Matematicas Avanzadas

Nombre: Jose Antonio Cantu González Matrícula: al02535750

Nombre del curso:

Matemáticas avanzadas I Nombre del profesor:

Edgar Melendrez Rivera

Módulo:

IV Actividad:

Actividad integradora IV

Fecha: 03/07/2013

Bibliografía:

Algebra lineal Stanley I Grossman Mc Graw Hill Quinta edicion

Demuestra en cada uno de los siguientes ejercicios el teorema de Fubini (realiza las integrales dobles intercambiando el orden de la integración).

y

sean a=-1,b=3,c=0,d=2

∫_c^d▒∫_a^b▒〖f(x,y)dxdy=∫_0^2▒∫_(-1)^3▒(3-2x^3 y^2 )dxdy〗

∫_0^2▒〖[3x-1/2 x^4 y^2 ]_(x=-1)^(x=3) dy〗

∫_0^2▒[(9-81/2 y^2 )-(-3-1/2 y^2)]dy

∫_0^2▒〖(12-40y^2 〗)dy=[12y-40/3 y^3 ]_(y=0)^(y=2)

24-320/3=-248/3

∫_a^b▒∫_c^d▒〖f(x,y)dydx=∫_(-1)^3▒∫_0^2▒(3-2x^3 y^2 )dydx〗

∫_(-1)^3▒〖[3y-2/3 x^3 y^3 ]_(y=0)^(y=2) dx〗

∫_(-1)^3▒(6-16/3 x^3 )dx

[6x-4/3 x^4 ]_(x=-1)^(x=3)

(18-108)-(-6-4/3)=90+22/3=-248/3

Se demuestra que:

∫_c^d▒∫_a^b▒〖f(x,y)dxdy=∫_a^b▒∫_c^d▒〖f(x,y)dydx=∬_r^ ▒f(x,y)da〗〗

y

sean a=-5,b=-1,c=1,d=5

∫_c^d▒∫_a^b▒〖f(x,y)dxdy=∫_1^5▒∫_(-5)^(-1)▒〖((3x^2)/y+x^3 y^(2/3) )dxdy=∫_1^5▒〖[x^3/y+1/4 x^4 y^(2/3) ]_(x=-5)^(x=-1) dy〗 〗〗

∫_1^5▒[(-1/y+y^(2/3)/4)-(125/y+(625y^(2/3))/4) ]dy

∫_1^5▒(124/y-156y^(2/3) )dy

[124 ln⁡(y)-468/5 y^(5/3) ]_(y=1)^(y=5)

[124 ln⁡(5)-468/5 (5)^(5/3) ]-[124 ln⁡(1)-468/5 (1)^(5/3) ]

124 ln⁡(5)-468∛25+468/5=-507.098

∫_a^b▒∫_c^d▒〖f(x,y)dydx=∫_(-5)^(-1)▒〖∫_1^5▒〖 ((3x^2)/y+x^3 y^(2/3) )dydx〗 〗〗

∫_(-5)^(-1)▒〖[3x^2 ln⁡(y)+3/5 x^3 y^(5/3) ]_(y=1)^(y=5) dx〗

∫_(-5)^(-1)▒{[3x^2 ln⁡(5)+3/5 x^3 (5)^(5/3) ]-[3x^2 ln⁡(1)+3/5 x^3 (1)^(5/3) ]}dx

∫_(-5)^(-1)▒[3x^2 ln⁡(5)+(3∛(25 )-3/5) x^3 ]dx

[x^3 ln⁡(5)+(3/4 ∛25-3/20) x^4 ]_(x=-5)^(x=-1)

(-ln⁡(5)+3/4 ∛25-3/20)

-(-125 ln⁡(5)+1875/4 ∛25-1875/20)

124 ln⁡(5)-468∛25+468/5=-507.098

Se demuestra que:

∫_c^d▒∫_a^b▒〖f(x,y)dxdy=∫_a^b▒∫_c^d▒〖f(x,y)dydx=∬_R^ ▒f(x,y)da〗〗

y

sean a=4,b=8,c=2,d=5

∫_c^d▒∫_a^b▒〖f(x,y)dxdy=∫_2^5▒∫_4^8▒〖[sen(4x)+cos⁡(3y) ] dxdy〗〗

∫_2^5▒[-1/4 cos⁡(4x)+xcos(3y)]_(x=4)^(x=8) dy

∫_2^5▒{[-1/4 cos⁡(32)+8 cos⁡(3y) ]-[-1/4 cos⁡(16)+4 cos⁡(3y) ]}dy

∫_2^5▒[-1/4 cos⁡(32)+1/4 cos⁡(16)+4cos⁡(3y)]dy

[-y/4 cos⁡(32)+y/4 cos⁡(16)+4/3 sen(3y)]_(y=2)^(y=5)

[-5/4 cos⁡(32)+5/4 cos⁡(16)+4/3 sen(15)]-[-2/4 cos⁡(32)+2/4 cos⁡(16)+4/3 sen(6)]

-3/4 cos⁡(32)+3/4 cos⁡(16)+4/3 sen(15)-4/3 sen(6)=0.2906

∫_a^b▒∫_c^d▒〖f(x,y)dydx=∫_4^8▒∫_2^5▒[sen(4x)+cos⁡(3y)]dydx〗

∫_4^8▒〖[y sen(4x)+1/3 sen(3y)]_(y=2)^(y=5) dx〗

∫_4^8▒{[5 sen (4x)+1/3 sen(15)]-[2sen(4x)+1/3 sen(6)]} dx

∫_4^8▒[3sen(4x)+1/3 sen(15)-1/3 sen(6)]dx

[-3/4 cos⁡(4x)+x/3 sen(15)-x/3 sen(6) ]_4^8

[-3/4 cos⁡(32)+8/3 sen(15-8/3 sen(6)]-[-3/4 cos⁡(16)sen(15)-4/3 sen(6)]

-3/4 cos⁡(32)+3/4 cos⁡(16)+4/3 sen(15)-4/3 sen(6)=0.2906

Se demuestra que:

∫_c^d▒∫_a^b▒〖f(x,y)dxdy=∫_a^b▒∫_c^d▒〖f(x,y)dydx=∬_R^ ▒f(x,y)da〗〗

y

sean a=2,b=4,c=-2,d=0

∫_c^d▒∫_a^b▒〖f(x,y)dxdy=∫_(-2)^0▒∫_2^4▒(xe^y )dxdy〗

∫_(-2)^0▒〖[1/2 x^2 e^y ]_(x=2)^(x=4) dy〗

∫_(-2)^0▒〖(8e^y-2e^y )dy=∫_(-2)^0▒〖6e^y dy〗〗

[6e^y ]_(y=-2 )^(y=0)=6-6e^(-2)=5.1879

∫_a^b▒∫_c^d▒〖f(x,y)dydx=∫_2^4▒∫_(-2)^0▒〖(xe^y )dydx=∫_2^4▒〖[xe^y ]_(y=-2)^(y=0) dx〗〗〗

∫_2^4▒〖(x-xe^(-2) )dx=[x^2/2-x^2/2 e^(-2) ]_(x=2)^(x=4) 〗

(8-8e^(-2) )-(2-2e^(-2) )=6-6e^(-2)=5.1879

Se demuestra que:

∫_c^d▒∫_a^b▒〖f9x,y)dxdy=∫_a^b▒∫_c^d▒〖f(x,y)dydx=∬_R^ ▒f(x,y)da〗〗

y

sean a=1,b=5,c=0,d=2x

∫_a^b▒∫_c^d▒〖f(x,y)dydx=∫_1^5▒∫_0^2x▒〖xy^3 dydx=∫_1^5▒〖[1/4 xy^4 ]_(y=0)^(y=2x) dx〗〗〗

∫_1^5▒〖4x^5 dx=[2/3 x^6 ]_(x=1)^(x=5)=31250/3-2/3=10416〗

∫_c^d▒∫_a^b▒〖f(x,y)dxdy=∫_0^2x▒∫_1^5▒〖xy^3 dxdy=∫_2^10▒∫_(y/2)^5▒〖xy^3 dxdy+∫_0^2▒∫_1^5▒〖xy^3 dxdy〗〗〗〗

∫_2^10▒〖[1/2 x^2 y^3 ]_(x=y/2)^(x=5) dy+∫_0^2▒〖[1/2 x^2 y^3 ]_(x=1)^(x=5) dy〗〗

∫_2^10▒〖(25/2 y63-1/8 y^5 )dy+∫_0^2▒(25/2 y^3-1/2 y^3 )dy〗

[25/8 y^4-1/48 y^6 ]_(y=2)^(y=10)+[3y^4 ]_0^2

31250-62500/3-50+4/3+48=10416

Se demuestra que

∫_c^d▒∫_a^b▒〖f(x,y)dxdy=∫_a^b▒∫_c^d▒f(x,y)dydx〗

y

sean a=3/2 π,b=2π,c=0,d=x/2

∫_a^b▒∫_c^d▒〖f(x,y)dydx=∫_(3/2 π)^2π▒∫_0^(x/2)▒〖sen(2x-y)dydx=∫_(3/2 π)^2π▒〖[cos⁡(2x-y) ]_(y=0 )^(y=x/2) dx〗〗〗

∫_(3/2 π)^2π▒〖[cos⁡(3/2 x)-cos⁡(2x)]dx=[2/3 sen(3/2 x)-1/2 sen(2x)]_(x=3/2 π)^(x=2π) 〗

2/3 sen(3π)-1/2 sen(4π)-2/3 sen(9/4 π)+1/2 sen(3π)

0-0-2/(3√2)=-0.471

∫_c^d▒∫_a^b▒〖f(x,y)dxdy=∫_0^(x/2)▒∫_(3/2 π)^2π▒sen(2x-y)dxdy〗

∫_(3/4 π)^π▒∫_2y^2π▒〖sen92x-y)dxdy+∫_0^(3/4 π)▒∫_(3/2 π)^2π▒sen(2x-y)dxdy〗

∫_(3/4 π)^π▒〖[-1/2 cos92x-y)]_(x=2y)^(x=2π) dy+∫_0^(3/4 π)▒[-1/2 cos⁡(2x-y)]_(x=3/2 π)^(x=2π) 〗+∫_0^(3/4 π)▒[-1/2 cos⁡(4π-y)]dy

[1/2 sen(4π-y)+1/6 sen(3y)]_(y=3/4 π)^(y=π)+[1/2 sen(4π-y)-1/2 sen(3π-y)]_(y=0)^(y=3/4 π)

1/2 sen(3π)-1/6 sen(3π)-1/2 sen(13/4 π)-1/6 sen(9/4 π)+1/2 sen(13/4 π)-1/2 sen(9/4 π)-1/2 sen(4π)+1/2 sen(3π)

0+0-2/(3√(2 ))-0+0=2/(3√2)=-0.471

Se cumple el teorema de fubini:

∫_c^d▒∫_a^b▒〖f(x,y)dxdy=∫_a^b▒∫_c^d▒〖f(x,y)dydx=∬_R^ ▒f(x,y)da〗〗

Calcula el área bajo las curvas utilizando el concepto de integración doble.

despejamos y de la segunda ecuacion se obtiene:

y=x/3-1/2 …(1)

y=x^2-4…(2)

graficando (1)y(2)

Igualando (1) y (2) y resolviendo:

x/3-1/2=x^2-4

x^2-1/3 x-7/2=0

x=(-b±√(b^2-4ac))/2a

x=(1/3±√(1/9+14))/2

De donde

x_1=(1+√127)/6=2.044

x_2=(1-√127)/6=-1.711

Sustituyendo en (1)

y_1=(-8-√127)/18=0.181

y_2=(-8-√127)/18=-1.07

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