High school calculus power series integral approximation problems with solutions

📚 Power Series and Integral Approximation: A Comprehensive Guide

Power series provide a powerful tool for approximating definite integrals, especially when the antiderivative of the integrand is not expressible in elementary functions. This technique leverages the ease with which power series can be integrated term by term. Let's dive in!

📜 History and Background

The use of infinite series to approximate functions dates back to the early days of calculus with mathematicians like Newton and Leibniz. The systematic development of power series and their applications, including integral approximation, emerged in the 18th and 19th centuries, providing solutions to problems in physics, engineering, and mathematics where closed-form solutions were elusive.

🔑 Key Principles

• ➕ Power Series Representation: Represent the integrand, $f(x)$, as a power series: $f(x) = \sum_{n=0}^{\infty} c_n x^n$.
• ∫ Term-by-Term Integration: Integrate the power series term by term: $\int f(x) dx = \int \sum_{n=0}^{\infty} c_n x^n dx = \sum_{n=0}^{\infty} c_n \int x^n dx = \sum_{n=0}^{\infty} c_n \frac{x^{n+1}}{n+1} + C$.
• 📏 Definite Integral Approximation: Evaluate the definite integral using the integrated power series: $\int_a^b f(x) dx = \left[ \sum_{n=0}^{\infty} c_n \frac{x^{n+1}}{n+1} \right]_a^b$.
• 📉 Error Estimation: Determine the number of terms needed to achieve the desired accuracy using error bounds or by observing the convergence of the series.

💡 Real-World Examples

Example 1: Approximating $\int_0^{0.2} e^{-x^2} dx$

Approximate the integral $\int_0^{0.2} e^{-x^2} dx$ using a power series.

1. Power Series Representation: We know that $e^u = \sum_{n=0}^{\infty} \frac{u^n}{n!}$. Therefore, $e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}$.
2. Term-by-Term Integration: $\int e^{-x^2} dx = \int \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!} dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \int x^{2n} dx = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{n!(2n+1)} + C$.
3. Definite Integral: $\int_0^{0.2} e^{-x^2} dx = \left[ \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{n!(2n+1)} \right]_0^{0.2}$.
4. Approximation: Let's use the first three terms (n = 0, 1, 2) to approximate the integral: $\frac{(0.2)^1}{0!(1)} - \frac{(0.2)^3}{1!(3)} + \frac{(0.2)^5}{2!(5)} = 0.2 - \frac{0.008}{3} + \frac{0.00032}{10} = 0.2 - 0.002667 + 0.000032 = 0.197365$. So, $\int_0^{0.2} e^{-x^2} dx \approx 0.197365$.

Example 2: Approximating $\int_0^{1} \frac{\sin(x)}{x} dx$

Approximate the integral $\int_0^{1} \frac{\sin(x)}{x} dx$ using a power series.

1. Power Series Representation: We know that $\sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$. Therefore, $\frac{\sin(x)}{x} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n+1)!}$.
2. Term-by-Term Integration: $\int \frac{\sin(x)}{x} dx = \int \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n+1)!} dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \int x^{2n} dx = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!(2n+1)} + C$.
3. Definite Integral: $\int_0^{1} \frac{\sin(x)}{x} dx = \left[ \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!(2n+1)} \right]_0^{1}$.
4. Approximation: Let's use the first three terms (n = 0, 1, 2) to approximate the integral: $\frac{(1)^1}{1!(1)} - \frac{(1)^3}{3!(3)} + \frac{(1)^5}{5!(5)} = 1 - \frac{1}{18} + \frac{1}{600} = 1 - 0.05555 + 0.001667 = 0.946117$. So, $\int_0^{1} \frac{\sin(x)}{x} dx \approx 0.946117$.

Example 3: Approximating $\int_0^{0.5} \frac{1}{1+x^4} dx$

Approximate the integral $\int_0^{0.5} \frac{1}{1+x^4} dx$ using a power series.

1. Power Series Representation: We know that $\frac{1}{1-u} = \sum_{n=0}^{\infty} u^n$ for $|u| < 1$. Therefore, $\frac{1}{1+x^4} = \frac{1}{1-(-x^4)} = \sum_{n=0}^{\infty} (-x^4)^n = \sum_{n=0}^{\infty} (-1)^n x^{4n}$.
2. Term-by-Term Integration: $\int \frac{1}{1+x^4} dx = \int \sum_{n=0}^{\infty} (-1)^n x^{4n} dx = \sum_{n=0}^{\infty} (-1)^n \int x^{4n} dx = \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+1}}{4n+1} + C$.
3. Definite Integral: $\int_0^{0.5} \frac{1}{1+x^4} dx = \left[ \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+1}}{4n+1} \right]_0^{0.5}$.
4. Approximation: Let's use the first three terms (n = 0, 1, 2) to approximate the integral: $\frac{(0.5)^1}{1} - \frac{(0.5)^5}{5} + \frac{(0.5)^9}{9} = 0.5 - \frac{0.03125}{5} + \frac{0.001953125}{9} = 0.5 - 0.00625 + 0.000217 = 0.494$. So, $\int_0^{0.5} \frac{1}{1+x^4} dx \approx 0.494$.

✍️ Practice Quiz

Question 1: Approximate $\int_0^{0.1} \cos(x^2) dx$ using the first three non-zero terms.

Question 2: Approximate $\int_0^{0.5} e^{x^3} dx$ using the first three non-zero terms.

Question 3: Approximate $\int_0^{0.2} \frac{1}{1-x^3} dx$ using the first three non-zero terms.

Question 4: Approximate $\int_0^{1} x^2 \cos(x) dx$ using the first three non-zero terms.

Question 5: Approximate $\int_0^{0.5} \frac{\arctan(x)}{x} dx$ using the first three non-zero terms.

Question 6: Approximate $\int_0^{0.1} \sqrt{1+x^2} dx$ using the first three non-zero terms.

Question 7: Approximate $\int_0^{0.5} \ln(1+x) dx$ using the first three non-zero terms.

✅ Conclusion

Power series offer a flexible and effective method for approximating integrals, particularly when dealing with functions lacking elementary antiderivatives. By understanding the underlying principles and practicing with various examples, you can master this technique and apply it to a wide range of problems. Happy calculating!