/*! 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 14 Converge. Evaluate the integrals... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Converge. Evaluate the integrals without using tables. $$\int_{-\infty}^{\infty} \frac{x d x}{\left(x^{2}+4\right)^{3 / 2}}$$

Short Answer

Expert verified
The integral is zero because the function is odd and the limits are symmetric.

Step by step solution

01

Identify the symmetry of the integrand

Notice that the integrand, \( \frac{x}{(x^2 + 4)^{3/2}} \), is an odd function because if you replace \( x \) with \( -x \), the integrand becomes \( -\frac{x}{(x^2 + 4)^{3/2}} \).
02

Apply the property of integration of odd functions

Given that the function is odd, and the limits of integration are symmetric about the origin (from \( -\infty \) to \( \infty \)), the integral evaluates to zero. In general, the integral of an odd function over a symmetric interval is zero.

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.

Odd function
An odd function is a crucial concept in calculus, particularly when dealing with integrals. A function is defined as odd if, for every point \( x \), the equation \( f(-x) = -f(x) \) holds true. This concept plays a significant role when evaluating integrals over symmetric intervals. Odd functions have fascinating properties, especially in integration, which often simplify the process.
For instance, consider the function \( f(x) = \frac{x}{(x^2 + 4)^{3/2}} \). By substituting \( -x \) for \( x \), the function transforms into \( -\frac{x}{(x^2 + 4)^{3/2}} \). This confirms that the function is odd. This property aids in reducing the complexity of calculating certain types of integrals over symmetric boundaries. Understanding this can streamline the approach to solving improper integrals.
Symmetrical integration
Symmetrical integration is a technique that leverages the symmetry of the function and the interval of integration to simplify calculations. When a function exhibits symmetry, it can make the integration process more straightforward, especially when dealing with odd and even functions.
Symmetrical integrals often appear in calculus problems with the interval \(-a\) to \(a\). If the function is odd, as in our example \(f(x) = \frac{x}{(x^2 + 4)^{3/2}}\), the integral over symmetric limits (e.g., \(-\infty\) to \(\infty\)) equals zero.
This occurs because the portions of the integral cancel out due to their symmetrical nature, simplifying to zero. Symmetrical integration serves as a powerful tool in evaluating integrals quickly, especially in cases of improper integrals.
Improper integrals
Improper integrals deal with integrals that have infinite limits or an unbounded integrand. They are called 'improper' because they extend the traditional definition of integrals. Calculating such integrals often involves limits, allowing us to analyze the area under curves that extend infinitely or have discontinuities.
In the context of our example, \( \int_{-\infty}^{\infty} \frac{x}{(x^2 + 4)^{3/2}} \, dx \), the improper nature comes from the infinite limits. To evaluate such integrals, it's crucial to analyze the function's behavior over its domain. Symmetry, especially the property of odd functions, can simplify solving these integrals tremendously. Using the symmetry of odd functions over symmetric limits often results in the integral equating to zero, as demonstrated here.
Function symmetry in calculus
Function symmetry in calculus refers to how functions behave when their input values are reversed. This symmetry falls into different categories: odd, even, and neither. Understanding these distinctions can significantly help in integrating functions, as different symmetries have implications on the ease of solving integrals.
Odd functions, as described earlier, mirror their output value when inputs flip signs. Even functions, by contrast, hold that \( f(x) = f(-x) \). These symmetries, especially when the integration limits are symmetric around the origin, reduce the complexity of evaluating integrals.
Utilizing symmetry, particularly in problems involving improper integrals or infinite limits, is a powerful strategy. It can turn otherwise complex calculations into straightforward evaluations, highlighting the elegance and utility of symmetry in calculus.

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

Find, to two decimal places, the \(x\) -coordinate of the centroid of the region in the first quadrant bounded by the \(x\) -axis, the curve \(y=\tan ^{-1} x,\) and the line \(x=\sqrt{3}\).

Use a CAS to explore the integrals for various values of \(p\) (include noninteger values). For what values of \(p\) does the integral converge? What is the value of the integral when it does converge? Plot the integrand for various values of \(p.\) $$\int_{e}^{\infty} x^{p} \ln x d x$$

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 \(b>1,\) $$\int_{1}^{b} 2 \pi \frac{1}{x} \sqrt{1+\frac{1}{x^{4}}} d x>2 \pi \int_{1}^{b} \frac{1}{x} d x$$ (Check your book to see image) However, the integral $$\int_{1}^{\infty} \pi\left(\frac{1}{x}\right)^{2} d x$$ for the volume of the solid converges. a. Calculate it. b. This solid of revolution is sometimes described as a can that does not hold enough paint to cover its own interior. Think about that for a moment. It is common sense that a finite amount of paint cannot cover an infinite surface. But if we fill the horn with paint (a finite amount), then we will have covered an infinite surface. Explain the apparent contradiction.

Evaluate the integrals in Exercises \(39-50\). $$\int \frac{1}{x^{6}\left(x^{5}+4\right)} d x$$

Integration by parts leads to a rule for integrating es that usually gives good results:$$\begin{aligned}\int f^{-1}(x) d x &=\int y f^{\prime}(y) d y \\\&=y f(y)-\int f(y) d y \\ &=x f^{-1}(x)-\int f(y) d y\end{aligned}$$. The idea is to take the most complicated part of the integral, in this case \(f^{-1}(x)\), and simplify it first. For the integral of \(\ln x,\) we get $$\begin{aligned}\int \ln x \, d x &=\int y e^{y} d y \\\&=y e^{y}-e^{y}+C \\\&=x \ln x-x+C \end{aligned}$$ For the integral of \(\cos ^{-1} x\) we get $$\begin{aligned}\int \cos ^{-1} x d x &=x \cos ^{-1} x-\int \cos y \, d y \\\&=x \cos ^{-1} x-\sin y+C \\\&=x \cos ^{-1} x-\sin \left(\cos ^{-1} x\right)+C\end{aligned}$$ Use the formula $$\int f^{-1}(x) d x=x f^{-1}(x)-\int f(y) d y$$ Express your answers in terms of \(x.\) $$\int \log _{2} x d x$$

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.