Problem 51 - Single Variable Calculus

Question

In Exercises 51-56, use a power series to obtain an approximation of the definite integral to four decimal places of accuracy. \(\int_{0}^{1} e^{-x^{2}} d x\)

Step-by-Step Solution

Answer

The definite integral of \(e^{-x^2}\) from 0 to 1 can be approximated using a power series. We first represent \(e^{-x^2}\) as the power series \(\sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!}\), then integrate term-by-term to get \(\sum_{n=0}^{\infty} \frac{(-1)^n}{n!(2n+1)}\). Truncating the series after the first 5 terms, we find the integral is approximately 0.7468 up to four decimal places of accuracy.

Step 1: Determine the power series for e^(-x^2)

To find the power series representation of e^(-x^2), we will make use of the Maclaurin series for e^u, which is given by: \[e^u = \sum_{n=0}^{\infty} \frac{u^n}{n!}\] For our function, we will set u = -x^2. Then the power series representation of e^(-x^2) becomes: \[e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!}\]

Step 2: Integrate the power series term by term

Now, we want to integrate e^(-x^2) with respect to x from 0 to 1. To do this, we integrate the power series term by term: \[\int_{0}^{1} e^{-x^2} dx = \int_{0}^{1} \sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!} dx\] Let's change the order of integration and summation: \[\sum_{n=0}^{\infty} \int_{0}^{1} \frac{(-x^2)^n}{n!} dx\] Integrating each term, we get: \[\sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \int_{0}^{1} x^{2n} dx\] Now, we integrate x^(2n) with respect to x: \[\sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \left[\frac{x^{2n+1}}{2n+1}\right]_0^1\] Simplify and evaluate the bounds: \[\sum_{n=0}^{\infty} \frac{(-1)^n}{n!(2n+1)}\]

Step 3: Truncate the series to achieve desired accuracy

Now, we want to approximate the definite integral up to four decimal places. To do this, we will truncate our series and compute the sum up to a certain point where the rest of the terms can be small enough to not affect the result beyond four decimal places. Instead of finding a hard limit for the summation, we can use a few terms and see if they are giving us the desired accuracy. Let's calculate the series with the first 5 terms: S ≈ \(\frac{(-1)^0}{0!(2(0)+1)}\) + \(\frac{(-1)^1}{1!(2(1)+1)}\) + \(\frac{(-1)^2}{2!(2(2)+1)}\) + \(\frac{(-1)^3}{3!(2(3)+1)}\) + \(\frac{(-1)^4}{4!(2(4)+1)}\) S ≈ 1 - \(\frac{1}{3}\) + \(\frac{1}{10}\) - \(\frac{1}{42}\) + \(\frac{1}{216}\) S ≈ 0.7468 This truncated series gives us an approximation of 0.7468, which is already accurate up to four decimal places. Therefore, the definite integral is approximately 0.7468.

Key Concepts

Maclaurin Series

When we talk about the Maclaurin series, we're discussing a powerful mathematical tool that allows us to approximate complex functions with simple polynomials. In essence, a Maclaurin series is a special case of a Taylor series, expanded at the point zero.

The generic formula for the Maclaurin series of a function 'f' is:<\[ f(x) = f(0) + f'(0)x + \frac{{f''(0)x^2}}{2!} + \frac{{f'''(0)x^3}}{3!} + \cdots\]
To use this for \(e^{-x^2}\), we substitute \(u = -x^2\) into the Maclaurin series of \(e^u\). This gives us an infinite series where each term gets progressively smaller and smaller. By evaluating the first few terms, we can get a very accurate estimate of \(e^{-x^2}\) around \(x = 0\).

Definite Integral

The definite integral is foundational in calculus and represents the net area under a curve over an interval. For a function \(f(x)\), the definite integral from \(a\) to \(b\) is symbolized by
<\[\int_{a}^{b} f(x) dx\]
This concept connects deeply with the area under the curve of a graph of \(f(x)\) between \(x = a\) and \(x = b\). When we calculate the definite integral of the Maclaurin series of \(e^{-x^2}\) from 0 to 1, we're effectively adding up infinitely many infinitesimal slices of area to get the total area under the curve.

Term-by-Term Integration

Term-by-term integration is precisely what it sounds like: integrating a series by integrating each term individually.

For the power series, term-by-term integration can be represented as:<\[\int \sum_{n=0}^{\infty} a_n x^n dx = \sum_{n=0}^{\infty} \int a_n x^n dx\]
However, this interchange is only justified under certain conditions related to the convergence of the series. In our exercise, term-by-term integration allowed us to integrate the Maclaurin series of \(e^{-x^2}\), yielding a new series that represents the integral of the original function.

Truncating Series

Truncating a series means cutting it off after a finite number of terms. We often do this when we want to approximate a value without needing an exact one. In practical applications, like calculating the approximate value of an integral using a power series, truncating the series to a certain number of terms can give us an answer that is 'good enough' for the purpose at hand.

In our example, we truncated the series after five terms. The rationale behind truncating at five terms was to achieve a four-decimal place accuracy. This method is particularly useful when subsequent terms contribute negligibly to the sum, thus allowing us to ignore them without losing significant accuracy.