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Taller de Calculo
• Integrantes• Luis Gómez• José Álvarez• Julia Rodríguez• José Delgado• Jhondervin Flórez• Walter Casamayor
• Sección: 3410
República Bolivariana de VenezuelaUniversidad Politecnica Territorial “Andres Eloy Blanco”
P.N.F. en Sistema de Calidad y Ambiente
Ejercicio #2
•
•
1
Ejercicio # 8
∫∞
dxx(x²+a²)⁵
Sug. u = x²+a²
Cambio de Variable
u= x²+a²du= 2x dxdu = x dx
2
Lim ⅟₂ du = Lim ⅟₂ u⁻⁵ du = Lim ⅟₂ u⁻⁵⁺ 1 b→∞ u⁵ b→∞ b→∞ -5+1 = Lim 1 u⁻⁴ = Lim - 1 = Lim - 1 b→∞ 2 -4 b→∞ 2.4 u⁴ b→∞ (x²+a²)⁴Aplicamos teorema fundamental del calculo
= - 1 - ( -1 ) 8 (∞ + a²)⁴ 8 (1² + a²)⁴ - 1 + 1 8 (∞)⁴ 8 (1 + a²)⁴ = - 1 + 1 = 1 ∞ 8 (1 + a²)⁴ 8 (1 + a²)⁴
∫b
1∫b
1
b
1
b
1
0
∫Ejercicio # 15
2
∞dx
x(x4+4x²)
Sug. Descomp. en Fracciones Parc.
∫2
∞dx
xx2(x2+4)
Ax2
+B
x+
Cx+D
(x2+4)
∫2
∞dx
x(x4+4x²)
= Lim ⅟₂b→∞
1
x2(x2+4)=
A (x2+4) + B x (x2+4) + x2 (Cx + Dx)x2 (x2+4)
1
x2(x2+4)=
Ax2+4A +Bx3+4Bx + Cx3 + Dx2
1
x2(x2+4)=
x2 (x2+4)
1 = Ax2+4A +Bx3+4Bx + Cx3 + Dx2
1 = Bx3+ Cx3 + Ax2+ Dx2 +4Bx+ 4A
1 = (B+C)x3 + (A+D)x2 +4Bx+ 4A
B+C= 0 → C= -B → C= 0
A+D= 0 → D= -A → D= -1/4
4 B=0 → B= 0
4 A = 1 → A= 1/4
dxA
x2+
B
x+
Cx- D
(x2+4)=∫
2
∞dx
xx2(x2+4)
Lim ⅟₂b→∞ ∫
2
∞Lim ⅟₂b→∞
= ∫2
∞Lim ⅟₂b→∞
1/4
x2
0
x++ dx(0x-1/4)
(x2+4)
b→∞
1
x2 -(x2+4)∫Lim
b→∞ =
41 dx
41 ∫ dx
-(x2+22)∫Lim
b→∞ =
41
X-2 dx 41 ∫ dxLim ⅟₂
b→∞
-Lim b→∞
=-2+1X-2 + 1
41 Lim
b→∞ 21 tan -1 x
2
=
2
b
2
b
1 - 81 tan -1 x
22
b
4 x
4
-
=1
(4 ∞)- -8
Lim b→∞
1 tan -1 ∞2
+ 18
+8
1 tan -1 22
b
2b b
Lim
2 2
2
b
= -8
1 . ∏2
+ 18
+8 4
0
1 . ∏ =16 ∏- + 1
8 ∏+8
=- ∏432
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