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Chapter 11: Problem 34

In Exercises \(33-36\) , use series to estimate the integrals' values with an error of magnitude less than \(10^{-3} .\) (The answer section gives the integrals' values rounded to five decimal places.) $$ \int_{0}^{0.2} \frac{e^{-x}-1}{x} d x $$

Short Answer

Expert verified
The integral value estimate is approximately -0.22133.

Step by step solution

01

Recognize the Series Representation

The integral involves a function that can be expanded as a series. Recognize that the exponential function \( e^{-x} \) can be represented as a Maclaurin series:\[e^{-x} = 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \cdots \]}},{
02

Simplify the Integrand

Use the series expansion of \( e^{-x} \) and simplify the integrand. The given function is \( \frac{e^{-x}-1}{x} \), so:\[ e^{-x} - 1 = -(x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots) \]Then, divide each term by \(x\), resulting in:\[ - \left( 1 + \frac{x}{2} + \frac{x^2}{6} + \cdots \right) \]
03

Integrate Term by Term

Integrate each term of the series from 0 to 0.2:\[\int_{0}^{0.2} - (1 + \frac{x}{2} + \frac{x^2}{6} + \cdots) \, dx = \int_{0}^{0.2} -1 \, dx - \int_{0}^{0.2} \frac{x}{2} \, dx - \int_{0}^{0.2} \frac{x^2}{6} \, dx -\cdots \]
04

Solve the Individual Integrals

Calculate each of the integrals:1. \( \int_{0}^{0.2} -1 \, dx = -[x]_{0}^{0.2} = -0.2 \)2. \( \int_{0}^{0.2} -\frac{x}{2} \, dx = -\frac{1}{2} [\frac{x^2}{2}]_{0}^{0.2} = -\frac{0.04}{2} = -0.02 \)3. \( \int_{0}^{0.2} -\frac{x^2}{6} \, dx = -\frac{1}{6} [\frac{x^3}{3}]_{0}^{0.2} = -0.00133 \)...These integrals are added term by term until the desired accuracy is reached.
05

Ensure Accuracy

Continue adding terms from the series until the cumulative error is less than \(10^{-3}\). For this example, the first three terms will already meet the required precision limit.
06

Sum the Results for the Final Estimate

Add together the results of the integrals to form the estimate:\[ -0.2 - 0.02 - 0.00133 = -0.22133 \]

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Key Concepts

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

Maclaurin series
The Maclaurin series is a special case of the Taylor series. It is used to represent functions as infinite sums of terms calculated from the derivatives of the function at zero. This makes it particularly powerful for estimating values and solving integrals when a function's exact expression is complex or unknown.

In the case of the function \( e^{-x} \), it can be expanded using its Maclaurin series:
  • Starts with the value 1.
  • Subtracts \( x \), which represents the first derivative term.
  • Further terms like \( \frac{x^2}{2} \) and \( -\frac{x^3}{6} \) are added sequentially.
This series expansion helps in expressing and simplifying otherwise complicated functions in a usable form for processes like integration, as seen in this exercise.
integral calculus
Integral calculus is a vast area of mathematics that focuses on the process of integration. Integration is used to find areas, volumes, central points, and many other useful things by summing continuous data points.

In this exercise, the objective is to calculate an integral:
  • The integral symbol \( \int \) indicates the process of finding the area under a curve.
  • The function \( \frac{e^{-x}-1}{x} \) is our target for the integration from 0 to 0.2.
  • After simplifying the function with the Maclaurin series, integrating term by term becomes manageable.
By meticulously integrating each term in the series, we accumulate a total sum that serves as an approximation of the integral's value up to a defined accuracy.
error analysis
Error analysis is crucial in numerical methods to ensure that the estimated result is adequately close to the exact value. When working with series approximations, each series term contributes to both the solution and potential errors.

In this problem, the strategy was to calculate terms until the size of added errors falls below a specified threshold (\(10^{-3}\)):
  • The first few series terms ( \(-0.2, -0.02, -0.00133\)) cumulatively match the accuracy criterion.
  • Each term has a diminishing impact, making it possible to stop the addition when terms become negligible.
  • Continue to add more terms if further precision is necessary but stop once acceptable accuracy is achieved.
With error analysis, we ensure that the final result is both accurate and efficient, avoiding unnecessary calculations beyond the desired precision.

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Most popular questions from this chapter

Find series solutions for the initial value problems in Exercises \(15-32\) . $$ y^{\prime \prime}-y=-x, \quad y^{\prime}(2)=-2 \text { and } y(2)=0 $$

Logistic difference equation The recursive relation $$ a_{n+1}=r a_{n}\left(1-a_{n}\right) $$ is called the logistic difference equation, and when the initial value \(a_{0}\) is given the equation defines the logistic sequence \(\left\\{a_{n}\right\\} .\) Throughout this exercise we choose \(a_{0}\) in the interval \(03.57\) . Choose \(r=3.65\) and calculate and plot the first 300 terms of \(\left\\{a_{n}\right\\} .\) Observe how the terms wander around in an unpredictable, chaotic fashion. You cannot predict the value of \(a_{n+1}\) from previous values of the sequence. g. For \(r=3.65\) choose two starting values of \(a_{0}\) that are close together, say, \(a_{0}=0.3\) and \(a_{0}=0.301 .\) Calculate and plot the first 300 values of the sequences determined by each starting value. Compare the behaviors observed in your plots. How far out do you go before the corresponding terms of your two sequences appear to depart from each other? Repeat the exploration for \(r=3.75 .\) Can you see how the plots look different depending on your choice of \(a_{0} ?\) We say that the logistic sequence is sensitive to the initial condition a_{0} .

How many terms of the Taylor series for \(\ln (1+x)\) should you add to be sure of calculating \(\ln (1.1)\) with an error of magnitude less than \(10^{-8} ?\) Give reasons for your answer.

Use a CAS to perform the following steps for the sequences in Exercises \(129-140 .\) a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L ?\) b. If the sequence converges, find an integer \(N\) such that \(\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$ a_{n}=\frac{n^{41}}{19^{n}} $$

Use series to evaluate the limits in Exercises \(47-56\) $$ \lim _{y \rightarrow 0} \frac{\tan ^{-1} y-\sin y}{y^{3} \cos y} $$
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