Integral triple de una función con 3 variables, URGENTE!!!
robsnake
∫dx∫dy∫(4x+5y²+z)dz;integras respecto a z manteniendo x e y constantes ∫dx∫dy[(4xz+5y²z+z²/2] limlites(1-0) ∫dx∫dy[(4x(1-0)+5y²(1-0)+(1²-0)/2] ∫dx∫[(4x+5y²+(1/2)]dy integras respecto a y manteniendo x constante ∫dx∫[(4xy+5y³/3+(1/2)y] limlites(1-0) ∫dx∫[(4x(1-0)+5(1³-0)/3+(1/2)(1-0)] ∫[(4x+5/3+(1/2)]dx integras respecto a x [(4x²/2+(5/3)x+(1/2)x] limlites(1-0) 2+5/3+1/2=(12+10+3)/6=25/6
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wolffire01
porque integras primero comenzando por Z ?, porque no comienzas por el X?
∫dx∫dy[(4xz+5y²z+z²/2] limlites(1-0)
∫dx∫dy[(4x(1-0)+5y²(1-0)+(1²-0)/2]
∫dx∫[(4x+5y²+(1/2)]dy integras respecto a y manteniendo x constante
∫dx∫[(4xy+5y³/3+(1/2)y] limlites(1-0)
∫dx∫[(4x(1-0)+5(1³-0)/3+(1/2)(1-0)]
∫[(4x+5/3+(1/2)]dx integras respecto a x
[(4x²/2+(5/3)x+(1/2)x] limlites(1-0)
2+5/3+1/2=(12+10+3)/6=25/6