 ν1 = ν {[0] β [ 3 5 1 8 β ( ) 5 β3 β 12 (β 1 8 4 β3 ) + 32 (β 1 8 )2] β β1 8 16 ( )3 3 } ν1= ν [ 3 160 + 3 4 β 1 96 β 1 2 ] = 31 120 ν ν2 β 0 β€ ν β€ 1 8 ; βν 3 β€ ν β€ 4ν 1 8 ν2 = ν β« (4ν β 8)2 β (βν 3 β 8)2 0 νν¦ 2 3 1 8 ν2 = ν β« 16 ν2 β 64 ν + 64 β ν 0 1 3 β 64 νν¦ β 16 ν 2 3 1 8 ν2 = ν β« 16 ν2 β 64 ν β ν 0 1 3 νν¦ β 16 ν ν2 = ν [16 ν3 3 β 64 ν2 2 β 5 3 5 3 ν + 16 4 3 4 3 ν ] β νν 0 ν 1 8 ν2 = ν { 1 8 16 ( )3 3 1 8 β 32 ( )2 β 3 5 1 8 ( 5 β3 ) 1 8 + 12 ( 4 β3 ) β (0)} ν2= ν [ 1 96 β 1 2 β 3 160 + 3 4 ] = 29 120 ν
- 9. νν= 31 120 ν + 29 120 ν = ν 2 ν’2 c) Hallar el volumen del sΓ³lido que se genera al rotar alrededor del eje x la elipse ν₯2 ν2 + ν¦2 ν2 = 1 Capa CilΓndricas Por SimetrΓa y b x -a a νν = 2 ν1 ννννν ν1 νν ν‘ν νννν ννν 0 β€ ν β€ ν ; 0 β€ ν β€ ν ν βν2 β ν2 ν·νν ννννννν ν: ν2 ν2 = 1 β ν2 ν2 ν2 = ν2 [ ν2 β ν2 ν2 ] ν2 ν2 (ν2 β ν2 )) = ν = β( ν ν β(ν2 β ν2) ν1 = 2ν β« ν¦ β ν ν ν 0 βν2 β ν2 νν¦ = 2 ν ν ν ν β« ν¦ 0 βν2 β ν2 νν¦ Cambio de Variable ν’ = ν2 β ν2 ; νν’ = β2ν νν¦ β β νν’ 2 = ννν¦
- 10. νν ν = ν β ν’ = 0 νν ν = 0 β ν’ = ν2 ν1 = 2 ν ν 1 2 ν ( ) β« ν’ 1 2 νν’ = 0 ν2 β ν ν ν [ 3 2 3 2 ν’ ] νν ν2 ν 0 ν½ν = β ν ν ν β 2 3 [β03 β β(ν2)3] = β ν ν ν β 2 3 [βν3] = 2 3 ν ν2ν ν½ν» = ν [ ν ν ν ν2ν] = 4 3 ν ν2ν d) Hallar el volumen del sΓ³lido que genera la regiΓ³n encerrada por. ν¦ = 4 β ν₯ 2, ννν ν₯, νν ννννν ννννννννν νν νν νννν‘ν ν₯ = 3 y x X=3 νν ν = 0 ; 4 β ν2 = 0 β ν2 = 4 β βν2 = β4 β |ν| = 2 β (ν = β2); (ν = 2) Radio ν (ν) = 3 β ν νΓ©ν‘ννν νν νννν‘νν§νν νΆννννννννν ν ν = 2 ν β« ν (ν) ν [νΉ(ν) β νΊ(ν)] νν₯ 2 ν = 2 ν β« (3 β ν) β2 [(4 β ν2) β 0] νν₯ 2 ν = 2 ν β« (3 β ν) (4 β ν2) β2 νν₯
- 11. 2 ν = 2 ν β« 12 β 3ν2 β 4ν + ν3 β2 νν₯ ν = 2 ν [12ν β 3 ν3 3 β 4 ν2 2 + ν4 4 ] νν β 2 ν 2 ν = 2 ν {[12(2) β 23 β 2(2)2 + (2)4 4 ] β [12(β2) β (β2) 3 β 2(β2)2 + (β2)4 4 ] } ν = 2 ν [12 β (β20)] ν = 64 ν ν’3 3/ Hallar la longitud de la curva dada a) ν¦ = ν₯3 6 + 1 2ν₯ , ννν νν ν₯ = 1 βνν ν‘ν ν₯ = 3 ν νΏ = β« β1 + νΉβ²ν₯2 ν νν₯ y 1 2 3 Derivando ν¦β² = 3ν₯2 6 + 1 2 β1 ν₯2 ) = ( ν₯2 2 β 1 2ν₯2 = 2ν₯4 β 2 4ν₯2 2ν₯4 β 2 4ν₯2 )2 3 νΏ = β« β1 + ( 1 νν₯
- 12. νΏ = β« β1 + 4ν₯8 β 8ν₯4 + 4 16ν₯4 3 1 νν₯ 16ν₯4 β 4ν₯8 β 8ν₯4 + 4 νΏ = β« β 16ν₯4 3 1 νν₯ νΏ = β« β4ν₯8 + 8ν₯4 + 4 4ν₯2 3 1 νν₯ νΏ = β« β(2ν₯4 + 2)2 4ν₯2 3 1 νν₯ = β« 2ν₯4 + 2 4ν₯2 3 1 νν₯ νΏ = β« 1 2 3 1 ν₯2 + 1 2 ν₯β2 νν₯ = 1 2 ν₯3 3 + 1 2 ν₯β1 β1 βνν 1 ν 3 νΏ = 1 6 ν₯3 β 1 2ν₯ β νν 1 ν 3 νΏ = [ 1 6 33 β 1 2(3) ] β [ 1 6 β 1 2 ] = 14 3 b) ν¦ = ννν ννν₯, ννν νν ν₯ = 0, βνν ν‘ν ν₯ = ν 3 ν¦ = ln(sec ν₯) ν·νν νν ν₯ = 0 βνν ν‘ν ν₯ = ν 3 y ν 3 νΏ = β« β1 + ((ln sec ν₯)β²)2 0 νν₯
- 13. νΏ = β« β1 + [ 1 sec ν₯ β (ννν ν₯ β ν‘νν₯)] 2 ν 3 0 νν₯ ν 3 νΏ = β« β1 + ν‘ν2ν₯ 0 νν₯ ν 3 νΏ = β« βsec 2 ν₯ 0 νν₯ ν 3 νΏ = β« sec 2 ν₯ 0 νν₯ = νν|ννν ν₯ β ν‘νν₯| νν ν ν ν 3 1 ννν νΏ = νν | ν 3 + ν νν ν 3 ννν ν 3 | β ln | 1 ννν 0 + ν νν 0 ννν 0 | νΏ = ln | 1 1 2 + β3 21 2 | β ln(1) νΏ = ln(2 + β3)