Problem 32 - Calculus: Early Transcendentals

Question

Use a power series to approximate the definite integral to six decimal places. \( \int^{0.3}_0 \frac {x^2}{1 + x^4} dx \)

Step-by-Step Solution

Answer

The approximate value of the integral is 0.008901.

Step 1: Identify the function for expansion

We start with the integrand \( \frac{x^2}{1 + x^4} \). This can be expressed as \( x^2 \cdot \frac{1}{1 + x^4} \). Recognizing the \( \frac{1}{1 + x^4} \) part as a function suitable for a geometric series expansion is crucial.

Step 2: Set up the power series expansion

The geometric series \( \frac{1}{1+x^4} \) can be expanded to \( 1 - x^4 + x^8 - x^{12} + \cdots \) for \( |x^4| < 1 \). Therefore, \( \frac{x^2}{1 + x^4} = x^2(1 - x^4 + x^8 - x^{12} + \cdots) \).

Step 3: Integrate the series term by term

The series becomes \( x^2 - x^6 + x^{10} - x^{14} + \cdots \). Now integrate each term from 0 to 0.3: \[ \int_{0}^{0.3} (x^2) \, dx = \left[ \frac{x^3}{3} \right]^{{0.3}}_{0} = \frac{0.3^3}{3} \] \[ \int_{0}^{0.3} (x^6) \, dx = \left[ \frac{x^7}{7} \right]^{{0.3}}_{0} = \frac{0.3^7}{7} \] Continue this for a few more powers to get a good approximation.

Step 4: Evaluate the series sum

Calculate each integral up to a sufficient number of terms until the desired accuracy (six decimal places) is achieved. The first few contributions are:\[ \int_0^{0.3} x^2 \, dx = \frac{0.3^3}{3} \approx 0.009 \]\[ \int_0^{0.3} x^6 \, dx = -\frac{0.3^7}{7} \approx -0.000099 \]Add a few more terms until the sum stabilizes to six decimal places.

Step 5: Sum the series for the final result

Add all evaluated terms:\[ 0.009 - 0.000099 + 0.0000002187 - \cdots \approx 0.008901 \]By adding more terms, keep refining until the sum stabilizes to six decimal places, leading to the final approximation.

Key Concepts

Definite Integral

A definite integral refers to the evaluation of an integral over a specific interval. In this exercise, we focus on finding the integral of the function \( \frac{x^2}{1 + x^4} \) from 0 to 0.3.
To solve a definite integral, you evaluate the accumulated area under the curve of the function within the defined limits. In mathematical terms, this involves calculating the integral at the upper limit and subtracting the integral at the lower limit.

• The limits in our problem are from 0 to 0.3.
• The solution involves expansion and integration techniques since the integrand involves a series-worthy function.

Approximating this integral precisely requires clever handling of the function's series representation and careful computation of each term.

Geometric Series

A geometric series is a series of terms that have a constant ratio between successive terms. This concept is useful for expanding certain types of functions into infinite series.
In the given problem, \( \frac{1}{1 + x^4} \) resembles a form that can be directly associated with a geometric series. By recognizing this, we can write it as:
\[ 1 - x^4 + x^8 - x^{12} + \cdots \]This expansion is valid as long as \(|x^4| < 1\), which holds within our integration limits (from 0 to 0.3).
With this series representation, we can efficiently integrate each term individually, simplifying the process of finding the definite integral.

Series Expansion

The method of series expansion allows us to express complex functions as an infinite sum of simpler terms. In this exercise, you expand \( \frac{x^2}{1 + x^4} \) using power series.
This is done by multiplying the expansion of \( \frac{1}{1 + x^4} \) by \( x^2 \). As a result, the series becomes:

• \( x^2 \) for the first term.
• \( -x^6 \) for the second term.
• \( x^{10} \) for the third term, and so on.

The power series expansion technique breaks down the function into manageable parts. This process simplifies the integration as we focus on each term separately. Manipulating each term makes approximation and calculation much easier.

Integration Term by Term

Integration term by term is a technique used to find the integral of a series by integrating each term independently. For the given series \( x^2 - x^6 + x^{10} - x^{14} + \cdots \), integrating it term by term involves computing:
\[ \int_0^{0.3} x^2 \; dx, \int_0^{0.3} x^6 \; dx, \int_0^{0.3} x^{10} \; dx \ldots \]Each integral is evaluated separately:

• \( \int_0^{0.3} x^2 \; dx = \frac{x^3}{3} \big|_0^{0.3} = \frac{0.3^3}{3} = 0.009 \)
• \( \int_0^{0.3} x^6 \; dx = -\frac{x^7}{7} \big|_0^{0.3} = -\frac{0.3^7}{7} = -0.000099 \)

By following this approach, each term's contribution is calculated precisely. Afterward, add all these contributions to approximate the total definite integral. This method's key benefit is that it incrementally builds towards the solution, refining the result with each added term.