Question

Let $a<b\text{and}c<d$be real numbers and $R$be rectangle defined by

$R=\left\{\right(x,y)\mid a\le x\le b\text{and}c\le y\le d\}$in $xy$plane. If $g\left(x\right)$is continuous in interval $[a,b]$and $h\left(y\right)$is continuous on $[c,d]$

Use Fubini's theorem to prove that${\iint }_{R}g\left(x\right)h\left(y\right)dA=\left({\int }_a^{b}g\left(x\right)dx\right)\left({\int }_c^{d}h\left(y\right)dy\right).$

Short answer

It can be proved using definition of Fubini's theorem

${\iint }_{R}g\left(x\right)h\left(y\right)dA={\int }_a^b{\int }_c^{d}g\left(x\right)h\left(y\right)dydx$.

Step-by-step solution

Given Information

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

Rectangle is defined by

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

Fubini's Theorem

It states that ${\iint }_{R}g\left(x\right)h\left(y\right)dA={\int }_a^b{\int }_c^{d}g\left(x\right)h\left(y\right)dydx$

Treating $x$as constant

${\iint }_{R}g\left(x\right)h\left(y\right)dA={\int }_a^b{\int }_c^{d}g\left(x\right)h\left(y\right)dydx$

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

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

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

$\Rightarrow {\iint }_{R}g\left(x\right)h\left(y\right)dA=\left({\int }_a^{b}g\left(x\right)dx\right)\left({\int }_c^{d}h\left(y\right)dy\right)$

Hence proved