Problem 18 - Thomas Calculus

Question

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}

Step-by-Step Solution

Answer

The integral is approximately 0.3514219.

Step 1: 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 \( n = \frac{1}{3} \).

Step 2: 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.

Step 3: 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:- \( \int_{0}^{0.35} 1 \, dx = 0.35 \)- \( \int_{0}^{0.35} \frac{1}{3}x^2 \, dx = \frac{1}{3} \left[ \frac{x^3}{3} \right]_0^{0.35} = 0.001428 \)- \( \int_{0}^{0.35} -\frac{1}{9}x^4 \, dx = -\frac{1}{9} \left[ \frac{x^5}{5} \right]_0^{0.35} = -0.00000679 \)- \( \int_{0}^{0.35} \frac{5}{81}x^6 \, dx = \frac{5}{81} \left[ \frac{x^7}{7} \right]_0^{0.35} = 0.00000066 \)

Step 4: Sum the Integrated Values

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

Step 5: 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.

Key Concepts

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.