Question

Let $a<b$and $c<d$be real numbers, and let $R$be the

rectangle in the $xy$-plane defined by

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

Prove that role="math" localid="1653851117688" ${\iint }_{R}dA=(b-a)(d-c)$, what is the relation between $R$and product of$(b-a)(d-c)$?

Short answer

This is evaluated using Fubini's theorem${\iint }_{R}dA={\int }_a^b{\int }_c^{d}dydx$

Step-by-step solution

Given Information

It is given that $a<b\text{and}c<d$, $a,b,c,d$are real numbers.

$R$ is rectangle in cartesian plane defined by$R=\left\{\right(x,y)\mid a\le x\le b\text{and}c\le y\le d\}$

Use Fubini's Theorem

By theorem ${\iint }_{R}dA={\int }_a^b{\int }_c^{d}dydx$

Treat $x$as constant

$={\int }_a^b\left({\int }_c^{d}dy\right)dx$

$={\int }_a^b[y{]}_{y=c}^{y=d}dx$

$=(d-c){\int }_a^{b}dx$

$=(d-c)[x{]}_{x=a}^{x=b}$

$=(d-c)(b-a)$

$\Rightarrow {\iint }_{R}dA=(d-c)(b-a)$

Hence proved.