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E J
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R C
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C I
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S
DE
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A
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A T
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C A
1. F(X)=X2-4 g(X)=X-4
X2-4=X-4
X2-4-X+4=0
X2-X=0
X(X-1)=0
X=0 X=1
0.5 (X4) = X2-4
g(X)=X- 4
- 4
b. Y = X3, Y = 4X
X3
= 4X
X3
- 4X = 0
X (X2
- 4) = 0
X = 0 X = -2 X = 2
Y = X3
0 2
A = ʃ (X3 –
4X)dx + ʃ ( 4X - X
3)dx
-2 0
2
A = 2 ʃ ( 4X - X
3)dx
0
2
A = 2 [ 4X² - X4
]
2 4 0
2
A = 2 [ 2X² - X4
]
4 0
2
A = 2 [ (2)² - (2)4
]
4 0
A = 2 [ 8 - 4 ]
c) X = 12 , X = 0, Y = 1, Y = е2
Y GRAFICO
A = 8 Ʋ2
е2
A = ʃ 12 dy
1 Y
е2
A = 12 ʃ dy
1
Y
е2
12 Ln Y
1
12 [Ln е2 - Ln 1]
12 [ 2 Ln е]
d) f (x) = tn x, el eje x y las rectas x = 0, x = 1 π
2 2
GRAFICO
A = 24 Ʋ2
π/2
A = ʃ tn (X/2)dx
0
Cambio de variable
μ = X/2
dμ = dx
2
2 dμ = dx
A = 2 ʃ tn μ dμ
A = 2 Ln [ sec μ ]
Devuelvo el cambio de variable
π/2
A = 2 Ln [ sec x ]
2 0
A = 2 Ln [ sec π -
sec 0
]
4
2) Hallar el volumen del solido de revolución generando por las región
encerrada por las curvas dadas (utilice el método del disco arandelas y corteza cilindrica)
A) un arco de y=cos2x, alrededor del eje x
A = 2 Ln [ √2 - 1]
b) x=4y, x =√y, alrededor de la recta x =8
b. x = 4y, x = 3√4 64 y
2 - 1= 0
4y = 3√4 64 y
2 = 1
(4y)3 = (
3√4)
3 y
2 = 1
64
64y3 = y
64y - y = 0
y (64 y -1) = 0
1/8
3
2 y = 0
3
3
y = -1 8
y = 1 8
3.- Hallar la longitud de la curva
a. Y = X3 + 1 X = 1 hasta X = 3
6 2X
b
L =ʃ 1 + [f´(x)]2 dx
a
Y´ = 3X2 - 1
6 2X2
Y´ = X2 - 1
2 2X2
Y´ = X4 - 1
2X2
3
L =ʃ 1+ X4 - 1
2 dx
1 2X2
3
L =ʃ 1+ [X4 - 1]
2 dx
1 4X4
3
L =ʃ 4X4 + X
8 - 2X
4 + 1
dx
1 4X4
3
L =ʃ X8 + 2X
4 + 1
dx
1 4X4
3
L =ʃ (X4 + 1)
2 dx
1 4X4
3
L =ʃ X4 + 1
dx
1 2X2
3
L = 1 ʃ (X2 + X
-2) dx
2 1
3
L = 1 ʃ X2
dx + ʃ X-2
dx
2 1
3
L = 1 33
- 1 - 1 + 1
2 3 3 3 1
3
L = 1 9 - 2 + 1
2 3 1
3
L = 1 10 - 2
2 3 1
L = 14
3
B)