Problem 35 - Calculus: One and Several Variables

Question

Estimate to within 0.01 by using series. $$\int_{0}^{1} \arctan x^{2} d x$$

Step-by-Step Solution

Answer

The estimated value of the integral \(\int_{0}^{1} \arctan x^{2} dx\) to within 0.01 would be obtained by summing the terms of series \(\sum_{n=0}^{\infty} \frac{(-1)^n}{(4n+3)(2n+1)}\) until the absolute value of a term becomes less than 0.01.

Step 1: Expressing arctan(x^2) as a power series

We know that the power series for \(\arctan(x)\) is \(\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}\) for \(|x|<1\). Substituting \(x^{2}\) in place of \(x\), we get the power series for \(\arctan(x^{2})\) as \(\sum_{n=0}^{\infty} \frac{(-1)^n (x^{2})^{2n+1}}{2n+1} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+2}}{2n+1}\).

Step 2: Integrating term by term

The next step is to integrate the power series term by term from 0 to 1. The result is \(\int_{0}^{1} \arctan(x^{2}) dx = \int_{0}^{1} \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+2}}{2n+1} dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \int_{0}^{1} x^{4n+2} dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{(4n+3)(2n+1)}\).

Step 3: Estimating to within 0.01

We need to calculate terms of the series until the absolute value of the (n+1)th term is less than 0.01. The absolute error in the integral is less than the first neglected term (Weierstrass M-test).

Key Concepts

Power Series

A power series is a representation of a function as an infinite sum of terms, each being a product of a coefficient and a power of a variable. It is generally written in the form \[ \sum_{n=0}^{\infty} a_n x^n \] where \( a_n \) represents the coefficient for the \( n \)-th term, and \( x \) is the variable.

Power series are particularly useful because they allow for functions to be written in a form that is easy to manipulate, especially for calculation purposes. These series have a radius of convergence, within which they converge to the function they represent. For the power series of the arctangent function (\

Term by Term Integration

Term by term integration is a technique that allows us to integrate a power series by integrating each term individually. This technique is handy when dealing with an infinite series representation of a function. Mathematically, if we have a power series \[ \sum_{n=0}^{\infty} a_n x^n \], we can integrate it term by term as follows: \[ \int \sum_{n=0}^{\infty} a_n x^n dx = \sum_{n=0}^{\infty} \int a_n x^n dx \].

The result is another power series where each new coefficient is the original coefficient multiplied by the reciprocal of one plus the exponent on the variable, reflecting the rule for integrating power functions. This technique can be used as long as the series converges on the interval of integration and the function represented by the series is continuous on that interval.

Applying this to our exercise, we can integrate the series representation of the arctangent function term by term to calculate the integral of \( \arctan(x^2) \), given by: \[ \int_{0}^{1} \arctan(x^2) dx \].