5 Special cases

If the integrand can be written as

$f\left(x,y\right)=g\left(x\right)h\left(y\right)$

then the double integral

$I={\int \; <!--nolimits\; -->}_a^b{\int \; <!--nolimits\; -->}_c^{d}g\left(x\right)h\left(y\right)dydx$

can be written as

$I={\int \; <!--nolimits\; -->}_a^{b}g\left(x\right)dx\times {\int \; <!--nolimits\; -->}_c^{d}h\left(y\right)dy$

i.e. the product of the two individual integrals. For example, the integral

$I={\int \; <!--nolimits\; -->}_{x=-1}^2{\int \; <!--nolimits\; -->}_{y=-2}^3{x}^{2}ydydx$

which was evaluated earlier can be written as

$I={\int \; <!--nolimits\; -->}_{x=-1}^2{x}^{2}dx\times {\int \; <!--nolimits\; -->}_{y=-2}^{3}ydy={\left[\frac{x^3}{3}\right]}_{-1}^2{\left[\frac{y^2}{2}\right]}_{-2}^3=\left[\frac{8}{3}-\frac{\left(-1\right)}{3}\right]\left[\frac{9}{2}-\frac{4}{2}\right]$

$=3\times \frac{5}{2}=\frac{15}{2}$

the same result as before.

Imagine the integral

$I={\int \; <!--nolimits\; -->}_{-1}^1{\int \; <!--nolimits\; -->}_0^{1}x{\text{e}}^{-y^2}dydx$

Approached directly, this would involve evaluating the integral ${\int \; <!--nolimits\; -->}_0^{1}x{\text{e}}^{-y^2}dy$ which cannot be done by algebraic means (i.e. it can only be determined numerically). $I={\int \; <!--nolimits\; -->}_{-1}^{1}xdx\times {\int \; <!--nolimits\; -->}_0^1{\text{e}}^{-y^2}dy={\left[\frac{1}{2}{x}^2\right]}_{-1}^1\times {\int \; <!--nolimits\; -->}_0^1{\text{e}}^{-y^2}dy=0\times {\int \; <!--nolimits\; -->}_0^1{\text{e}}^{-y^2}dy=0$

and the result can be found without the need to evaluate the difficult integral.

If the integrand is independent of one of the variables and is simply a function of the other variable, then only one integration need be carried out.

The integral $I_1={\int \; <!--nolimits\; -->}_a^b{\int \; <!--nolimits\; -->}_c^{d}h\left(y\right)dydx$ may be written as $I_1=\left(b-a\right){\int \; <!--nolimits\; -->}_c^{d}h\left(y\right)dy$ and the integral $I_2={\int \; <!--nolimits\; -->}_a^b{\int \; <!--nolimits\; -->}_c^{d}g\left(x\right)dydx$ may be written as $I_2=\left(d-c\right){\int \; <!--nolimits\; -->}_a^{b}g\left(x\right)dx$ i.e. the integral in the variable upon which the integrand depends multiplied by the length of the range of integration for the other variable.

Example 5

Evaluate the double integral

$I={\int \; <!--nolimits\; -->}_0^2{\int \; <!--nolimits\; -->}_{-1}^2{y}^{2}dydx$

Solution

As the integral in $y$ can be multiplied by the range of integration in $x$ , the double integral will equal

$I=\left(2-0\right){\int \; <!--nolimits\; -->}_{-1}^2{y}^{2}dy=2{\left[\frac{y^3}{3}\right]}_{-1}^2=2\left[\frac{2^3}{3}-\frac{{\left(-1\right)}^3}{3}\right]=6$

Note that the two integrations can be carried out in either order as long as the limits are associated with the correct variable. For example

$I={\int \; <!--nolimits\; -->}_{x=0}^1{\int \; <!--nolimits\; -->}_{y=-1}^2{x}^{4}ydydx={\int \; <!--nolimits\; -->}_{x=0}^1{\left[\frac{x^4{y}^2}{2}\right]}_{-1}^{2}dx={\int \; <!--nolimits\; -->}_{x=0}^1\left[2x^4-\frac{1}{2}{x}^4\right]dx$

$={\int \; <!--nolimits\; -->}_0^1\frac{3}{2}{x}^{4}dx={\left[\frac{3}{10}{x}^5\right]}_0^1=\frac{3}{10}\times 1-\frac{3}{10}\times 0=\frac{3}{10}$

and

$I={\int \; <!--nolimits\; -->}_{y=-1}^2{\int \; <!--nolimits\; -->}_{x=0}^1{x}^{4}ydxdy={\int \; <!--nolimits\; -->}_{y=-1}^2{\left[\frac{x^{5}y}{5}\right]}_0^{1}dy={\int \; <!--nolimits\; -->}_{-1}^2\left[\frac{y}{5}-0\right]dy$

$={\int \; <!--nolimits\; -->}_{-1}^2\frac{y}{5}dy={\left[\frac{y^2}{10}\right]}_{-1}^2=\frac{4}{10}-\frac{1}{10}=\frac{3}{10}$

Task!

Evaluate the following integral:

$I={\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_{-1}^{1}z\left(w+1\right)dwdz$ .

Answer

Exercises

1. Evaluate the following integrals:
   1. $I={\int \; <!--nolimits\; -->}_0^{\pi ∕2}{\int \; <!--nolimits\; -->}_0^1\left(ycosx\right)dydx$
   2. $I={\int \; <!--nolimits\; -->}_{-8}^3{\int \; <!--nolimits\; -->}_{-1}^1{y}^{2}dydx$
   3. $I={\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^5{\left(s+1\right)}^{4}dtds$
2. Evaluate the integrals ${\int \; <!--nolimits\; -->}_{-1}^3{\int \; <!--nolimits\; -->}_0^2{x}^{3}ydydx$ and ${\int \; <!--nolimits\; -->}_0^2{\int \; <!--nolimits\; -->}_{-1}^3{x}^{3}ydxdy$ and show that they are equal. As explained in the text, the order in which these integrations are carried out does not matter for integrations over rectangular areas.

Answer