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 estimated value of the integral is approximately 0.3513250.

Step 1: Determine the Series Expansion

First, expand the function \( \sqrt[3]{1+x^2} \) as a binomial series. The expansion is given by the formula for binomial series: \((1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \cdots\). Here, \( n = \frac{1}{3} \) and \( x = x^2 \). Thus, we have \( (1+x^2)^{\frac{1}{3}} = 1 + \frac{1}{3}x^2 - \frac{1}{9}x^4 + \frac{5}{81}x^6 - \cdots \).

Step 2: Integrate the Series Term-by-Term

Integrate each term of the series from 0 to 0.35. Integrating individually, we get: \[ \int_{0}^{0.35} \left( 1 + \frac{1}{3}x^2 - \frac{1}{9}x^4 + \frac{5}{81}x^6 - \cdots \right) \, dx = \left[ x + \frac{1}{9}x^3 - \frac{1}{45}x^5 + \frac{5}{567}x^7 - \cdots \right]_{0}^{0.35} \].

Step 3: Set the Error Condition

To ensure the error is less than \(10^{-5}\), decide how many terms of the series to keep. The error from the neglected terms must be smaller than \(10^{-5}\). Use the next term in the series to evaluate this. For \( x=0.35 \), estimate the next term's value after the last included term, ensuring its integral contribution is significantly less than the error threshold.

Step 4: Compute the Integral Contribution from Each Term

Calculate each integral: For the first term, \( \left[ 0.35 - 0 \right] = 0.35 \), for the second term, \( \left[ \frac{1}{9}(0.35)^3 - 0 \right] \approx 0.001352 \), for the third term, \( \left[ -\frac{1}{45}(0.35)^5 - 0 \right] \approx -0.000027 \).

Step 5: Sum the Contributions and Verify Error

Sum the terms calculated: \( 0.35 + 0.001352 - 0.000027 = 0.351325 \). Check the accuracy by calculating the first disregarded term to ensure the total error does not exceed \(10^{-5}\).

Step 6: Final Estimate

The estimate of the integral using the series expansion with an acceptable error threshold is approximately \(0.3513\). Rounded to seven decimal places, the answer is \(0.3513250\).

Key Concepts

Binomial Series

A binomial series is a powerful tool used to expand expressions of the form \((1+x)^n\), where \(n\) can be any real number. This expansion allows us to represent the expression as an infinite series:

• \((1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots\)

For our problem, we need to expand \((1+x^2)^{\frac{1}{3}}\). In this case, \(n = \frac{1}{3}\) and \(x = x^2\). The series then becomes:

• \((1+x^2)^{\frac{1}{3}} = 1 + \frac{1}{3}x^2 - \frac{1}{9}x^4 + \frac{5}{81}x^6 - \cdots\)

This expansion is useful because it breaks down a complex function into simpler polynomial terms that are easier to integrate or approximate numerically.

Integral Estimation

Estimating integrals using a series involves integrating each term of the series separately. The target is to approximate the definite integral:

• \( \int_{0}^{0.35} \sqrt[3]{1+x^{2}} \, dx \)

To do this, we integrate each term from the series expansion derived in the binomial series section. This is done term-by-term over the given limits of integration (from 0 to 0.35 in our example):

• \[ \int_{0}^{0.35} \left( 1 + \frac{1}{3}x^2 - \frac{1}{9}x^4 + \frac{5}{81}x^6 - \cdots \right) \, dx \]
• Evaluate each term separately and sum them to get the integral estimate.
• For instance, the integral of 1 is \(x\), while that of \(x^2\) is \(\frac{x^3}{3}\).

This method can provide a highly accurate approximation if enough terms are included, balanced by computational feasibility.

Error Analysis

Error analysis is crucial when approximating numerical results, especially in series approximations. In this context, we aim to ensure the error is less than a specified threshold, such as \(10^{-5}\) in our example.

• The error in a series estimation is primarily due to the neglected terms.
• To control the error, evaluate the next term in the series after those included in the approximation.
• If the contribution of this next term (when integrated) is less than the allowed error threshold, the error is deemed acceptable.

For instance, if the included terms give us an integral value of approximately 0.3513, we check the next term's contribution over the interval \(0\) to \(0.35\). If this value is smaller than \(10^{-5}\), our approximation is acceptable.

Term-by-Term Integration

Term-by-term integration is a straightforward method used with polynomial expressions derived from series expansions. This technique involves integrating each term of the expanded series separately over the specified bounds.

• First, identify each polynomial term in the series.
• Next, integrate these terms one by one from the lower to the upper limit.
• Adjust each integral according to the power of \(x\) in the term.
• Sum the results of these individual integrals to find the total integrated value.

For instance, integrating the term \(\frac{1}{3}x^2\) from 0 to 0.35 involves solving \(\frac{1}{9}(0.35)^3\). Doing this for all terms provides a cumulative, approximate value of the original integral. This approach simplifies complex integrals into manageable calculations, allowing approximation with controlled error.