/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 The integrals converge. Evaluate... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The integrals converge. Evaluate the integrals without using tables. $$\int_{0}^{\infty} 2 e^{-\theta} \sin \theta d \theta$$

Short Answer

Expert verified
The integral evaluates to -1.

Step by step solution

01

Recognize the Integral Type

The given problem is to evaluate the integral \( \int_{0}^{\infty} 2 e^{-\theta} \sin \theta \, d\theta \). Notice that this is an integral of the form \( \int_{0}^{\infty} e^{-ax} \sin(bx) \, dx \), which often uses integration by parts.
02

Choose Substitution for Integration by Parts

Integration by parts is given by \( \int u \, dv = uv - \int v \, du \). Let \( u = \sin \theta \) and \( dv = 2e^{-\theta} \, d\theta \). This choice allows \( du = \cos \theta \, d\theta \) and \( v = -2e^{-\theta} \).
03

Apply Integration by Parts Formula

Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula: \( \int 2e^{-\theta} \sin \theta \, d\theta = -2e^{-\theta} \sin \theta \Big|_{0}^{\infty} + \int 2e^{-\theta} \cos \theta \, d\theta \).
04

Evaluate the Boundary Terms

The boundary term \( -2e^{-\theta} \sin \theta \Big|_{0}^{\infty} \) results in 0 when evaluated: \[ \lim_{b \to \infty} -2e^{-b} \sin b - (-2e^{0} \sin 0) = 0 \].
05

Evaluate the Second Integral with Integration by Parts Again

For \( \int 2e^{-\theta} \cos \theta \, d\theta \), use integration by parts again with \( u = \cos \theta \), \( dv = 2e^{-\theta} \, d\theta \), \( du = -\sin \theta \, d\theta \), \( v = -2e^{-\theta} \). This gives: \( -2e^{-\theta} \cos \theta \Big|_{0}^{\infty} - \int (-2)(-\sin \theta)e^{-\theta} \, d\theta \).
06

Substitute and Simplify

The boundary term \( -2e^{-\theta} \cos \theta \Big|_{0}^{\infty} \) results in: \( \lim_{b \to \infty} -2e^{-b} \cos b - (-2e^{0} \cos 0) = -2 \). The entire expression simplifies as \( \int 2e^{-\theta} \cos \theta \, d\theta = 0 \) when summed with the previous boundary terms.
07

Calculate the Final Result

Summing all contributions, the integral evaluates to \( -2 + 0 = -2 \). Hence, \( \int_{0}^{\infty} 2 e^{-\theta} \sin \theta \, d\theta = -1 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Integration by Parts
Integration by parts is a powerful technique used to solve integrals involving a product of functions. This method is based on the product rule for differentiation and is particularly useful when dealing with exponential and trigonometric functions, as seen in our exercise.
To apply integration by parts, we use the formula:
  • \( \int u \, dv = uv - \int v \, du \)
Here, it's crucial to choose \( u \) and \( dv \) in such a way that\( du\) and \(v \) are easier to integrate or differentiate.
In our problem, we start by identifying \(u = \sin \theta\) and \( dv = 2e^{-\theta} \, d\theta \). This choice helps us by yielding \( du = \cos \theta \, d\theta \) and \( v = -2e^{-\theta} \).
Thus, we systematically use integration by parts twice, as re-application is often necessary when the resulting integral resembles the original. This recursive process allows us to simplify the integral step by step until a solution is reached.
Convergence of Integrals
Understanding the convergence of integrals, especially improper ones extending to infinity, is a key aspect of calculus. Integrals like \( \int_{0}^{\infty} 2 e^{-\theta} \sin \theta \, d\theta \) require careful evaluation limits to determine convergence or divergence.
Improper integrals are approached by considering limits. Here, we evaluate the integral \( \int_{0}^{b} 2 e^{-\theta} \sin \theta \, d\theta \) as \( b \) approaches infinity.
It is essential to look at the boundary terms after applying integration methods, as they can greatly influence the integral's behavior. In our case, as \( b \to \infty \), the boundary parts like \( -2e^{-b} \sin b \) vanish due to exponential decay. This ensures the integral converges, which is affirmed as accidental infinite values do not affect the final result.
Exponential and Trigonometric Functions
The interplay between exponential and trigonometric functions often appears in calculus problems, owing to their unique properties. These functions complement each other with characteristics like exponential growth/decay and periodicity.
In our integral, \( 2 e^{-\theta} \) is an exponential function representing decay. Its role is significant because it diminishes terms as \( \theta \to \infty \), aiding in convergence.
The trigonometric function \( \sin \theta \) introduces periodicity. When matched with an exponential function, the resulting behavior can make evaluating the integral straightforward if one carefully considers the functions' properties.
Together, these functions necessitate the use of techniques like integration by parts to handle their product effectively. The exponential function ensures that, over infinite bounds, possible oscillations in trigonometric functions diminish, further securing the convergence of the integral.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The infinite paint can or Gabriel's horn As Example 3 shows, the integral \(\int_{1}^{\infty}(d x / x)\) diverges. This means that the integral $$\int_{1}^{\infty} 2 \pi \frac{1}{x} \sqrt{1+\frac{1}{x^{4}}} d x, $$ which measures the surface area of the solid of revolution traced out by revolving the curve \(y=1 / x, 1 \leq x,\) about the \(x\) -axis, diverges also. By comparing the two integrals, we see that, for every finite value \(b>1\) , $$\int_{1}^{b} 2 \pi \frac{1}{x} \sqrt{1+\frac{1}{x^{4}}} d x>2 \pi \int_{1}^{b} \frac{1}{x} d x.$$ However, the integral $$\int_{1}^{\infty} \pi\left(\frac{1}{x}\right)^{2} d x$$ for the volume of the solid converges. \begin{equation} \begin{array}{l}{\text { a. Calculate it. }} \\ {\text { b. This solid of revolution is sometimes described as a can that }} \\ \quad {\text { does not hold enough paint to cover its own interior. Think }} \\ \quad {\text { about that for a moment. It is common sense that a finite }} \\ \quad {\text { amount of paint cannot cover an infinite surface. But if we fill }} \\ \quad {\text { the horn with paint (a finite amount), then we will have covered }} \\\ \quad {\text { an infinite surface. Explain the apparent contradiction. }} \end{array} \end{equation}

About the infinite region in the first quadrant between the curve \(y=e^{-x}\) and the \(x\) -axis. Find the centroid of the region.

Elliptic integrals The length of the ellipse $$ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $$ turns out to be $$Length =4 a \int_{0}^{\pi / 2} \sqrt{1-e^{2} \cos ^{2} t} d t$$ where \(e=\sqrt{a^{2}-b^{2}} / a\) is the ellipse's eccentricity. The integral in this formula, called an elliptic integral, is nonelementary except when \(e=0\) or \(1 .\) a. Use the Trapezoidal Rule with \(n=10\) to estimate the length of the ellipse when \(a=1\) and \(e=1 / 2\) . b. Use the fact that the absolute value of the second derivative of \(f(t)=\sqrt{1-e^{2} \cos ^{2} t}\) is less than 1 to find an upper bound for the error in the estimate you obtained in part (a).

Cholesterol levels The serum cholesterol levels of children aged 12 to 14 years follows a normal distribution with mean \(\mu=162\) mg/dl and standard deviation \(\sigma=28 \mathrm{mg} / \mathrm{dl} .\) In a population of 1000 of these children, how many would you expect to have serum cholesterol levels between 165 and 193\(?\) between 148 and 167\(?\)

Pollinating flowers A biologist models the time in minutes until a bee arrives at a flowering plant with an exponential distribution having a mean of 4 minutes. If 1000 flowers are in a field, how many can be expected to be pollinated within 5 minutes?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.