Prove Theorem 13.10 (b). That is, show that if f(x,y)and g(x,y)are integrable functions on the general region ,then

∬Ω(f(x,y)+g(x,y))dA=∬Ωf(x,y)dA+∬Ωg(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

It is given that ∬Ω[f(x,y)+g(x,y)]dA=∬Ωf(x,y)dA+∬Ωg(x,y)dA

R={(x,y)∣a≤x≤bandc≤y≤d},that is,Ω⊆Randc is real number.

02

Using Property of Double Integrals

For any functionf that is continuous over regionΩ

∬Ωf(x,y)dA=∬RF(x,y)dA

and Fx,y=f(x,y),if(x,y)∈Ωand role="math" localid="1653944271832" Fx,y=0,if(x,y)∉Ω

03

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

Writing as double Riemann sum, we get

∬R(F(x,y)+G(x,y))dA=limΔ→0∑i=1m∑j=1nFxi*,yj*+Gxi*,yj*ΔA

and

Δ=(Δx)2+(Δy)2

Simplify RHS

∬R(F(x,y)+G(x,y))dA=limΔ→0∑i=1m∑j=1nFxi*,yj*ΔA+∑i=1m∑j=1nGxi*,yj*ΔA

=limΔ→0∑i=1m∑j=1nFxi*,yj*ΔA+limΔ→0∑i=1m∑j=1nGxi*,yj*ΔA

=∬RF(x,y)dA+∬RG(x,y)dA

Using same property again

∬Ω[f(x,y)+g(x,y)]dA=∬Ωf(x,y)dA+∬Ωg(x,y)dA

The equation is true.

04

Simplification

Changing order of sum

∬R(F(x,y)+G(x,y))dA=limΔ→0∑j=1n∑i=1mFxi*,yj*+Gxi*,yj*ΔA

Simplify RHS

∬R(F(x,y)+G(x,y))dA=limΔ→0∑j=1n∑i=1mFxi*,yj*ΔA+∑j=1n∑i=1mGxi*,yj*ΔA

=limΔ→0∑j=1n∑j=1mFxi*,yj*ΔA+limΔ→0∑j=1n∑j=1mGxi*,yj*ΔA

=∬RF(x,y)dA+∬RG(x,y)dA

Using same property again

∬Ω[f(x,y)+g(x,y)]dA=∬Ωf(x,y)dA+∬Ωg(x,y)dA

Hence, equation is true.

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