91Ó°ÊÓ

A geometric series is a type of series with a constant ratio between successive terms. For a geometric series, the general form can be written as:g_0 + g_0r + g_0r^2 + g_0r^3 + ...where:

• g_0 is the first term
• r is the common ratio

In this problem, we use a special form of the geometric series expansion for the function \( \frac{1}{1+x^4} \). We rewrite this function to make it fit the geometry series form \( \frac{1}{1-(-x^4)} \). By doing this, we can use the geometric series formula:\[ \frac{1}{1-(-x^4)} = \sum_{n=0}^{\infty} (-x^4)^n = \sum_{n=0}^{\infty} (-1)^n x^{4n} \]This series expansion allows us to simplify our integral by expressing it as an infinite sum of simpler terms.

Term-by-term integration is a method which allows us to integrate an infinite series term by term. Once we've expressed our integrand as a power series, we can rewrite our integral as follows:\[ \int_{0}^{1/3} \frac{dx}{1+x^4} = \int_{0}^{1/3} \sum_{n=0}^{\infty} (-1)^n x^{4n} dx \]According to the properties of integrals, we can switch the order of summation and integration:\[ \int_{0}^{1/3} \sum_{n=0}^{\infty} (-1)^n x^{4n} dx = \sum_{n=0}^{\infty} \int_{0}^{1/3} (-1)^n x^{4n} dx \]This is remarkably convenient because integrating each term of \( (-1)^n x^{4n} \) separately is much easier. We use the following integration rule for power functions:\[ \int x^k dx = \frac{x^{k+1}}{k+1} \]Applying this to each term, we get:\[ \int_{0}^{1/3} (-1)^n x^{4n} dx = (-1)^n \int_{0}^{1/3} x^{4n} dx = (-1)^n \frac{x^{4n+1}}{4n+1} \Bigg|_{0}^{1/3} \]

The last part of the solution involves evaluating the series to the required accuracy. Once we have the individual integrals:\[ (-1)^n \frac{(1/3)^{4n+1}}{4n+1} \]we need to sum these terms until the series converges to the desired accuracy—four decimal places in this case. The first few terms provide a good approximation:

• \(n=0\): \( \frac{1/3}{1} = 0.3333 \)
• \(n=1\): \( - \frac{(1/3)^5}{5} = -0.0009 \)
• \(n=2\): \( + \frac{(1/3)^9}{9} = 0.00000865 \)

Adding these values gives:\[ \sum_{n=0}^{\infty} (-1)^n \frac{(1/3)^{4n+1}}{4n+1} ≈ 0.3333 - 0.0009 + 0.00000865 = 0.3324\]Thus, the integral \[ \int_{0}^{1/3} \frac{dx}{1+x^4} \approx 0.3324 \] accurate to four decimal places. By evaluating more terms, the approximation becomes more precise, ensuring the accuracy required.