Question

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

${\iint }_R\alpha f(x,y)dA=\alpha {\iint }_{R}f(x,y)dA.$

Short answer

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

Step-by-step solution

Given Information

It is given that ${\iint }_R\alpha f(x,y)dA=\alpha {\iint }_{R}f(x,y)dA$and $R=\left\{\right(x,y)\mid a\le x\le b\text{and}c\le y\le d\}$

$\alpha$is real number.

Proof

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

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

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

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

The equation is true.

Changing order of sum

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

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

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

Equation is true.