Problem 52 - Single Variable Calculus

Question

Use a power series to obtain an approximation of the definite integral to four decimal places of accuracy. \(\int_{0}^{0.5} x^{2} e^{-x^{2}} d x\)

Step-by-Step Solution

Answer

To approximate the definite integral \(\int_{0}^{0.5} x^{2} e^{-x^{2}} d x\) to four decimal places, we start by finding the Maclaurin series of the function \(x^2 e^{-x^2}\), which is \(\sum_{n=0}^\infty \frac{(-1)^n x^{2n+2}}{n!}\). We then integrate the power series term by term and evaluate it within the given range, leading to a 10-term approximation: \[I(0.5) \approx \sum_{n=0}^9 \frac{(-1)^n}{n!} \frac{(0.5)^{2n+3}}{2n+3} \approx 0.1174\] Therefore, \(\int_{0}^{0.5} x^{2} e^{-x^{2}} d x \approx 0.1174\).

Step 1: Find the Maclaurin series of the function

To find the Maclaurin series of the function, we'll express it as \(f(x) = x^2 e^{-x^2}\). We know that the Maclaurin series of \(e^u\) is \(\sum_{n=0}^\infty \frac{u^n}{n!}\). In our case, \(u = -x^2\), so we have: \[e^{-x^2} = \sum_{n=0}^\infty \frac{(-x^2)^n}{n!}\] Now, multiply by \(x^2\) to get the Maclaurin series for \(x^2 e^{-x^2}\): \[f(x) = x^2 e^{-x^2} = x^2 \sum_{n=0}^\infty \frac{(-x^2)^n}{n!} = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+2}}{n!}\]

Step 2: Integrate the power series term by term

Now, we integrate the power series term by term for the function \(f(x)\) from 0 to x: \[\int f(x) dx = \int \sum_{n=0}^\infty \frac{(-1)^n x^{2n+2}}{n!} dx = \sum_{n=0}^\infty \frac{(-1)^n}{n!} \int x^{2n+2} dx\] Integrating the term inside the sum, we get: \[\sum_{n=0}^\infty \frac{(-1)^n}{n!} \frac{x^{2n+3}}{2n+3} + C\]

Step 3: Approximate the definite integral to four decimal places

To approximate the definite integral to four decimal places, we will evaluate the integrated power series from 0 to 0.5 and truncate it to a sufficiently large number of terms. Let's start with a 10-term approximation: \[I(0.5) \approx \sum_{n=0}^9 \frac{(-1)^n}{n!} \frac{(0.5)^{2n+3}}{2n+3} - \sum_{n=0}^9\frac{(-1)^n}{n!} \frac{(0)^{2n+3}}{2n+3}\] The second term of the right side is zero since it involves multiplying by 0. Now, let's calculate the value of the first term: \[\sum_{n=0}^9 \frac{(-1)^n}{n!} \frac{(0.5)^{2n+3}}{2n+3} \approx 0.1174\] Thus, the definite integral to four decimal places of accuracy is approximately: \[\int_{0}^{0.5} x^{2} e^{-x^{2}} dx \approx 0.1174\]

Key Concepts

Maclaurin Series

The Maclaurin series is a specific type of Taylor series centered at zero. It provides us with a way to express functions as infinite polynomials, which makes them easier to manipulate, differentiate, or integrate term by term. For a function like \(e^{-x^2}\), the expansion begins with the known series of \(e^u\), which is \(\sum_{n=0}^\infty \frac{u^n}{n!}\). By substituting \(u = -x^2\), it becomes \(\sum_{n=0}^\infty \frac{(-x^2)^n}{n!}\). This power series representation allows the original function \(x^2 e^{-x^2}\) to be expressed by multiplying \(x^2\) with the series term by term, resulting in: \[f(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+2}}{n!}\] This expansion is essential for approximation as it opens up the function for easy term-by-term integration.

Definite Integral

The concept of a definite integral involves calculating the net area under a curve between two given limits – here, between \(0\) and \(0.5\). In this problem, after converting \(x^2 e^{-x^2}\) into a series, the task is to integrate each term's contribution over the interval. This gives us an approximation of the original integral. The integrated power series has the expression: \[\sum_{n=0}^\infty \frac{(-1)^n}{n!} \int x^{2n+2} dx\] This operation transforms each term into a new series with terms like \(\frac{x^{2n+3}}{2n+3}\). The run of integration from \(0\) to \(0.5\) helps compute the integral's value within this specific range, aiding in the approximation of the original integral function.

Term by Term Integration

Term by term integration is a powerful technique that allows us to integrate series one term at a time. Each term of the Maclaurin series sum becomes a simple integral of an elementary function. For example, the term \(\frac{(-1)^n x^{2n+2}}{n!}\) simplifies the integration process considerably because it turns into basic polynomial integrals: - Integration carries out individually for terms from \(x^{2n+2}\)- Results in \(\frac{x^{2n+3}}{2n+3}\) This individual handling makes even complex functions like \(x^2 e^{-x^2}\) manageable by breaking them into smaller, simpler pieces. It transforms lots of small calculations into a cohesive solution, and sums these new integrated terms to approximate the definite integral.

Truncation for Accuracy

In practical applications, infinite series (like Maclaurin) can't be calculated up to infinity. Instead, we truncate the series after a certain number of terms to ensure the result is both computationally feasible and accurate to a desired decimal place. Truncating the series is balancing act – including more terms can increase the accuracy, but also requires more computational effort. For the problem at hand, we approximate the integral value to four decimal places by including the first ten terms. Each additional term ensures that the result is closer to the true value. The series was truncated after ten terms, resulting in the approximation value of \(0.1174\). By analyzing how additional terms affect the sum, we ensure the desired accuracy level is achieved without unnecessary computations.