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Let f(x,y)and g(x,y)be integrable functions on the rectangle R={(x,y)axbandcyd}. Use the definition of the double integral to prove that

role="math" localid="1653940239011" R(f(x,y)+g(x,y))dA=Rf(x,y)dA+Rg(x,y)dA

Short Answer

Expert verified

To prove this, write the double integral on left hand side as double Reimann sum.

Step by step solution

01

Given Information

Equality is given as R(f(x,y)+g(x,y))dA=Rf(x,y)dA+Rg(x,y)dA

R={(x,y)axbandcyd}

02

Proof

Write the double integral on left hand side as double Reimann sum.

R(f(x,y)+g(x,y))dA=limΔ0i=1mj=1nfxi*,yj*+gxi*,yj*ΔA

and Δ=(Δx)2+(Δy)2

Simplifying

R(f(x,y)+g(x,y))dA=limΔ0i=1mj=1nfxi*,yj*ΔA+i=1mj=1ngxi*,yj*ΔA

=limΔ0i=1mj=1nfxi*,yj*ΔA+limΔ0i=1mj=1ngxi*,yj*ΔA

=Rf(x,y)dA+Rg(x,y)dA

The equation is true.

03

Changing order of sum

It givesR(f(x,y)+g(x,y))dA=limΔ0j=1ni=1mfxi*,yj*+gxi*,yj*ΔA

R(f(x,y)+g(x,y))dA=limΔ0j=1ni=1mfxi*,yj*ΔA+j=1ni=1mgxi*,yj*ΔA

=limΔ0j=1ni=1mfxi*,yj*ΔA+limΔ0j=1ni=1mgxi*,yj*ΔA

=Rf(x,y)dA+Rg(x,y)dA

The equation is true.

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