/*! 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 47 Use series to approximate the de... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use series to approximate the definite integral to within the indicated accuracy. $$\int_{0}^{1} x \cos \left(x^{3}\right) d x \quad(\text { three decimal places })$$

Short Answer

Expert verified
The integral evaluates to approximately 0.441.

Step by step solution

01

Understand the Problem

We need to approximate the integral \( \int_{0}^{1} x \cos\left(x^{3}\right) \, dx \) to an accuracy of three decimal places. The function inside the integral, \( x \cos\left(x^{3}\right) \), suggests that we can use the Taylor series for \( \cos(u) \), where \( u = x^3 \), to simplify the integration.
02

Expand the Function Using Taylor Series

The cosine function can be expanded using the Taylor series: \( \cos(u) = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \cdots \). Setting \( u = x^3 \), the series becomes: \( \cos(x^3) = 1 - \frac{(x^3)^2}{2} + \frac{(x^3)^4}{4!} - \cdots \).
03

Insert the Series into the Integral

Substituting the Taylor series into the integral gives: \( \int_{0}^{1} x \left( 1 - \frac{x^6}{2} + \frac{x^{12}}{4!} - \cdots \right) \, dx \). This simplifies to: \( \int_{0}^{1} \left( x - \frac{x^7}{2} + \frac{x^{13}}{4!} - \cdots \right) \, dx \).
04

Compute the Integral Term-by-Term

Integrate each term separately:- \( \int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1}{2} \).- \( \int_{0}^{1} \frac{x^7}{2} \, dx = \left[ \frac{x^8}{16} \right]_{0}^{1} = \frac{1}{16} \).- \( \int_{0}^{1} \frac{x^{13}}{24} \, dx = \left[ \frac{x^{14}}{336} \right]_{0}^{1} = \frac{1}{336} \).
05

Sum the Series to Approximated Value

Approximate the value using the first few terms:\[ \frac{1}{2} - \frac{1}{16} + \frac{1}{336} \approx 0.5000 - 0.0625 + 0.0030 \approx 0.4405 \]. Adding more terms if needed, check that further terms decrease the value negligibly for three decimal accuracy.

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.

Taylor Series
The Taylor Series is a powerful mathematical tool that allows us to approximate functions using an infinite sum of terms calculated from the values of the function's derivatives at a single point. In the context of this problem, we use the Taylor series to express the function \(\cos(u)\) where \(u = x^3\). By expanding \(\cos(u)\) as a Taylor series, we obtain:
  • \( \cos(u) = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \cdots \).
For our integral, this becomes \(\cos(x^3) = 1 - \frac{(x^3)^2}{2} + \frac{(x^3)^4}{4!} - \cdots \).
Substituting this series into the integral allows each term to be integrated separately, providing an approximation of the integral's value.
The key advantage here is that we can choose the number of terms in the series based on the desired level of accuracy. The more terms we use, the closer we get to the true value, but even a few terms can offer a good approximation for many practical purposes.
Definite Integral
A Definite Integral represents the area under the curve of a function over a specific interval. It is represented by:
  • \( \int_{a}^{b} f(x) \, dx \), where \(a\) and \(b\) are the limits of integration.
In this exercise, the definite integral is \( \int_{0}^{1} x \cos(x^3) \, dx \).
We are not looking for an exact symbolic answer here, but an approximation to three decimal places.
By using the Taylor series to simplify the function inside the integral, this problem becomes manageable. Each term in the series forms its own simpler integral, which can be calculated independently.
The result of each of these integrals will then be summed to approximate the value of the original integral.
Integration
Integration is the process of finding the integral of a function, which can be thought of as finding the total accumulation, or area under the curve, of that function. In a practical sense, integration undoes differentiation.
  • When approaching integration with a series, such as in this problem, each term of the series is integrated separately.
  • This term-by-term integration takes advantage of the linearity of the integral and simplifies complicated functions to series of basic polynomials.
Here, each term from the Taylor series expansion of \( \cos(x^3) \) is integrated separately over the interval from 0 to 1.
If we consider \( x - \frac{x^7}{2} + \frac{x^{13}}{24} - \cdots \), each component is essentially a polynomial function that can be integrated using known rules.
For example, integrating the first term gives \( \frac{x^2}{2} \), evaluated from 0 to 1. Each higher-degree polynomial term is similarly integrated and then substituted back into the original definite integral limits.
This approach drastically simplifies the integration and allows us to find an approximate numerical result to the desired level of accuracy.

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

In Section 4.6 we considered Newton's method for approxi- mating a root \(r\) of the equation \(f(x)=0,\) and from an initial approximation \(x_{1}\) we obtained successive approximations \(x_{2}, x_{3}, \ldots,\) where $$x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}$$ Use Taylor's Formula with \(n=1, a=x_{n},\) and \(x=r\) to show that if \(f^{\prime \prime}(x)\) exists on an interval \(I\) containing \(r, x_{n}\) and \(x_{n+1},\) and \(\left|f^{\prime \prime}(x)\right| \leqslant M,\left|f^{\prime}(x)\right| \geqslant K\) for all \(x \in I,\) then $$ \left|x_{n+1}-r\right| \leqslant \frac{M}{2 K}\left|x_{n}-r\right|^{2}$$ [This means that if \(x_{n}\) is accurate to \(d\) decimal places, then \(x_{n+1}\) is accurate to about 2\(d\) decimal places. More precisely,if the error at stage \(n\) is at most \(10^{-m}\) , then the error at stage \(n+1\) is at most \((M / 2 K) 10^{-2 m} . ]\)

If the radius of convergence of the power series \(\Sigma_{n-0}^{\infty} c_{n} x^{n}\) is \(10,\) what is the radius of convergence of the series \(\sum_{n=1}^{\infty} n c_{n} x^{n-1} ?\) Why?

The resistivity \(\rho\) of a conducting wire is the reciprocal of the conductivity and is measured in units of ohm-meters \((\Omega-m) .\) The resistivity of a given metal depends on the temperature according to the equation $$\rho(t)=\rho_{20} e^{\alpha(t-20)}$$ where \(t\) is the temperature in \(^{\circ} \mathrm{C} .\) There are tables that list the values of \(\alpha\) (called the temperature coefficient) and \(\rho_{20}\) (the resistivity at \(20^{\circ} \mathrm{C} )\) for various metals. Except at very low temperatures, the resistivity varies almost linearly with temperature and so it is common to approximate the expression for \(\rho(t)\) by its first- or second-degree Taylor polynomial at \(t=20\) . (a) Find expressions for these linear and quadratic approximations. (b) For copper, the tables give \(\alpha=0.0039 /^{\circ} \mathrm{C}\) and \(\rho_{20}=1.7 \times 10^{-8} \Omega-\mathrm{m} .\) Graph the resistivity of copper and the linear and quadratic approximations for \(-250^{\circ} \mathrm{C} \leqslant t \leqslant 1000^{\circ} \mathrm{C}\) (c) For what values of \(t\) does the linear approximation agree with the exponential expression to within one percent?

Find all positive values of \(b\) for which the series \(\sum_{n=1}^{\infty} b^{\ln n}\) converges.

Evaluate the indefinite integral as a power series. What is the radius of convergence? $$ \int x^{2} \ln (1+x) d x $$

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.