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

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

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

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

Comments

Post a Comment

Popular posts from this blog

Open sets and closed sets in \(\mathbb{R}^n\)

-
Image
Open sets In \(\mathbb{R}\), we have the notion of an open interval such as \((a, b) = \{x \in \mathbb{R} | a < x < b\}\). We want to extend this idea to apply to \(\mathbb{R}^n\). We also introduce the notions of bounded sets and closed sets in \(\mathbb{R}^n\). Recall that the \(\varepsilon\)-neighbor of a point \(x\in\mathbb{R}^n\) is defined as \(N_{\varepsilon}(x) = \{y \in \mathbb{R}^n | d(x, y) < \varepsilon \}\) where \(d(x,y)\) is the distance between \(x\) and \(y\). Definition (Open set) A subset \(U\) of \(\mathbb{R}^n\) is said to be an  open set  if the following holds: \[\forall x \in U ~ \exists \delta > 0 ~ (N_{\delta}(x) \subset U).\tag{Eq:OpenSet}\] That is, for every point in an open set \(U\), we can always find an open ball centered at that point, that is included in \(U\). See the following figure. Perhaps, it is instructive to see what is  not  an open set. Negating (Eq:OpenSet), we have \[\exists x \in U ~ \forall \delta > 0 ~ (N_{\delta}(x) \not
Read more

Euclidean spaces

-
Image
We would like to study multivariate functions (i.e., functions of many variables), continuous multivariate functions in particular. To define continuity, we need a measure of "closeness" between points. One measure of closeness is the Euclidean distance. The set \(\mathbb{R}^n\) (with \(n \in \mathbb{N}\)) with the Euclidean distance function is called a Euclidean space. This is the space where our functions of interest live. The real line is a geometric representation of \(\mathbb{R}\), the set of all real numbers. That is, each \(a \in \mathbb{R}\) is represented as the point \(a\) on the real line. The coordinate plane , or the \(x\)-\(y\) plane , is a geometric representation of \(\mathbb{R}^2\), the set of all pairs of real numbers. Each pair of real numbers \((a, b)\) is visualized as the point \((a, b)\) in the plane. Remark . Recall that \(\mathbb{R}^2 = \mathbb{R}\times\mathbb{R} = \{(x, y) | x, y \in \mathbb{R}\}\) is the Cartesian product of \(\mathbb{R}\) with i
Read more

Newton's method

-
Image
Newton's method (the Newton-Raphson method) is a very powerful numerical method for solving nonlinear equations.  Suppose we'd like to solve a non-linear equation \(f(x) = 0\), where \(f(x)\) is a (twice) differentiable non-linear function. Newton's method generates a sequence of numbers \(c_1, c_2, c_3, \cdots\) that converges to a solution of the equation. That is, if \(\alpha\) is a solution (i.e., \(f(\alpha) = 0\)) then,  \[\lim_{n\to\infty}c_n = \alpha,\] and this sequence \(\{c_n\}\) is generated by a series of linear approximations of the function \(f(x)\). Theorem (Newton's method) Let \(f(x)\) be a function that is twice differentiable on an open interval \(I\) that contains the closed interval \([a, b]\) (i.e., \([a,b]\subset I\)) and satisfy the following conditions: \(f(a) < 0\) and \(f(b) > 0\); For all \(x\in [a, b]\), \(f'(x) > 0\) and \(f''(x) > 0\). Let us define the sequence \(\{c_n\}\) by \[ \begin{eqnarray} c_1 &
Read more