/*! 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 44 (a) Let \(g(x)\) be the sum of t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) Let \(g(x)\) be the sum of the first two nonzero terms of the Maclaurin series for \(f(x) .\) Use \(g(x)\) to approximate \(\int_{0}^{1} f(x) d x\) and \(\int_{1}^{2} f(x) d x .\) (b) Graph, on the same coordinate axes, \(f\) and \(g\) for \(0 \leq x \leq 2,\) and then use the graphs to compare the accuracy of the approximations in (a). \(f(x)=\sinh x\)

Short Answer

Expert verified
The approximation is more accurate for \(\int_{0}^{1} f(x) \, dx\) than for \(\int_{1}^{2} f(x) \, dx\).

Step by step solution

01

Find the Maclaurin Series for \(f(x)\)

The Maclaurin series for \(f(x) = \sinh x\) is given by:\[\sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots\]To identify the first two non-zero terms, we consider:\[g(x) = x + \frac{x^3}{6}.\]
02

Approximate \(\int_{0}^{1} f(x) \, dx\) Using \(g(x)\)

We approximate the integral \(\int_{0}^{1} f(x) \, dx\) using:\[g(x) = x + \frac{x^3}{6}.\]Thus, the integral becomes:\[\int_{0}^{1} \left(x + \frac{x^3}{6}\right) \, dx = \left[ \frac{x^2}{2} + \frac{x^4}{24} \right]_{0}^{1} = \frac{1}{2} + \frac{1}{24} = \frac{12}{24} + \frac{1}{24} = \frac{13}{24}.\]
03

Approximate \(\int_{1}^{2} f(x) \, dx\) Using \(g(x)\)

Similarly, approximate the integral \(\int_{1}^{2} f(x) \, dx\) using:\[g(x) = x + \frac{x^3}{6}.\]Calculate:\[\int_{1}^{2} \left(x + \frac{x^3}{6}\right) \, dx = \left[ \frac{x^2}{2} + \frac{x^4}{24} \right]_{1}^{2}.\]Evaluate the limits:\[= \left(\frac{2^2}{2} + \frac{2^4}{24}\right) - \left(\frac{1^2}{2} + \frac{1^4}{24}\right) = \left(2 + \frac{16}{24}\right) - \left(\frac{1}{2} + \frac{1}{24}\right) \]\[= \left(2 + \frac{2}{3}\right) - \left(\frac{1}{2} + \frac{1}{24}\right) = \frac{20}{12} - \frac{13}{24} = \frac{40}{24} - \frac{13}{24} = \frac{27}{24} = \frac{9}{8}.\]
04

Graph \(f(x)=\sinh x\) and \(g(x)\)

Graph \(f(x) = \sinh x\) and \(g(x) = x + \frac{x^3}{6}\) on the same axes for the interval \(0 \leq x \leq 2\). Observe how \(g(x)\) approximates \(f(x)\) best near the origin (x = 0) but deviates as x increases past 1. This indicates the approximation is more accurate for \(\int_{0}^{1} f(x) \, dx\) compared to \(\int_{1}^{2} f(x) \, dx\).

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.

Understanding the Sinh Function
The sinh function, or hyperbolic sine function, is an important mathematical function appearing frequently in calculus and related fields. It is defined as \[ \sinh x = \frac{e^x - e^{-x}}{2}, \]where \( e \) is the base of the natural logarithm. This function is quite similar to the more familiar sine function in trigonometry, but instead of using angles, it applies to real numbers. The function has the following properties:
  • It is odd, meaning that \( \sinh(-x) = -\sinh(x) \).
  • Its graph is unbounded, continuously increasing from negative to positive infinity as \( x \) grows.
  • It has a critical point at the origin where \( \sinh(0) = 0 \).
Beyond its basic definition, the sinh function appears in many areas like solving differential equations, especially when the physical context involves hyperbolic geometry or certain exponential growth models. It is also valuable in function approximation using series such as the Maclaurin series.
Calculating and Understanding Definite Integrals
Definite integrals allow us to calculate the accumulated value of a function over a specified interval. When we set up an integral such as \[ \int_{a}^{b} f(x) \, dx, \]it represents the area under the curve of the function \( f(x) \) from \( x = a \) to \( x = b \). Here's how definite integrals work:
  • The limits of integration, \( a \) and \( b \), define the interval over which we compute the area.
  • We evaluate the antiderivative of \( f(x) \), known as a primary 'integrating' function.
  • The antiderivative is applied at the upper and lower bounds, and their difference provides the integral's value.
In our problem, we approximate integrals using the simpler approximation \( g(x) \) instead of the original, more complex function \( \sinh x \). This method facilitates calculations, especially when precisely evaluating the sinh function's integral directly can be challenging or requires numerical methods.
Exploring Function Approximation with Maclaurin Series
Function approximation is an essential tool when dealing with complex functions, which can be difficult to evaluate directly. The Maclaurin series offers one such approach. It expresses a function as a power series, expanding around zero. For \( \sinh x \), the Maclaurin series is given by:\[ \sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots \]To approximate the function over a small range, we only consider the leading terms, usually discarding higher-order components. This yields a simpler function that is easier to integrate or manage, particularly when precision constraints are not strict.
  • This approach is powerful as it enables practical calculations of integrals or derivatives for intricate functions.
  • In this exercise, using just the first two non-zero terms, \( x + \frac{x^3}{6} \), offers sufficient accuracy for integration purposes over a modest interval.
  • The closer x is to 0, the better this approximation reflects the true function, often graphically observed as well.
Function approximation like this is crucial in mathematical modeling, simplifying calculations while providing practical results for engineering, physics, and computational applications.

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

Determine whether the series converges or diverges. \(\sum_{n=1}^{x} 3^{1 / n}\)

If a dosage of \(Q\) units of a certain drug is administered to an individual, then the amount remaining in the bloodstream at the end of \(t\) minutes is given by \(Q e^{-c t}\) where \(c>0\). Suppose this same dosage is given at successive \(T\) -minute intervals. (a) Show that the amount \(A(k)\) of the drug in the bloodstream immediately after the \(k\) th dose is given by \(A(k)=\sum_{n=0}^{k-1} Q e^{-n c T}\) (b) Find an upper bound for the amount of the drug in the bloodstream after any number of doses. (c) Find the smallest time between doses that will ensure that \(A(k)\) does not exceed a certain level \(M\) for \(M>Q\)

If the series is positive-term, determine whether it is convergent or divergent; if the series contains negative terms, determine whether it is absolutely convergent, conditionally convergent, or divergent. $$ \sum_{n=1}^{\infty}(-1)^{n} \frac{2 n+3}{n !} $$

If the series is positive-term, determine whether it is convergent or divergent; if the series contains negative terms, determine whether it is absolutely convergent, conditionally convergent, or divergent. $$ \sum_{n=0}^{x} \frac{1}{2+\left(\frac{1}{2}\right)^{n}} $$

Suppose that each dollar introduced into the economy recirculates as follows: \(85 \%\) of the original dollar is spent, then \(85 \%\) of that \(\$ 0.85\) is spent, and so on. Find the economic impact (the total amount spent) if \(\$ 1,000,000\) is introduced into the economy.

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.