Problem 2 - The Calculus with Analytic Geometry

Question

Compute the value of the given integral, accurate to four decimal places, by using series.\(\int_{0}^{1 / 3} \frac{d x}{1+x^{4}}\)

Step-by-Step Solution

Answer

The integral evaluates to approximately 0.3324.

Step 1 - Write down the integral

Consider the integral \(\int_{0}^{1 / 3} \frac{d x}{1 + x^{4}}\).

Step 2 - Use the series expansion

Use the geometric series expansion for \(\frac{1}{1 - (-x^{4})}\) which is \(\frac{1}{1 - (-x^{4})} = \sum_{n=0}^{\infty} (-x^{4})^{n} = \sum_{n=0}^{\infty} (-1)^{n} x^{4n} \).

Step 3 - Plug the series expansion into the integral

Now we have: \(\int_{0}^{1 / 3} \frac{d x}{1 + x^{4}} = \int_{0}^{1 / 3} \sum_{n=0}^{\infty} (-1)^{n} x^{4n} d x\).

Step 4 - Integrate term by term

Integrate term by term: \(\int_{0}^{1 / 3} \sum_{n=0}^{\infty} (-1)^{n} x^{4n} d x = \sum_{n=0}^{\infty} \int_{0}^{1 / 3} (-1)^{n} x^{4n} d x\).

Step 5 - Evaluate the individual integrals

Evaluate each term in the series: \(\int_{0}^{1 / 3} (-1)^{n} x^{4n} d x = (-1)^{n} \int_{0}^{1 / 3} x^{4n} d x = (-1)^{n} \frac{x^{4n+1}}{4n+1} \Bigg|_{0}^{1 / 3} = (-1)^{n} \frac{(1/3)^{4n+1}}{4n+1}\).

Step 6 - Sum the series to the required accuracy

Sum the series \(\sum_{n=0}^{\infty} (-1)^{n} \frac{(1/3)^{4n+1}}{4n+1}\) to achieve an accuracy of four decimal places. The first few terms up to where the series converges to four decimal places should be computed as: \( \frac{1}{3}\ - \frac{(1/3)^5}{5}\ + \frac{(1/3)^9}{9}\ - ... = 0.3333 - 0.0009 + 0.00000865 - ... ≈ 0.3324 ... \).

Step 7 - Final value

The final value of the integral \(\int_{0}^{1 / 3} \frac{d x}{1 + x^{4}}\text{ accurate to four decimal places is } 0.3324.\)

Key Concepts

geometric series

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

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:\[ \int_{0}^{1/3} (-1)^n x^{4n} dx = (-1)^n \int_{0}^{1/3} x^{4n} dx = (-1)^n \frac{x^{4n+1}}{4n+1} \Bigg|_{0}^{1/3} \]

evaluating series

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.