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Use series to estimate the integrals' values with an error of magnitude less than \(10^{-5}\) . (The answer section gives the integrals' values rounded to seven decimal places.) \begin{equation} \int_{0}^{0.35} \sqrt[3]{1+x^{2}} d x \end{equation}

Short Answer

Expert verified
The integral is approximately 0.3514219.

Step by step solution

01

Expand the Function

We begin by expanding the function \( \sqrt[3]{1+x^2} \) into a series using the binomial series expansion. For a function of the form \( (1 + u)^n \), the expansion is \( 1 + nu + \frac{n(n-1)}{2!}u^2 + \cdots \). Here, set \( u = x^2 \) and \( n = \frac{1}{3} \).
02

Simplify the Series

With \( u = x^2 \) and \( n = \frac{1}{3} \), the expansion becomes:\[ 1 + \frac{1}{3}x^2 - \frac{1}{9}x^4 + \frac{5}{81}x^6 - \cdots \] We'll use this series to approximate the integral.
03

Integrate the Series

Integrate term by term within the given limits \( 0 \) to \( 0.35 \):\[ \int_{0}^{0.35} \left( 1 + \frac{1}{3}x^2 - \frac{1}{9}x^4 + \frac{5}{81}x^6 \right) dx \] Calculating each term's integral gives:- \( \int_{0}^{0.35} 1 \, dx = 0.35 \)- \( \int_{0}^{0.35} \frac{1}{3}x^2 \, dx = \frac{1}{3} \left[ \frac{x^3}{3} \right]_0^{0.35} = 0.001428 \)- \( \int_{0}^{0.35} -\frac{1}{9}x^4 \, dx = -\frac{1}{9} \left[ \frac{x^5}{5} \right]_0^{0.35} = -0.00000679 \)- \( \int_{0}^{0.35} \frac{5}{81}x^6 \, dx = \frac{5}{81} \left[ \frac{x^7}{7} \right]_0^{0.35} = 0.00000066 \)
04

Sum the Integrated Values

Add the results of the integrations to approximate the integral:\[ 0.35 + 0.001428 - 0.00000679 + 0.00000066 \approx 0.35142187 \]
05

Check Error Magnitude

The magnitude of the last term \( 0.00000066 \) dictates the approximation error. Since it is less than \( 1 \times 10^{-5} \), the current series approximation satisfies the error condition determined in the problem.

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Key Concepts

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

Binomial Series Expansion
The binomial series expansion is a powerful tool for approximating functions of the form \((1+u)^n\), where \(u\) is a term involving the variable, and \(n\) does not have to be an integer. This is particularly useful for functions that are not easily integrable using standard methods. In our exercise, we have the function \(\sqrt[3]{1+x^2}\). By setting \(u = x^2\) and \(n = \frac{1}{3}\), we create a series expansion:
  • First term: 1
  • Second term: \(\frac{1}{3}x^2\)
  • Third term: \(-\frac{1}{9}x^4\)
  • Fourth term: \(\frac{5}{81}x^6\)
This series gives us a polynomial that acts as an approximate representation of the original function. Each additional term in the series provides a finer approximation of \(\sqrt[3]{1 + x^2}\), but increasing the number of terms also increases computation. We must balance both accuracy and computational efficiency.
Error Estimation
When approximating integrals using series, understanding and managing the error is crucial. The error arises because a series only uses a finite number of terms to represent the original function, leading to small discrepancies. In practical terms, error estimation helps us determine how many terms we need to include to achieve a desired level of accuracy. In the context of the exercise, the error magnitude needs to be less than \(10^{-5}\). The potential error is primarily determined by the magnitude of the first term we neglect. For our series:
  • We calculated the integral of each term and noted the error imposed by omitting further terms.
  • The magnitude of the last included term, \(0.00000066\), met the assumption because it is well below \(10^{-5}\).
Thus, careful monitoring and calculating when adding terms ensures that our result is both precise and efficient.
Definite Integral
A definite integral represents the area under a curve between two points, yielding a numerical value that describes this area. For the given exercise, we are evaluating the integral:\[ \int_{0}^{0.35} \sqrt[3]{1+x^{2}} \, dx \]Since the function \(\sqrt[3]{1+x^2}\) does not have an elementary antiderivative, we resorted to series expansion to express this function as a polynomial, making it possible to integrate term by term:
  • For each term of the series, we performed an individual definite integration, calculated between the specified bounds \(0\) and \(0.35\).
  • Then, we summed the individual results to obtain the approximate value of the integral: \(0.35142187\).
  • The sequence of these simple integrations helped in neatly organizing and solving what might initially seem complex.
  • Summing all partial integrals provided us an accurate estimation of the whole integral's value within the prescribed error tolerance.
Definite integrals like these are not only fundamental in calculus but are crucial in applications across physics, engineering, and beyond.

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Most popular questions from this chapter

The sum of the series \(\sum_{n=0}^{\infty}\left(n^{2} / 2^{n}\right)\) . To find the sum of this series, express 1\(/(1-x)\) as a geometric series, differentiate both sides of the resulting equation with respect to \(x,\) multiply both sides of the result by \(x\) , differentiate again, multiply by \(x\) again, and set \(x\) equal to 1\(/ 2 .\) What do you get?

Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation}\sum_{n=1}^{\infty} \frac{\ln n}{\sqrt{n} e^{n}}\end{equation}

Approximation properties of Taylor polynomials Suppose that \(f(x)\) is differentiable on an interval centered at \(x=a\) and that \(g(x)=b_{0}+b_{1}(x-a)+\cdots+b_{n}(x-a)^{n}\) is a polynomial of degree \(n\) with constant coefficients \(b_{0}, \ldots, b_{n}\) . Let \(E(x)=\) \(f(x)-g(x) .\) Show that if we impose on \(g\) the conditions i) \(E(a)=0\) ii) $$\lim _{x \rightarrow a} \frac{E(x)}{(x-a)^{n}}=0$$ then $$\begin{array}{r}{g(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\cdots} \\ {+\frac{f^{(n)}(a)}{n !}(x-a)^{n}}\end{array}$$ Thus, the Taylor polynomial \(P_{n}(x)\) is the only polynomial of degree less than or equal to \(n\) whose error is both zero at \(x=a\) and negligible when compared with \((x-a)^{n}.\)

Use series to estimate the integrals' values with an error of magnitude less than \(10^{-5}\) . (The answer section gives the integrals' values rounded to seven decimal places.) \begin{equation} \int_{0}^{0.4} \frac{e^{-x}-1}{x} d x \end{equation}

In the alternating harmonic series, suppose the goal is to arrange the terms to get a new series that converges to \(-1 / 2 .\) Start the new arrangement with the first negative term, which is \(-1 / 2 .\) Whenever you have a sum that is less than or equal to \(-1 / 2,\) start introducing positive terms, taken in order, until the new total is greater than \(-1 / 2 .\) Then add negative terms until the total is less than or equal to \(-1 / 2\) again. Continue this process until your partial sums have been above the target at least three times and finish at or below it. If \(s_{n}\) is the sum of the first \(n\) terms of your new series, plot the points \(\left(n, s_{n}\right)\) to illustrate how the sums are behaving.

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