Problem 57 - Calculus: Early Transcendentals

Question

Use series to approximate the definite integral to within the indicated accuracy. \( \int^{1/2}_0 x^3 \arctan x dx \) \( \text { (four decimal places)} \)

Step-by-Step Solution

Answer

The integral is approximately 0.0037.

Step 1: Identify the series expansion

First, we need the series expansion of \( \arctan x \). The expansion is given by:\[arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots\]We will use this series to approximate \( \arctan x \) in the integral.

Step 2: Set up the integral with the series

Substitute the series expansion of \( \arctan x \) into the integral:\[\int_0^{1/2} x^3 (x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots) dx\]Distribute \( x^3 \) into the series:\[\int_0^{1/2} (x^4 - \frac{x^6}{3} + \frac{x^8}{5} - \cdots) dx\]

Step 3: Integrate term-by-term

Integrate each term separately:\[\int_0^{1/2} x^4 dx = \frac{x^5}{5}\Big|_0^{1/2} = \frac{1}{32 imes5} = \frac{1}{160}\]\[\int_0^{1/2} \frac{x^6}{3} dx = \frac{1}{3} \cdot \frac{x^7}{7}\Big|_0^{1/2} = \frac{1}{672} \]\[\int_0^{1/2} \frac{x^8}{5} dx = \frac{1}{5} \cdot \frac{x^9}{9}\Big|_0^{1/2} = \frac{1}{23040}\]

Step 4: Add the values and check for accuracy

Sum up the integrated terms up to enough terms to meet the accuracy requirement of four decimal places. \[\frac{1}{160} - \frac{1}{672} + \frac{1}{23040} \approx 0.003660714\]

Step 5: Determine the number of terms for accuracy

Compute additional terms until the sum converges to four decimal place accuracy. The next term:\[\int_0^{1/2} \frac{x^{10}}{7} dx = \frac{1}{7} \cdot \frac{x^{11}}{11}\Big|_0^{1/2} = \approx \frac{1}{309760}\]Adding this to the previous sum does not change the fourth decimal place of 0.003660714, confirming the accuracy.

Key Concepts

Series Expansion

When approaching the integral of a function like \( x^3 \arctan x \), one effective method to approximate the result is through a series expansion. A series expansion is a way of expressing a function as an infinite sum of terms. These expansions help simplify complex functions into easily manageable parts. For \( \arctan x \), the series expansion is expressed as:\[arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots\]When dealing with definite integrals, this approximation provides a more straightforward way to tackle the integral by converting a potentially complicated function into an infinite polynomial sum. This transformation is valuable because it opens up pathways for simple operations like term-by-term integration, greatly simplifying calculations.

Term-by-Term Integration

Once a function is represented as a series, integrating term-by-term allows you to treat each term in the series independently. This simplifies the process of integrating complex expressions. Taking each term separately can reduce mistakes and enhance comprehension.Let's consider integrating the series:\[\int_0^{1/2} x^3 (x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots) dx\]By distributing the \( x^3 \) across each term of the series expansion, you rewrite the integrand as:\[x^4 - \frac{x^6}{3} + \frac{x^8}{5} - \cdots\]Each term in this new series is integrated separately over the interval \([0, 1/2]\). Through this approach, integration can be performed with ease, ensuring each step is clear and follows the rules of calculus directly.

Arctan Series

The arctan series plays a pivotal role in approximating integrals involving the arctangent function. The power series for \( \arctan x \), derived through calculus, is:\[arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots\]This expansion is particularly useful for integration purposes due to its alternating nature and quick convergence for values \( |x| \leq 1 \). By breaking \( \arctan x \) into a sum of polynomial terms, it becomes more manageable mathematically. This series is alternating and decreases progressively in value, making it an efficient tool for approximation to desired precision levels. It's essential to include enough terms to ensure the accuracy of approximations derived from this series.

Convergence to Accuracy

When approximating a definite integral to a specified accuracy, such as to four decimal places, convergence is a critical consideration. Convergence involves deciding how many terms of the series are necessary to achieve the desired level of precision.For the integral \( \int_0^{1/2} x^3 \arctan x dx \), terms from the series expansion of \( \arctan x \) are summed until the required precision of four decimal places is attained.Through calculations, each term is added sequentially:

1. The term \( \int_0^{1/2} x^4 dx \) contributes 0.00625.
2. The subsequent term \( -\frac{1}{3} \int_0^{1/2} x^6 dx \) reduces the sum to 0.0036607.
3. Additional terms are calculated, with each new term being smaller and affecting the result less as more terms are included.

You continue adding terms until the change is negligible relative to the specified accuracy. This process ensures that the approximation remains within the desired precision, demonstrating careful attention to the principle of convergence, essential in series approximations.