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Chapter 15: Problem 72

Improper double integrals can often be computed similarly to improper integrals of one variable. The first iteration of the following improper integrals is conducted just as if they were proper integrals. One then evaluates an improper integral of a single variable by taking appropriate limits, as in Section \(8.8 .\) Evaluate the improper integrals as iterated integrals. $$\int_{0}^{\infty} \int_{0}^{\infty} x e^{-(x+2 y)} d x d y$$

Short Answer

Expert verified
The integral evaluates to \( \frac{1}{2} \).

Step by step solution

01

Set Up the Iterated Integral

First, let's set up the given double integral as an iterated integral. We are given \( \int_{0}^{\infty} \int_{0}^{\infty} x e^{-(x+2y)} \, dx \, dy \). This means we will first integrate with respect to \(x\) while treating \(y\) as a constant, and then integrate the resulting expression with respect to \(y\).
02

Integrate with respect to x

We start by integrating \( x e^{-(x+2y)} \) with respect to \(x\). This becomes:\[\int_{0}^{\infty} x e^{-(x+2y)} \, dx\]Use integration by parts where \( u = x \) and \( dv = e^{-(x+2y)} \, dx \). Then \( du = dx \) and \( v = -\frac{1}{1} e^{-(x+2y)} \). The integration by parts formula \( \int u \, dv = uv - \int v \, du \) gives:\[-x e^{-(x+2y)} \bigg|_0^\infty + \int_0^\infty e^{-(x+2y)} \, dx\]The boundary term \(-x e^{-(x+2y)} \bigg|_0^\infty = 0\) because as \(x \to \infty\), the exponential decay makes \(x e^{-(x+2y)} \to 0\), and when \(x = 0\), it is also zero. Thus, this reduces to:\[\int_0^\infty e^{-(x+2y)} \, dx = \frac{1}{1} e^{-(x+2y)} \bigg|_0^\infty\]Evaluating this integral from 0 to \(\infty\) gives \(\frac{1}{1} (0 - e^{-2y}) \), which simplifies to \(-e^{-2y}\).
03

Integrate with respect to y

Now, substitute the result \(-e^{-2y}\) back into the outer integral:\[\int_{0}^{\infty} -e^{-2y} \, dy\]Elementarily integrate \(-e^{-2y}\) with respect to \(y\), which becomes:\[-\int_{0}^{\infty} e^{-2y} \, dy = \left[-\frac{1}{2} e^{-2y} \right]_{0}^{\infty}\]When we evaluate this from 0 to \(\infty\), it results in:\[-0 - \left(-\frac{1}{2}\right) = \frac{1}{2}\].
04

Write the Final Result

The final value of the improper double integral \( \int_{0}^{\infty} \int_{0}^{\infty} x e^{-(x+2y)} \, dx \, dy \) is \( \frac{1}{2} \). Thus, the integral converges to \( \frac{1}{2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Iterated Integration
Iterated integration is a method used when dealing with multiple integrals, where you break them down into simpler, single-variable integrals. In this exercise, we are dealing with a double integral of the form \( \int_{0}^{\infty} \int_{0}^{\infty} xe^{-(x+2y)} \, dx \, dy \). Here, the goal is to evaluate the inner integral with respect to \(x\) first, treating \(y\) as a constant.
The process involves treating one variable as a constant and performing the integration, thereby transforming the double integral into a series of single-variable integrations.
  • First, integrate with respect to \(x\), while \(y\) remains constant.
  • Then, use the result from the inner integral to perform the integration with respect to \(y\).
Each step of integration simplifies the problem until all variables have been integrated. It's a systematic approach to solving multi-variable integrals.
Integration by Parts
Integration by parts is a technique derived from the product rule for differentiation. It's very useful when you need to integrate products of functions. The formula is \( \int u \, dv = uv - \int v \, du \).

In this context, we applied integration by parts to the inner integral \( \int_{0}^{\infty} x e^{-(x+2y)} \, dx \). We need to choose the functions appropriately:
  • Let \( u = x \) which means \( du = dx \).
  • Let \( dv = e^{-(x+2y)} \, dx \) which gives \( v = -e^{-(x+2y)} \).
By substituting these into the formula, we perform the integration of each part. The choice of \(u\) and \(dv\) is important because it simplifies the integration process. Once you've set up the parts, you reduce the problem to simpler integrals and apply limits.
Exponential Functions
Exponential functions often appear in calculus and are characterized by expressions of the form \(e^{u}\), where \(e\) is the base of the natural logarithm. These functions need special attention when integrating, especially when they span infinite limits like the improper integral in this exercise.

The integral \( \int_{0}^{\infty} e^{-(x+2y)} \, dx \) deals with an exponential decay function. Such functions approach zero as the exponent increases, which is crucial for evaluating limits in improper integrals. Here, evaluating \( \lim_{x \to \infty} x e^{-(x+2y)} = 0 \) helps in simplifying the integral boundaries.
  • Exponential decay, \(e^{-u}\), approaches zero as \(u\) increases.
  • It's important in improper integrals since it ensures convergence for evaluation over infinite intervals.
This behavior of exponential functions is exploited in improper integrals to reach a definitive limiting value.

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Most popular questions from this chapter

Use a CAS to change the Cartesian integrals into an equivalent polar integral and evaluate the polar integral. Perform the following steps in each exercise. a. Plot the Cartesian region of integration in the \(x y\) -plane. b. Change each boundary curve of the Cartesian region in part (a) to its polar representation by solving its Cartesian equation for \(r\) and \(\theta\) c. Using the results in part (b), plot the polar region of integration in the \(r \theta\) -plane. d. Change the integrand from Cartesian to polar coordinates. Determine the limits of integration from your plot in part (c) and evaluate the polar integral using the CAS integration utility. $$\int_{0}^{1} \int_{y}^{2-y} \sqrt{x+y} d x d y$$

Sketch the region of integration and write an equivalent double integral with the order of integration reversed. $$\int_{0}^{1} \int_{1-x}^{1-x^{2}} d y d x$$

Sketch the region of integration, reverse the order of integration, and evaluate the integral. $$\int_{0}^{2} \int_{x}^{2} 2 y^{2} \sin x y d y d x$$

Let \(P_{0}\) be a point inside a circle of radius \(a\) and let \(h\) denote the distance from \(P_{0}\) to the center of the circle. Let \(d\) denote the distance from an arbitrary point \(P\) to \(P_{0} .\) Find the average value of \(d^{2}\) over the region enclosed by the circle. (Hint: Simplify your work by placing the center of the circle at the origin and \(P_{0}\) on the \(x\) -axis.)

Sketch the region of integration and the solid whose volume is given by the double integral. $$\int_{0}^{4} \int_{-\sqrt{16-y^{2}}}^{\sqrt{16-y^{2}}} \sqrt{25-x^{2}-y^{2}} d x d y$$
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