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Similar to Wilder ejercicios unidad 5 (18) 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)
![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)