Problem 29 - Calculus: Early Transcendentals

Question

Use a power series to approximate the definite integral to six decimal places. \( \int^{0.3}_0 \frac {x}{1 + x^3} dx \)

Step-by-Step Solution

Answer

The integral is approximately 0.026919.

Step 1: Identify the function and express it as a power series

The function to integrate is \( f(x) = \frac{x}{1 + x^3} \). First, express \( \frac{1}{1 + x^3} \) as a geometric series. Note that \( \frac{1}{1 + x^3} = \sum_{n=0}^{\infty} (-1)^n x^{3n} \) for \(|x^3| < 1\).

Step 2: Formulate the power series for \( f(x) = \frac{x}{1 + x^3} \)

Multiply each term in the series by \(x\) to get the series for \(f(x)\). Hence, \( f(x) = x \sum_{n=0}^{\infty} (-1)^n x^{3n} = \sum_{n=0}^{\infty} (-1)^n x^{3n+1} \).

Step 3: Integrate the power series term by term

The integral becomes \( \int_0^{0.3} \sum_{n=0}^{\infty} (-1)^n x^{3n+1} \, dx = \sum_{n=0}^{\infty} (-1)^n \int_0^{0.3} x^{3n+1} \, dx \). The integral of each term is \( \int x^{3n+1} \, dx = \frac{x^{3n+2}}{3n+2} \).

Step 4: Evaluate the integrated terms with bounds

Substitute the bounds into each term: \( \sum_{n=0}^{\infty} (-1)^n \left( \frac{(0.3)^{3n+2}}{3n+2} - \frac{0^{3n+2}}{3n+2} \right) = \sum_{n=0}^{\infty} (-1)^n \frac{(0.3)^{3n+2}}{3n+2} \).

Step 5: Compute terms until the desired accuracy is reached

Calculate each term \( (-1)^n \frac{(0.3)^{3n+2}}{3n+2} \) until the absolute value of the next term is less than \(10^{-7}\). Sum these terms to approximate the integral. Evaluate the terms until the sum converges to six decimal places accuracy.

Key Concepts

definite integral

A definite integral is a fundamental concept in calculus, which calculates the accumulation of quantities. It's like finding the area under a curve between two points on the x-axis. In our exercise, we aim to find the definite integral of the function \( \frac{x}{1 + x^3} \) between 0 and 0.3. This will give us the precise area under the curve in this interval.

Definite integrals have upper and lower limits, unlike indefinite integrals which do not. The expression \( \int^{0.3}_0 \frac{x}{1 + x^3} dx \) specifies:

• The function \( \frac{x}{1 + x^3} \) to be integrated,
• 0 and 0.3 as the limits of integration, defining the interval.

They are usually approached through techniques like substitution, integration by parts, or, as in our case, by approximating with a power series. This method, especially useful for complicated functions, helps in evaluating the integral without solving it directly.

geometric series

A geometric series is a sequence of terms where each term after the first is found by multiplying the previous term by a constant factor. This type of series is integral to solving the exercise because the function \( \frac{1}{1 + x^3} \) can be expressed as a geometric series.

The generic formula for a geometric series is:

• \( \sum_{n=0}^{\infty} ar^n \)

where \( a \) is the first term and \( r \) is the common ratio. For our function, we rewrite it as \( \sum_{n=0}^{\infty} (-1)^n x^{3n} \). The negative sign alternates the sign of each term, while \( x^{3n} \) captures each power of \( x \) in the series.

This transformation allows us to use a series, instead of a complex rational function, to approximate the area under the curve. Geometric series are powerful in calculus for breaking down complex expressions for easier manipulation.

term-by-term integration

Term-by-term integration involves breaking down a series into simpler terms, integrating them individually. This procedure turns an otherwise complex integral into a series of manageable calculations.

In the exercise, once we expressed \( f(x) = \frac{x}{1 + x^3} \) as a power series \( \sum_{n=0}^{\infty} (-1)^n x^{3n+1} \), we integrate each term separately. This approach makes it possible to find the integral of:

• \( \int x^{3n+1} \, dx = \frac{x^{3n+2}}{3n+2} \)

We then substitute the bounds (0 to 0.3) into each integrated term. By doing so, we convert the original, challenging problem into a form where we can easily compute each term of the series and sum them up for the final result. The advantage of this method lies in its simplicity and ease of computational steps.

accuracy in approximation

Accuracy in approximation is key when using power series to estimate integrals. Since the power series provides an infinite sum, we need to decide how many terms will give a satisfactory approximation of the original function.

The solution strategy involves evaluating terms in the series \( (-1)^n \frac{(0.3)^{3n+2}}{3n+2} \) until adding more terms changes the result negligibly. For practical reasons, we maintain a threshold for accuracy, often making the absolute value of the next term less than \(10^{-7}\).

• This ensures the cumulative sum approaches the true value of the integral.

It's crucial to balance the number of terms for desired precision against computational effort. Such approximations are incredibly useful when exact solutions are cumbersome or impossible to attain analytically. They play a vital role in real-world applications where speed and efficiency are as important as precision.