Question

Let $f(x,y)$and $g(x,y)$be integrable functions on the rectangle $R=\left\{\right(x,y)\mid a\le x\le b\text{and}c\le y\le d\}$. Use the definition of the double integral to prove that

role="math" localid="1653940239011" ${\iint }_R\left(f\right(x,y)+g(x,y\left)\right)dA={\iint }_{R}f(x,y)dA+{\iint }_{R}g(x,y)dA$

Short answer

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

Step-by-step solution

Given Information

Equality is given as ${\iint }_R\left(f\right(x,y)+g(x,y\left)\right)dA={\iint }_{R}f(x,y)dA+{\iint }_{R}g(x,y)dA$

$R=\left\{\right(x,y)\mid a\le x\le b\text{and}c\le y\le d\}$

Proof

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

${\iint }_R\left(f\right(x,y)+g(x,y\left)\right)dA=\underset{\Delta \to 0}{\lim }\sum _{i=1}^m\sum _{j=1}^n\left[f\left(x_i^{*},y_j^{*}\right)+g\left(x_i^{*},y_j^{*}\right)\right]\Delta A$

and $\Delta =\sqrt{(\Delta x{)}^2+(\Delta y{)}^2}$

Simplifying

${\iint }_R\left(f\right(x,y)+g(x,y\left)\right)dA=\underset{\Delta \to 0}{\lim }\left\{\sum _{i=1}^m\sum _{j=1}^{n}f\left(x_i^{*},y_j^{*}\right)\Delta A+\sum _{i=1}^m\sum _{j=1}^{n}g\left(x_i^{*},y_j^{*}\right)\Delta A\right\}$

$=\underset{\Delta \to 0}{\lim }\sum _{i=1}^m\sum _{j=1}^{n}f\left(x_i^{*},y_j^{*}\right)\Delta A+\underset{\Delta \to 0}{\lim }\sum _{i=1}^m\sum _{j=1}^{n}g\left(x_i^{*},y_j^{*}\right)\Delta A$

$={\iint }_{R}f(x,y)dA+{\iint }_{R}g(x,y)dA$

The equation is true.

Changing order of sum

It gives${\iint }_R\left(f\right(x,y)+g(x,y\left)\right)dA=\underset{\Delta \to 0}{\lim }\sum _{j=1}^n\sum _{i=1}^m\left[f\left(x_i^{*},y_j^{*}\right)+g\left(x_i^{*},y_j^{*}\right)\right]\Delta A$

$\Rightarrow {\iint }_R\left(f\right(x,y)+g(x,y\left)\right)dA=\underset{\Delta \to 0}{\lim }\left\{\sum _{j=1}^n\sum _{i=1}^{m}f\left(x_i^{*},y_j^{*}\right)\Delta A+\sum _{j=1}^n\sum _{i=1}^{m}g\left(x_i^{*},y_j^{*}\right)\Delta A\right\}$

$=\underset{\Delta \to 0}{\lim }\sum _{j=1}^n\sum _{i=1}^{m}f\left(x_i^{*},y_j^{*}\right)\Delta A+\underset{\Delta \to 0}{\lim }\sum _{j=1}^n\sum _{i=1}^{m}g\left(x_i^{*},y_j^{*}\right)\Delta A$

$={\iint }_{R}f(x,y)dA+{\iint }_{R}g(x,y)dA$

The equation is true.