Iterated integral on a rectangular region

Defining the multiple integral is one thing; calculating it is another.

The iterated integral is a technique to calculate a multiple integral. Simply put, an iterated integral is a technique where we apply one-variable integration iteratively, thereby reducing a multiple integral to one-variable integrals. Here, we consider an iterated integral over a rectangular region.

Let $f(x,y)$ be a continuous function on the rectangular region $D=[a,b]\times [c,d]$. Fix an arbitrary $x_0\in [a,b]$. Then, $f(x_0,y)$ can be regarded as a univariate function of $y$. $f(x_0,y)$ is continuous on $[c,d]$ so it is integrable.

The integral

$${\int }_c^{d}f(x_0,y)dy$$

contains $x_0$ as a parameter. Replacing $x_0$ with $x$, let us define the function $F_1(x)$ of $x$ on $[a,b]$ by

$$F_1(x)={\int }_c^{d}f(x,y)dy.$$

We have the following lemma:

Lemma

The function $F_1(x)$ defined above is continuous on $[a,b]$.

Proof. The proof will be given in another post when we prove a more general lemma. ■

See also: Iterated integral on a bounded closed set.

Since $F_1(x)$ is continuous on $[a,b]$, it is integrable on $[a,b]$ so that the integral

$$\begin{array}{}\text{(Eq:iintyx)}& {\int }_a^b{F}_1(x)dx={\int }_a^b({\int }_c^{d}f(x,y)dy)dx\end{array}$$

is well-defined. The integral obtained by this procedure of iterating one-variable integration is called an iterated integral.

Remark. The iterated integral in (Eq:iintyx) is also denoted as

$${\int }_a^{b}dx{\int }_c^{d}f(x,y)dy.$$

This does not mean the product of two integrals ${\int }_a^{b}dx$ and ${\int }_c^{d}f(x,y)dy$. □

We can also integrate $f(x,y)$ by $x$ first, then by $y$. That is, we first calculate

$$F_2(y)={\int }_a^{b}f(x,y)dx,$$

and then calculate

$$\begin{array}{}\text{(Eq:iintxy)}& {\int }_c^d{F}_2(y)dy={\int }_c^d({\int }_a^{b}f(x,y)dx)dy.\end{array}$$

The integrals (Eq:iintyx) and (Eq:iintxy) yield the same result and they are equal to the double integral of $f(x,y)$ on $D=[a,b]\times [c,d]$. We summarize this fact as the following theorem.

Theorem

Let $f(x,y)$ be a continuous function on the rectangular region $D=[a,b]\times [c,d]$. Then, the following holds:

$${\iint }_{D}f(x,y)dxdy={\int }_a^b({\int }_c^{d}f(x,y)dy)dx={\int }_c^d({\int }_a^{b}f(x,y)dx)dy.$$

Proof. See another post. ■

Example. Let us find ${\iint }_D(2x+y)xdxdy$ where $D=[0,1]\times [0,1]$.

$$\begin{array}{rcl}{\iint }_D(2x+y)xdxdy& =& {\int }_0^1({\int }_0^1(2x^2+xy)dy)dx\\ & =& {\int }_0^1{[2x^{2}y+\frac{1}{2}x{y}^2]}_{y=0}^{y=1}dx\\ & =& {\int }_0^1(2x^2+\frac{1}{2}x)dx\\ & =& {[\frac{2}{3}{x}^3+\frac{1}{4}{x}^2]}_{x=0}^{x=1}\\ & =& \frac{2}{3}+\frac{1}{4}=\frac{11}{12}.\end{array}$$

Alternatively,

$$\begin{array}{rcl}{\iint }_D(2x+y)xdxdy& =& {\int }_0^1({\int }_0^1(2x^2+xy)dx)dy\\ & =& {\int }_0^1{[\frac{2}{3}{x}^3+\frac{1}{2}{x}^{2}y]}_{x=0}^{x=1}dy\\ & =& {\int }_0^1(\frac{2}{3}+\frac{1}{2}y)dy\\ & =& {[\frac{2}{3}y+\frac{1}{4}{y}^2]}_{y=0}^{y=1}\\ & =& \frac{2}{3}+\frac{1}{4}=\frac{11}{12}.\end{array}$$

□

If the function $f(x,y)$ can be expressed in the form of $g(x)h(y)$ (i.e., the product of univariate functions), then its integral can be simplified.

Corollary

Let $f(x,y)$ be a continuous function on $D=[a,b]\times [c,d]$. Suppose $f(x,y)=g(x)h(y)$ where $g(x)$ and $h(y)$ are continuous functions on $[a,b]$ and $[c,d]$, respectively. Then,

$${\iint }_{D}f(x,y)dxdy=({\int }_a^{b}g(x)dx)\cdot ({\int }_c^{d}h(y)dy).$$

Proof. Exercise. ■

Example. $$\begin{array}{rcl}{\iint }_{[0,1]\times [0,1]}{e}^{x+y}dxdy& =& {\int }_0^1\left({\int }_0^1{e}^{x+y}dx\right)dy\\ & =& \left({\int }_0^1{e}^{x}dx\right)\cdot \left({\int }_0^1{e}^{y}dy\right)\\ & =& (e-1{)}^2.\end{array}$$ □

The iterative integral of the triple integral

$${\iiint }_{D}f(x,y,z)dxdydz,D=[a_1,a_2]\times [b_1,b_2]\times [c_1,c_2]$$

is denoted as

$${\int }_{a_1}^{a_2}\left({\int }_{b_1}^{b_2}({\int }_{c_1}^{c_2}f(x,y,z)dz)dy\right)dx={\int }_{a_1}^{a_2}dx{\int }_{b_1}^{b_2}dy{\int }_{c_1}^{c_2}f(x,y,z)dz.$$

Example. The integral of $f(x,y,z)=\sin (x+y)\cos z$ on  $D=[0,\frac{\pi }{2}]\times [0,\frac{\pi }{2}]\times [0,\frac{\pi }{6}]$ is calculated as

$$\begin{array}{rcl}{\iiint }_D\sin (x+y)\cos zdxdydz& =& ({\int }_0^{\frac{\pi }{6}}\cos zdz)\cdot ({\int }_0^{\frac{\pi }{2}}dy{\int }_0^{\frac{\pi }{2}}\sin (x+y)dx)\\ & =& {[\sin z]}_{z=0}^{z=\frac{\pi }{6}}{\int }_0^{\frac{\pi }{2}}{[-\cos (x+y)]}_{x=0}^{x=\frac{\pi }{2}}dy\\ & =& \frac{1}{2}{\int }_0^{\frac{\pi }{2}}(\cos y+\sin y)dy\\ & =& \frac{1}{2}{[\sin y-\cos y]}_{y=0}^{y=\frac{\pi }{2}}=1.\end{array}$$