Title | Tarea Calificada 2 |
---|---|
Course | MatemΓ‘tica para Ingenieros 2 |
Institution | Universidad TecnolΓ³gica del PerΓΊ |
Pages | 6 |
File Size | 306.8 KB |
File Type | |
Total Downloads | 170 |
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TAREA CALIFICADA 2 β ECV TAREA: Determine las coordenadas del centro de gravedad de la regiΓ³n limitada por π¦ = 2π₯ β π₯2 , π¦ = 6π₯ π¦ = 2π₯ β π₯2 π¦ = 6π₯ β 2x β π₯2 π₯2 + 4π₯ = 0 X = 0; X = -4 0 (2π₯ β π₯ 2 ) β 6π₯ ππ₯ A = β«β4 0 = β«β4 β π₯ 2 β 4π₯ ππ₯ βπ₯3 0 2 = 3 β 2π₯ β 64 β4 =β{ 3 32 β 32 = 3 0 1 2 = 32 β«β4π₯ {2π₯ β ...
TAREA:
TAREA CALIFICADA 2 β ECV
Determine las coordenadas del centro de gravedad de la regiΓ³n limitada por π¦ = 2π₯ β π₯2 , π¦ = 6π₯
π¦ = 2π₯ β π₯2 π¦ = 6π₯ β 2x β π₯2 π₯2 + 4π₯ = 0 X = 0; X = -4
0 (2π₯ β π₯2 ) β 6π₯ππ₯ A = β«β4
0
= β«β4 β π₯2 β 4π₯ ππ₯ βπ₯3
0
2
ο²
= 3 β 2π₯
64
β4
=β{
3
32
β 32 = 3
2 = 32 β«β4π₯ {2π₯ β π₯ β 6π₯}ππ₯ = 0
1 3
3
=
4
π₯4
{β
32
4
-
π₯3 ο²
3
0
=
β4
3
32
3
32
{β
3 2 β«β4βπ₯ β 4π₯ ππ₯
64
0
} = β2
4
2 2 2 3 Θ³ = 2(132 β«β4 {2π₯ β π₯ ο² β {6π₯ο² ππ₯ = 64 0
3
=
3 64
π₯ 5
5
{
- π₯ 4β
32 3
β΄ ( , Θ³) = ( -2, β 552
π₯3 ο²
0 = 3{β 64 β4
3328
} 15
π₯4 β 4π₯3 β 32π₯2ππ₯ β«β4 0
=β
52 5
1. Determine el Γ‘rea de la regiΓ³n limitada por las curvas y = x3 β 7x + 6 e y = 6 β3x π¦ = π₯3 β 7π₯ + 6 π¦ = 6 β 3π₯ β π₯3 β 7x + 6 = 6 β 3x π₯3 β 4π₯ = 0 β π₯ (π₯2 β 4) = 0 X = -2; X = 0; X=2 A = β«0 (π₯3 β 7π₯ + 6) β (6 β 3π₯) ππ₯ + β«2(6 β 3π₯) β (π₯3 β 7π₯ + 6)ππ₯ β2
= β«0 π₯ β 4π₯ ππ₯ + β« 3
β2
= 2 { 2 π₯2 β
0
2(4π₯
β π₯ )ππ₯ = 2 β«2 4π₯ β π₯3ππ₯ 3
0 4
π₯
4
2 ο² =2{8β4} 2 8π’ 0
0
2. Calcule el Γ‘rea de la regiΓ³n limitada por la cardioide π = 1 + π ππ π πππ π y la circunferencia π = 3π π ππ πππ π 1 + π πππ = 3 π πππ 1 = 2 π πππ 1 = π πππ 2 π = π/6 Por SimetrΓa
1 π/2 2 2 π΄π‘ = 2 ( β« (3π πππ ) β (1 + π πππ ) ππ ) 2 π/6 π/2
π΄π‘ = β«
π/6
9π ππ2π β 1 β 2π πππ β π ππ2π ππ
π/2
π΄π‘ = β«
π/6
8π ππ2π β 2π πππ β 1 ππ
π/2
π΄π‘ = β«
π/6
3 β 2π πππ β 4πππ 2π ππ
π΄π‘ = [ 3π + 2πππ π β 2π πππ]πβ2 π π π 2β3 2β3 ) π΄π‘ = 3 ( ) β (3 ( ) + β 2 6 2 2
π π π΄π‘ = 3 ( ) β 2 2 π΄π‘ = π π’2
3. Calcule la longitud del arco de la cuerda definida por la funciΓ³n π±+πβπ± π²= π , y las rectas x = 1 , x=3 π
ππ¦ ππ₯
1 =
π₯
βπ₯
(π β π
2
3
πΏ=β« 1
β 1+ (
)
ex β eβπ₯ 2 2
) ππ₯
e2x β 2 + eβ2π₯ πΏ = β« β1 + ) ππ₯ 4 ( 3
1 3
πΏ=β«
β
e2x + 2 + eβ2π₯ ππ₯
4
1 3
πΏ= β«
β
ex + eβπ₯ 2 ) ππ₯
2
( 1
ex + eβπ₯ ππ₯ πΏ=β« 2 1 3
1
3 x
βπ₯
πΏ = β« (e + e 2 1 1 x βπ₯
) ππ₯ 3 1
πΏ=
1 2
3
β3
(e β e )
β1
πΏ= πΏ=
2
[(e β e ) β (π β π )] 1 (17.68)
2 πΏ = 8.84 π’
π₯ , π¦ = 3βπ₯ π₯y 4. Calcule el Γ‘rea de la regiΓ³n definida por las curvas π¦ = βπ₯ las rectas verticales x = 1 , x = 64 64
π΄ = β« ( βπ₯ β βπ₯ ) ππ₯ 3
1
64
π΄= β«
1
(π₯2
1 3 π₯ )
ππ₯
3 π₯43 )
64
β
1
π΄= (
2π₯32 3
β
4
3
1
3
2(β64) 3( β64)4 2 3 β π΄= ( ) β( β ) 3 4 3 4
2 β 512 3 β 256 1 ) β ) β (β π΄= ( 12 3 4 448 1 + π΄= 3 12 π΄=
1793 12
π΄ = 149.42 π’2...