91Ó°ÊÓ

To understand how to find the series expansion of a function, it's important to get familiar with the Maclaurin series. The Maclaurin series is a special case of the Taylor series centered at zero. For a function \(f(x)\), its Maclaurin series is given by:
\[f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \]
In simpler terms, the Maclaurin series expands a function as a sum of its derivatives at zero, each multiplied by a power of \(x\) and divided by the factorial of the power. This allows us to express complicated functions in a polynomial form, making them easier to work with for approximation and integration.

• The more terms you include, the more accurate the approximation.
• Common functions have well-known Maclaurin series, like \(e^x\).

By substituting \(-x^2\) in the Maclaurin series for \(e^x\), you can find the series expansion for \(e^{-x^2}\). This involves replacing every instance of \(x\) in the series with \(-x^2\), resulting in a new series that approximates the function \(e^{-x^2}\).

Once you have the series expansion, you can use it to approximate integrals. In this case, we approximate \(\int_{0}^{1} e^{-x^2} \, dx \) by integrating the series expansion term by term. Here are the steps to do it:

• Substitute the first few terms of the series expansion into the integral.
• Integrate each term individually.
• Sum the results to get the approximate value of the original integral.

Integrating term by term simplifies the process because integrating polynomials is straightforward. For example, integrating \(x^n\) results in \(\frac{x^{n+1}}{n+1}\). After integrating all terms from the series expansion of \(e^{-x^2}\), you add them up to approximate the integral.

In this problem, we used the first four non-zero terms of the series for \(e^{-x^2}\), which are:
\[1, -x^2, \frac{x^4}{2}, \text{and} -\frac{x^6}{6} \]
Substituting these into the integral and then integrating term by term gives us an approximation that simplifies the otherwise complex integral.

Series expansion integration is a technique where you use a series to represent a function and then integrate that series term by term. This method is particularly useful when dealing with functions that are difficult to integrate directly, like \(e^{-x^2}\).

Instead, we approximate \(e^{-x^2}\) by its series expansion \(1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6}\), and then integrate each term:
\[ \int_{0}^{1} \left( 1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6} \right) \, dx \]
This results in a simpler sum of integrals of polynomials:

• \( \int_{0}^{1} 1 \, dx = 1 \)
• \( \int_{0}^{1} x^2 \, dx = \frac{1}{3} \)
• \( \int_{0}^{1} \frac{x^4}{2} \, dx = \frac{1}{10} \)
• \( \int_{0}^{1} \frac{x^6}{6} \, dx = \frac{1}{42} \)

After integrating, we sum these values to get a final approximation:
\[1 - \frac{1}{3} + \frac{1}{10} - \frac{1}{42} \approx 0.743 \]

This result shows how series expansion integration allows you to approximate complex integrals in a manageable way, often to a high degree of accuracy.