Problem 60 - Calculus: Early Transcendentals

Question

Use series to approximate the definite integral to within the indicated accuracy. \( \int^{0.5}_0 x^2 e^{-x^{2}} dx \) \( \left( \mid \text {error} \mid < 0.001 \right) \)

Step-by-Step Solution

Answer

Approximate integral is 0.01795, with error < 0.001.

Step 1: Express the Function as a Series

The given function is the product of \( x^2 \) and \( e^{-x^2} \). We start by expressing \( e^{-x^2} \) as a power series. The exponential function \( e^{-x^2} \) can be expanded using Taylor series:\[e^{-x^2} = 1 - x^2 + \frac{(x^2)^2}{2!} - \frac{(x^2)^3}{3!} + \cdots.\]

Step 2: Multiply Series by \(x^2\)

To approximate the integral of \( x^2 e^{-x^2} \), multiply each term of the series \( e^{-x^2} \) by \( x^2 \):\[x^2 e^{-x^2} = x^2 \left(1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6} + \cdots\right) = x^2 - x^4 + \frac{x^6}{2} - \frac{x^8}{6} + \cdots.\]

Step 3: Integrate Term by Term Over \([0, 0.5]\)

Integrate each term individually from 0 to 0.5:\[\int_0^{0.5} x^2 \: dx = \frac{x^3}{3}\Bigg|_0^{0.5} = \frac{0.5^3}{3} - 0 = \frac{1}{24},\]\[\int_0^{0.5} x^4 \: dx = \frac{x^5}{5}\Bigg|_0^{0.5} = \frac{0.5^5}{5} - 0 = \frac{1}{320},\]\[\int_0^{0.5} \frac{x^6}{2} \: dx = \frac{x^7}{14}\Bigg|_0^{0.5} = \frac{0.5^7}{14} - 0 = \frac{1}{3584},\]\[\int_0^{0.5} \frac{x^8}{6} \: dx = \frac{x^9}{54}\Bigg|_0^{0.5} = \frac{0.5^9}{54} - 0 = \frac{1}{55296}.\]

Step 4: Sum the Integrated Series

Add the results of the integration up to the point where the error is less than 0.001:\[\int_0^{0.5} x^2 e^{-x^2} \: dx \approx \left(\frac{1}{24} - \frac{1}{320} + \frac{1}{3584} - \cdots\right).\] Calculate the sum up until the absolute value of the terms becomes smaller than the desired error threshold of 0.001:\[\frac{1}{24} - \frac{1}{320} + \frac{1}{3584} \approx 0.0208 - 0.003125 + 0.000279 = 0.017954.\]

Step 5: Verify the Error

Calculate the next term \( \int_0^{0.5} \frac{x^8}{6} \: dx \), which is \( \frac{1}{55296} \approx 0.000018 \), smaller than 0.001. Hence, stopping after the \( x^6 \) term gives a sufficient approximation.

Key Concepts

Taylor Series

The Taylor Series is a powerful mathematical tool that allows us to approximate complex functions using polynomials. This is particularly useful when the function itself is challenging to work with directly. In this exercise, we're dealing with the exponential function \(e^{-x^2}\). By expanding \(e^{-x^2}\) into a Taylor series around 0, we can express it as:

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

The beauty of the Taylor Series lies in its simplicity. We transform a potentially difficult function into a series of polynomial terms that are much easier to handle. This approximation provides a way to estimate the function's value at any given point, achieving higher accuracy by including more terms.

Definite Integral

A definite integral calculates the area under a curve within a specified range, offering a numerical understanding of the function over the interval. In our problem, we're looking at the integral

• \( \int_0^{0.5} x^2 e^{-x^2} \, dx \)

This integral cannot be solved with elementary antiderivatives due to the combination of polynomial and exponential components. However, by breaking down the exponential into its power series, we can integrate term-by-term over the interval \([0, 0.5]\). Each polynomial term of the series is integrated separately, allowing us to approximate the total area efficiently.

Error Estimation

Error estimation is crucial when using series approximations to ensure the result falls within the desired accuracy. For this exercise, the condition is:

• \(| \text{error} | < 0.001\)

After integrating the series terms and summing them to approximate our integral, each subsequent term's contribution diminishes. We stop summing the series once the next term is negligible (i.e., less than the error threshold of 0.001). In our case, the term \(\frac{x^8}{6}\), when evaluated, reveals an impact of approximately 0.000018, small enough to respect our accuracy requirement.

Power Series Expansion

The power series expansion is transformed from a function into an infinite series. This exercise showcases that using a power series expansion facilitates computing definite integrals of complex functions like \(x^2 e^{-x^2}\). The expansion leads to a series of simpler polynomial terms:

• \(x^2 e^{-x^2} = x^2 - x^4 + \frac{x^6}{2} - \frac{x^8}{6} + \cdots \)

Each term of this power series is integrated individually over the range \([0, 0.5]\). This approach is manageable and computationally less intensive, providing a clear pathway to approximate the integral with the accuracy we're aiming for. The power series expansion thus aids in breaking down the problem, enabling precise mathematical approximations.