Problem 3 - The Calculus with Analytic Geometry

Question

Compute the value of the given integral, accurate to four decimal places, by using series.\(\int_{0}^{1} e^{-x^{2}} d x\)

Step-by-Step Solution

Answer

The value of the integral is approximately 0.7468.

Step 1: Recall the Taylor Series expansion for \(e^{-x^2}\)

The Taylor Series expansion for \(e^x\) is given by \[ e^x = \frac{x^0}{0!} + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} + \frac{x^7}{7!} + \frac{x^8}{8!} + \frac{x^9}{9!} + O(x^{10}) \] Therefore, \(e^{-x^2}\) is given by substituting \(x^2\) for \(x\): \[ e^{-x^2} = \frac{(-x^2)^0}{0!} + \frac{(-x^2)^1}{1!} + \frac{(-x^2)^2}{2!} + \frac{(-x^2)^3}{3!} + \frac{(-x^2)^4}{4!} + O(x^{10}) \]

Step 2: Simplify the obtained Taylor Series

Simplify each term of the Taylor Series: \[ e^{-x^2} = 1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6} + \frac{x^8}{24} + O(x^{10}) \]

Step 3: Integrate the Series term by term

Integrate the simplified Taylor Series term by term within the limits from 0 to 1: \[ I = \frac{1}{1} - \frac{1^3}{3} + \frac{1^5}{10} - \frac{1^7}{42} + \frac{1^9}{216} + O(1^{10}) \]

Step 4: Compute the value of the integral

Combine all the terms to get the value of the integral: \[ I = 1 - \frac{1}{3} + \frac{1}{10} - \frac{1}{42} + \frac{1}{216} = 1 - 0.3333 + 0.1 - 0.0238 + 0.0046 \] Adding these values: \[ I \thickapprox 0.7468 \]

Key Concepts

Taylor series expansion

To understand how we can use the Taylor Series expansion for approximating integrals, let's break it down. The Taylor Series is a way to represent a function as an infinite sum of terms calculated from the values of its derivatives at a single point. For instance, the Taylor Series for a function \( f(x) \) at \( x = 0 \) is given by:
display style display math display style f(x) = f(0) + f'(0) x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + \frac{f''''(0)}{4!} x^4 + \frac{f'''''(0)}{5!} x^5 + \frac{f''''''(0)}{6!} x^6 + ...$$ This\; series\; continues\; indefinitely,\; but\; usually,\; we\; only\; need\; a\; few\; terms\; to\; get\; a\; good\; approximation.$$

In practical applications:

• We take just the first few terms of the series.
• We use these terms to approximate functions and calculate integrals more easily.

For example, the series for \( e^x \) is given by:
$e^x\; =\; 1\; +\; x\; +\; \backslash frac\{x^2\}\{2!\}\; +\; \backslash frac\{x^3\}\{3!\}\; +\; \backslash frac\{x^4\}\{4!\}\; +\; \backslash frac\{x^5\}\{5!\}\; +\; \backslash frac\{x^6\}\{6!\}\; +\; ...display\; style\; To\; approximate\; \backslash (\; e^\{-x^2\}\; \backslash ),\; we\; substitute\; \backslash (\; -x^2\; \backslash )\; in\; place\; of\; \backslash (\; x\; \backslash ).\; That\; is,$ e^\{-x^2\}\; =\; 1\; -\; x^2\; +\; \backslash frac\{x^4\}\{2!\}\; -\; \backslash frac\{x^6\}\{3!\}\; +\; \backslash frac\{x^8\}\{4!\}\; +\; ...$.$

In summary, understanding how to expand functions with the Taylor Series helps us approximate functions that might be difficult to integrate directly.

Definite integral

A definite integral is a way to calculate the area under a curve between two specific points. When you see an integral written like \( \int_{0}^{1} e^{-x^{2}} dx \) it means the area under the curve \( e^{-x^2} \) from \( x = 0 \) to \( x = 1 \).

To solve a definite integral:

• You find the antiderivative (also known as the indefinite integral) of the function.
• Evaluate this antiderivative at the upper and lower limits.
• Subtract the value of the antiderivative at the lower limit from its value at the upper limit.

For example, consider the integral \( \int_{0}^{1} e^{-x^{2}} dx \). Since finding the antiderivative of \( e^{-x^2} \) directly is very difficult, we use the Taylor Series expansion to approximate the function within the limits.

This method involves integrating the series term by term, which simplifies the computation. As shown in the solution, we approximate: display I = \frac{1}{1} - \frac{1}{3} + \frac{1}{10} - \frac{1}{42} + \frac{1}{216} = 0.7468 @math@

Exponential function

The exponential function, denoted as \( e^x \), is fundamental in mathematics. It is defined as the function whose value is always the natural exponent raised to the power of its input \(x\). The function has unique characteristics:

• It grows faster than polynomial functions in positive inputs.
• It decays faster than any polynomial function in negative inputs.
• It is always positive, never zero.
• Its derivative is equal to the function itself, \( \frac{d}{dx}e^x = e^x \).

These properties make it very useful in many fields such as biology, economics, and physics.

In the context of the given exercise, the function we are dealing with is \( e^{-x^2} \).

Given the definition of exponential functions and the complications in finding their antiderivatives directly, we use series expansions to simplify our work.
The fact that we can represent \( e^{-x^2} \) as a Taylor series expansion allows us to approximate the integral numerically.
This highlights how powerful the exponential function is, especially when working with integrals that don't have simple antiderivatives.
Understanding this function's properties and behavior is important for solving many mathematical and real-world problems.

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