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Let f(x, y, z) and g(x, y, z) be integrable functions on the rectangular solid R=x,y,z|a1xa2,b1yb2,c1zc2. . Use the definition of the triple integral to prove that :

Rf(x,y,z)+g(x,y,z)dV=Rf(x,y,z)dV+Rg(x,y,z)dV.

Short Answer

Expert verified

Rf(x,y,z)dV=lim0i=1lj=1mk=1nf(xi*,yj*,zk*)dV.Replacef(x,y,z)+g(x,y,z)withf(x,y,z)inabove,=lim0i=1lj=1mk=1nf(xi*,yj*,zk*)dV+lim0i=1lj=1mk=1ng(xi*,yj*,zk*)dVTherefore,Rf(x,y,z)+g(x,y,z)dV=Rf(x,y,z)dV+Rg(x,y,z)dV.Hence,proved.

Step by step solution

01

Step 1. Given Information.

Given:R=x,y,z|a1xa2,b1yb2,c1zc2

02

Step 2. Proof.

Asweknowfromthedefinitionoftripleintegrals:Rf(x,y,z)dV=lim0i=1lj=1mk=1nf(xi*,yj*,zk*)dV.Replacef(x,y,z)+g(x,y,z)withf(x,y,z)inabove,Rf(x,y,z)+g(x,y,z)dV=lim0i=1lj=1mk=1nf+g(xi*,yj*,zk*)dV=lim0i=1lj=1mk=1nf(xi*,yj*,zk*)+g(xi*,yj*,zk*)dV=lim0i=1lj=1mk=1nf(xi*,yj*,zk*)dV+lim0i=1lj=1mk=1ng(xi*,yj*,zk*)dV=Rf(x,y,z)dV+Rg(x,y,z)dV=RHSTherefore,Rf(x,y,z)+g(x,y,z)dV=Rf(x,y,z)dV+Rg(x,y,z)dV.Hence,proved.

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