Iterated integral on a rectangular region

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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

\[\int_c^df(x_0,y)dy\]

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

\[F_1(x) =   \int_c^df(x, y)dy.\]

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

\[\int_a^bF_1(x)dx = \int_a^b\left(\int_c^df(x,y)dy\right)dx\tag{Eq:iintyx}\]

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

\[\int_a^bdx\int_c^df(x,y)dy.\]

This does not mean the product of two integrals \(\int_a^bdx\) and \(\int_c^df(x,y)dy\). □

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

\[F_2(y) = \int_a^bf(x,y)dx,\]

and then calculate

\[\int_c^dF_2(y)dy = \int_c^d\left(\int_a^bf(x,y)dx\right)dy.\tag{Eq:iintxy}\]

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:

\[\iint_Df(x,y)dxdy = \int_a^b\left(\int_c^df(x,y)dy\right)dx= \int_c^d\left(\int_a^bf(x,y)dx\right)dy.\]

Proof. See another post. ■

Example. Let us find \(\iint_D(2x + y)x dxdy\) where \(D = [0,1]\times[0,1]\).

\[\begin{eqnarray*} \iint_D(2x + y)xdxdy &=& \int_0^1\left(\int_0^1(2x^2 + xy)dy\right)dx\\ &=&\int_0^1\left[2x^2y + \frac{1}{2}xy^2\right]_{y=0}^{y=1}dx\\ &=&\int_0^1\left(2x^2 + \frac{1}{2}x\right)dx\\ &=&\left[\frac{2}{3}x^3 + \frac{1}{4}x^2\right]_{x=0}^{x=1}\\ &=& \frac{2}{3} + \frac{1}{4} = \frac{11}{12}. \end{eqnarray*}\]

Alternatively,

\[\begin{eqnarray*} \iint_D(2x + y)xdxdy &=& \int_0^1\left(\int_0^1(2x^2 + xy)dx\right)dy\\ &=&\int_0^1\left[\frac{2}{3}x^3 + \frac{1}{2}x^2y\right]_{x=0}^{x=1}dy\\ &=&\int_0^1\left(\frac{2}{3} + \frac{1}{2}y\right)dy\\ &=&\left[\frac{2}{3}y + \frac{1}{4}y^2\right]_{y=0}^{y=1}\\ &=& \frac{2}{3} + \frac{1}{4} = \frac{11}{12}. \end{eqnarray*}\]

□

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,

\[\iint_Df(x,y)dxdy = \left(\int_a^bg(x)dx\right)\cdot \left(\int_c^dh(y)dy\right).\]

Proof. Exercise. ■

Example. \[\begin{eqnarray*} \iint_{[0,1]\times[0,1]}e^{x+y}dxdy &=& \int_0^1\left(\int_0^1e^{x+y}dx\right)dy\\ &=& \left(\int_0^1e^xdx\right)\cdot\left(\int_0^1e^ydy\right)\\ &=&(e - 1)^2. \end{eqnarray*}\] □

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

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

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

\[\begin{eqnarray*} \iiint_D\sin(x+y)\cos z dxdydz &=&\left(\int_{0}^{\frac{\pi}{6}}\cos z dz\right)\cdot \left(\int_0^{\frac{\pi}{2}}dy\int_0^{\frac{\pi}{2}}\sin(x+y)dx\right)\\ &=&\left[\sin z\right]_{z=0}^{z=\frac{\pi}{6}}\int_0^{\frac{\pi}{2}}\left[-\cos(x+y)\right]_{x=0}^{x = \frac{\pi}{2}}dy\\ &=&\frac{1}{2}\int_0^{\frac{\pi}{2}}(\cos y + \sin y)dy\\ &=&\frac{1}{2}\left[\sin y - \cos y\right]_{y=0}^{y = \frac{\pi}{2}} = 1. \end{eqnarray*}\]

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