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C 
I 
C 
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DE 
M 
A 
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E 
M 
A 
T 
I 
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 - X3)dx 
-2 0 
2 
A = 2 ʃ ( 4X - X3)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 y2 - 1= 0 
4y = 3√4 64 y2 = 1 
(4y)3 = (3√4)3 y2 = 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 + X8 - 2X4 + 1 dx 
1 4X4 
3 
L =ʃ X8 + 2X4 + 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)

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Ejercicios radhames ultima unidad

  • 1. E J E R C I C I O S DE M A T E M A T I C A
  • 2. 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
  • 3. b. Y = X3, Y = 4X X3 = 4X X3 - 4X = 0 X (X2 - 4) = 0 X = 0 X = -2 X = 2 Y = X3
  • 4. 0 2 A = ʃ (X3 – 4X)dx + ʃ ( 4X - X3)dx -2 0 2 A = 2 ʃ ( 4X - X3)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
  • 5. 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
  • 6. 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]
  • 7.
  • 8. b) x=4y, x =√y, alrededor de la recta x =8 b. x = 4y, x = 3√4 64 y2 - 1= 0 4y = 3√4 64 y2 = 1 (4y)3 = (3√4)3 y2 = 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
  • 9.
  • 10. 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 + X8 - 2X4 + 1 dx 1 4X4 3 L =ʃ X8 + 2X4 + 1 dx 1 4X4 3 L =ʃ (X4 + 1)2 dx
  • 11. 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
  • 12. B)