/*! 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 53 Evaluate the following definite ... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the following definite integrals. $$\int_{0}^{1 / 3} \frac{d x}{\left(9 x^{2}+1\right)^{3 / 2}}$$

Short Answer

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Question: Evaluate the definite integral $$\int_{0}^{1 / 3} \frac{d x}{\left(9 x^{2}+1\right)^{3 / 2}}$$. Answer: The value of the definite integral is \(\frac{\sqrt{2}}{6}\).

Step by step solution

01

Trigonometric substitution

Let \(x = \frac{1}{3} \tan{\theta}\). Then, \(dx = \frac{1}{3} \sec^2{\theta} d\theta\). Now substitute this into the integral and simplify: $$\int_{0}^{1 / 3} \frac{d x}{\left(9 x^{2}+1\right)^{3 / 2}} = \int_{0}^{\pi / 4} \frac{\frac{1}{3}\sec^2{\theta}\, d\theta}{\left(9(\frac{1}{9}\tan^2{\theta})+1\right)^{3 / 2}}$$
02

Simplify the integrand

Observe that the \((9 x^2 + 1)\) simplifies to \((1 + \tan^2{\theta})\): $$\int_{0}^{\pi / 4} \frac{\frac{1}{3}\sec^2{\theta}\, d\theta}{\left(1+\tan^2{\theta}\right)^{3 / 2}}$$ Now, note that \(\sec^2{\theta} = 1 + \tan^2{\theta}\). So the integrand simplifies even further: $$\int_{0}^{\pi / 4} \frac{\frac{1}{3}(1+\tan^2{\theta})\, d\theta}{(1+\tan^2{\theta})^{3 / 2}} = \frac{1}{3}\int_{0}^{\pi / 4} \frac{d\theta}{\sqrt{1+\tan^2{\theta}}}$$
03

Evaluate the integral

We now have the integral in a simpler form that we can evaluate. Recall that \(\sqrt{1+\tan^2{\theta}} = \sec{\theta}\): $$\frac{1}{3}\int_{0}^{\pi / 4} \frac{d\theta}{\sec{\theta}} = \frac{1}{3}\int_{0}^{\pi / 4}\cos{\theta}\, d\theta$$ Now, integrate with respect to \(\theta\): $$\frac{1}{3}[\sin{\theta}]_{0}^{\pi / 4} = \frac{1}{3}\left(\sin{\frac{\pi}{4}} - \sin{0}\right)$$
04

Calculate the definite integral's value

Evaluate the trigonometric functions and simplify: $$\frac{1}{3}\left(\sin{\frac{\pi}{4}} - 0\right) = \frac{1}{3}\left(\frac{\sqrt{2}}{2}\right) = \boxed{\frac{\sqrt{2}}{6}}$$ Therefore, the value of the definite integral is \(\frac{\sqrt{2}}{6}\).

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Key Concepts

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

Trigonometric Substitution
Trigonometric substitution is a technique in integral calculus used to simplify and evaluate integrals involving expressions like \(\sqrt{a^2 - x^2}\), \(\sqrt{a^2 + x^2}\), or \(\sqrt{x^2 - a^2}\). It involves substituting one trigonometric function for another variable to transform a seemingly complex integrand into a more manageable form.

In this exercise, we employed trigonometric substitution by setting \(x = \frac{1}{3} \tan{\theta}\). This choice was made because the integrand has the form \(\sqrt{9x^2 + 1}\), which fits the \(\sqrt{a^2 + x^2}\) pattern well when using tangent substitution. As a result, we were able to reduce the complexity of the integral by transforming variables into trigonometric ones.

To implement trigonometric substitution successfully:
  • Choose the appropriate trigonometric function based on the form of the expression under the square root.
  • Determine \(dx\) in terms of \(d\theta\) by differentiating the substitution.
  • Adjust the limits of integration due to the new variable \(\theta\).
These steps enable the simplification of the original integral, making it easier to solve.
Integral Calculus
Integral calculus focuses on the concept of integration, which is finding the whole from the parts. It is the reverse process of differentiation, assigning an accumulated quantity to the rate of change. It is frequently used to calculate areas under curves, volumes, and other quantities described by continuous functions.

In the example problem, the goal was to evaluate a definite integral, which represents the actual computation of the integral's value over the specified interval \([0, 1/3]\). Integral calculus uses techniques like substitution and integration by parts to handle different types of integrands effectively.

Key principles in integral calculus that could be useful include:
  • The Fundamental Theorem of Calculus, linking differentiation and integration.
  • Substitution methods, like the trigonometric substitution used here, to simplify integrals.
  • Integration techniques used for evaluating specific types of integrals.
Understanding these principles helps in tackling various integral problems, whether indefinite or definite, by breaking them down into steps that are easier to manage.
Evaluating Integrals
Evaluating integrals is the process of finding the definite or indefinite integral of a function. In the context of definite integrals, it involves calculating a numerical value representing the area under a curve within given bounds.

For the integral \( \int_{0}^{1/3} \frac{d x}{(9x^2 + 1)^{3/2}} \), after employing trigonometric substitution and simplification, we transformed it into an integral involving \( \cos{\theta} \) that can be directly integrated. This allowed us to compute it as an elementary integral:
  • The substitution changed the limits of integration to correspond to the new variable \( \theta \), resulting in bounds from 0 to \( \pi/4 \).
  • The simplification resulted in the integral \( \frac{1}{3}\int_{0}^{\pi / 4} \cos{\theta} \, d\theta \).
  • The definite evaluation \( \frac{1}{3}[\sin{\theta}]_{0}^{\pi / 4} \) gives the result by directly substituting the bounds.
Becoming comfortable with evaluating integrals involves practice in recognizing common patterns (like trigonometric forms) and utilizing appropriate methods to simplify the task. In the end, this skill aids in the accurate computation of definite areas, changes, or any represented quantities of interest.

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Most popular questions from this chapter

The work required to launch an object from the surface of Earth to outer space is given by \(W=\int_{R}^{\infty} F(x) d x,\) where \(R=6370 \mathrm{km}\) is the approximate radius of Earth, \(F(x)=G M m / x^{2}\) is the gravitational force between Earth and the object, \(G\) is the gravitational constant, \(M\) is the mass of Earth, \(m\) is the mass of the object, and \(G M=4 \times 10^{14} \mathrm{m}^{3} / \mathrm{s}^{2}.\) a. Find the work required to launch an object in terms of \(m.\) b. What escape velocity \(v_{e}\) is required to give the object a kinetic energy \(\frac{1}{2} m v_{e}^{2}\) equal to \(W ?\) c. The French scientist Laplace anticipated the existence of black holes in the 18th century with the following argument: If a body has an escape velocity that equals or exceeds the speed of light, \(c=300,000 \mathrm{km} / \mathrm{s},\) then light cannot escape the body and it cannot be seen. Show that such a body has a radius \(R \leq 2 G M / c^{2} .\) For Earth to be a black hole, what would its radius need to be?

Find the volume of the described solid of revolution or state that it does not exist. The region bounded by \(f(x)=\left(x^{2}-1\right)^{-1 / 4}\) and the \(x\) -axis on the interval (1,2] is revolved about the \(y\) -axis.

The nucleus of an atom is positively charged because it consists of positively charged protons and uncharged neutrons. To bring a free proton toward a nucleus, a repulsive force \(F(r)=k q Q / r^{2}\) must be overcome, where \(q=1.6 \times 10^{-19} \mathrm{C}\) is the charge on the proton, \(k=9 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}, Q\) is the charge on the nucleus, and \(r\) is the distance between the center of the nucleus and the proton. Find the work required to bring a free proton (assumed to be a point mass) from a large distance \((r \rightarrow \infty)\) to the edge of a nucleus that has a charge \(Q=50 q\) and a radius of \(6 \times 10^{-11} \mathrm{m}.\)

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