Problem 56 - Single Variable Calculus

Question

Use a power series to obtain an approximation of the definite integral to four decimal places of accuracy. \(\int_{0}^{0.5} \tan ^{-1} x^{3} d x\)

Step-by-Step Solution

Answer

The approximated integral of \(\int_{0}^{0.5} \arctan{x^3} dx\) to four decimal places is 0.0625, obtained using a power series and taking the sum up to \(N = 2\).

Step 1: Recall the power series expansion for arctan

Recall the power series expansion for \(\arctan{x}\): \[\arctan{x} = \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}x^{2n+1}\]

Step 2: Substitute x^3 instead of x in the power series

We need to approximate the definite integral of \(\arctan{x^3}\), so substitute \(x^3\) into the power series expansion: \[\arctan{x^3} = \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}(x^3)^{2n+1}\]

Step 3: Simplify the power series expression

Simplify the sum: \[\arctan{x^3} = \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}x^{6n+3}\]

Step 4: Integrate the power series term by term

Integrate the sum term by term with respect to \(x\) from 0 to 0.5: \[\int_{0}^{0.5} \arctan{x^3} dx = \int_{0}^{0.5}\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}x^{6n+3} dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\int_{0}^{0.5}x^{6n+3} dx\]

Step 5: Find the general term for the integrated series

Calculate the general term for the integrated series: \[\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\int_{0}^{0.5}x^{6n+3} dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\left[\frac{x^{6n+4}}{6n+4}\right]_0^{0.5}\]

Step 6: Calculate the approximation

We need to find an approximation to four decimal places, which means we need to find a sum that has an error of less than 0.0001. Start by taking some terms of the series and calculate the error of the remainder: Approximation: \[\sum_{n=0}^{N}\frac{(-1)^n}{2n+1}\frac{(0.5)^{6n+4}}{6n+4}\] Error for the remainder: \[\frac{(0.5)^{6(N+1)+4}}{6(N+1)+4}\] Use a calculator or computer program to find the smallest \(N\) that satisfies the error criterion of less than 0.0001. This process could require some computational work, but you should find that: \(N = 2\)

Step 7: Calculate the approximated integral

Using the results from Step 6, calculate the approximated integral with \(N = 2\): \[\int_{0}^{0.5} \arctan{x^3} dx \approx \sum_{n=0}^{2}\frac{(-1)^n}{2n+1}\frac{(0.5)^{6n+4}}{6n+4} = 0.06249\] Therefore, the approximated integral of \(\int_{0}^{0.5} \arctan{x^3} dx\) to four decimal places is 0.0625.

Key Concepts

Definite Integral

A definite integral is a way to calculate the area under a curve between two defined points. In this exercise, we are focused on the integral from 0 to 0.5 of the function \( \arctan(x^3) \). The process involves evaluating the integral using limits, which gives an exact area under the curve. Here, though, instead of using standard techniques, we leverage the power series expansion for approximation, making the process efficient for functions like \( \arctan(x^3) \) where direct integration is complex.

Arctan Series Expansion

Understanding the series expansion for \( \arctan(x) \) is crucial. The power series is given by:

• \( \arctan{x} = \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}x^{2n+1} \)

By substituting \( x^3 \) into this series, we obtain:

• \( \arctan{x^3} = \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}(x^3)^{2n+1} \)

This conversion allows us to simplify and work with the integral more constructively, transforming a complex expression into an approachable series of terms.

Numerical Approximation

Numerical approximation helps us achieve precise results when exact solutions are challenging. Here, we apply the series term-by-term integration method. After substituting and simplifying the series, we integrate term by term:

• \( \int_{0}^{0.5} \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}x^{6n+3} dx \)

This process requires evaluating each term and summing them up to a certain point, typically until further terms become negligible.

Error Estimation

Estimating the error is vital in achieving an accurate approximation. The task is to find \( N \) such that the remainder's error is beneath a desired threshold \(< 0.0001 \). Calculation of error results from considering the remaining terms after truncation:

• Approximation: \( \sum_{n=0}^{N}\frac{(-1)^n}{2n+1}\frac{(0.5)^{6n+4}}{6n+4} \)
• Error: \( \frac{(0.5)^{6(N+1)+4}}{6(N+1)+4} \)

Through computation, you find the appropriate \( N \), ensuring the approximation's precision, ultimately rounding to the nearest necessary decimal.