Problem 16 - Thomas Calculus

Question

Use series to estimate the integrals' values with an error of magnitude less than \(10^{-5}\) . (The answer section gives the integrals' values rounded to seven decimal places.) \begin{equation} \int_{0}^{0.4} \frac{e^{-x}-1}{x} d x \end{equation}

Step-by-Step Solution

Answer

Integrate via Taylor series expansion until error is < \( 10^{-5} \).

Step 1: Understanding the Integral

The integral to solve is \( \int_{0}^{0.4} \frac{e^{-x}-1}{x} \, dx \). The integrand is \( \frac{e^{-x}-1}{x} \), which is problematic at \( x = 0 \). Expanding \( e^{-x} \) into its Taylor series will help in estimating the integral.

Step 2: Expand \( e^{-x} \) in Taylor Series

The Taylor series for \( e^{-x} \) about \( x = 0 \) is given by:\[ e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \cdots \]Thus, \( e^{-x} - 1 = -x + \frac{x^2}{2!} - \frac{x^3}{3!} + \cdots \). Substitute this into the original integrand.

Step 3: Simplify the Integrand

Substitute \( e^{-x} - 1 \) with its series approximation and divide by \( x \):\[ \frac{e^{-x} - 1}{x} = \frac{-x + \frac{x^2}{2!} - \frac{x^3}{3!} + \cdots}{x} = -1 + \frac{x}{2!} - \frac{x^2}{3!} + \cdots \]This simplification allows us to consider the series term-by-term.

Step 4: Integrate Term by Term

Integrate each term separately from 0 to 0.4:- \( \int_{0}^{0.4} -1 \, dx = -0.4 \)- \( \int_{0}^{0.4} \frac{x}{2} \, dx = \frac{0.4^2}{4} = 0.04 \)- \( \int_{0}^{0.4} -\frac{x^2}{6} \, dx = -\frac{0.4^3}{18} \approx -0.0021333 \)Continue until the magnitude of the terms is less than \( 10^{-5} \).

Step 5: Sum the Evaluated Terms

Sum the evaluated terms to approximate the integral:\[ I \approx -0.4 + 0.04 - 0.0021333 + \cdots \]Add more terms if necessary to ensure that the error is within \( 10^{-5} \). The series will eventually converge towards the true integral value.

Step 6: Verify Error Magnitude

Check if the magnitude of the last added term is less than \( 10^{-5} \). In this exercise, continue adding terms until reaching four decimal places precision to keep overall error within the specified tolerance.

Key Concepts

Taylor Series

In calculus, Taylor Series is a powerful tool for approximating functions. It allows us to express complex functions as infinite sums of polynomials. A Taylor Series centered around a point "a" for a function \( f(x) \) is given by:

• \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots \)

This series expands infinitely, but for practical applications, we often only calculate the first few terms. That provides a good approximation of the function near the point "a".
The Taylor Series for \( e^{-x} \), expanded around \( x = 0 \), simplifies to:

• \( e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \cdots \)

Here, each term adds more precision to the approximation of \( e^{-x} \). As we add more terms, our estimate gets closer to the actual value of the function.
This series expansion makes it easier to integrate, especially when handling complex functions in calculus.

Integration

In the world of calculus, integration is the process of calculating the area under a curve. Specifically, it's crucial for finding definite integrals, which specify the accumulation of quantities. To find the integral of a function, we calculate its antiderivative or indefinite integral, and then use limits to solve the definite integral.
In the given exercise, we integrate:

• \( \int_{0}^{0.4} \frac{e^{-x} - 1}{x} \, dx \)

The challenge with this integral is its complexity at \( x = 0 \), where the function becomes undefined. By using the series expansion of \( e^{-x} - 1 \) and then dividing by \( x \), we simplify the function:

• \( \frac{e^{-x} - 1}{x} = -1 + \frac{x}{2!} - \frac{x^2}{3!} + \cdots \)

We then integrate each term independently across the defined limits (from 0 to 0.4). This term-by-term integration is straightforward because polynomial terms allow simple antiderivatives.
Summing these integrated terms provides an approximation of the definite integral.

Error Estimation

Error estimation is critical while approximating integrals using series. It determines how close the series sum is to the actual integral value. The goal is to ensure that the error remains below a defined tolerance level, often an arbitrary small value like \( 10^{-5} \).
In this exercise, to minimize the error, we truncate the series once the magnitude of the added term falls under the \( 10^{-5} \) threshold. This ensures that subsequent terms won't significantly change the integral's approximation.
To implement this, continue adding terms from the series into your sum. Calculate each term's integral until its value becomes insignificant (i.e., less than \( 10^{-5} \)). Once reached, the sum of all included terms is close enough to the integral's true value with acceptable precision.
This method of error control ensures that our solution is both accurate and practical for real-world applications, where exact calculations might be computationally intensive or impossible.