Problem 53 - Essential Calculus

Question

Use a power series to approximate the value of the integral with an error of less than \(0.0001 .\) (In Exercises 51 and \(52,\) assume that the integrand is defined as 1 when \(x=0 .)\) $$ \int_{0.1}^{0.3} \sqrt{1+x^{3}} d x $$

Step-by-Step Solution

Answer

The integral \(\int_{0.1}^{0.3} \sqrt{1+x^{3}} d x\) can be approximated by the value of the integral of the power series. The exact value will depend on how many terms in the power series are included to ensure the approximation error is less than \(0.0001\).

Step 1: Develop a Taylor series for the integrand

The first step is to develop a Taylor series for the function \(f(x) =\sqrt{1+x^{3}}\). We can expand this using the Binomial theorem, which gives \( (1+x)^{n} = 1 + nx + \frac{n(n-1)}{2!} x^{2} + \frac{n(n-1)(n-2)}{3!} x^{3} + ...\), where \(n = \frac{1}{2}\) and \(x = x^{3}\). This gives us the following series: \(f(x) = 1 + \frac{1}{2} x^{3} - \frac{1}{8} x^{6} + \frac{1}{16} x^{9} - ...\). This series will converge for \(|x| < 1\), so it is valid in the interval [0.1, 0.3].

Step 2: Approximate the integral

The second step is to approximate the integral using the power series expansion. We want the error to be less than \(0.0001\), so we need to include enough terms in the series. The integral will be \(\int_{0.1}^{0.3} \sqrt{1+x^{3}} d x \approx \int_{0.1}^{0.3} \left( 1 + \frac{1}{2} x^{3} - \frac{1}{8} x^{6} + \frac{1}{16} x^{9} \right)d x\). Calculate the integral for each term in the series to get the approximate value of the integral.

Step 3: Compute the error

The final step is to compute the error of the approximation and ensure it is indeed less than \(0.0001\). This can be done by comparing the exact value of the integral (calculated using high-precision numerical methods) with the approximate value we obtained. If the error is still too high, more terms need to be included in the power series.

Key Concepts

Understanding Taylor Series

A Taylor series is a powerful tool mathematicians use to approximate complex functions with a series of polynomial terms. It's based on the idea that you can represent a smooth function as an infinite sum of its derivatives at a single point. To create a Taylor series for a function, we take derivatives of the function at a point, typically denoted as 'a', and construct the series by adding progressively higher-degree terms multiplied by appropriate coefficients and a power of \(x-a\).For any function \(f(x)\), its Taylor series expansion around \(a\) is given by:\[ f(x) = f(a) + \frac{f'(a)}{1!} (x - a) + \frac{f''(a)}{2!} (x - a)^{2} + \cdots \]The more terms we include in the Taylor series, the closer our approximation gets to the true value of the function. The exercise involves finding the Taylor series expansion of \(f(x) = \sqrt{1+x^{3}}\) and then using it to approximate the value of an integral.

Integral Approximation with Power Series

Approximating integrals involves a strategy where we replace a complex function within an integral with a simpler one that closely matches the original function's behavior. Using the Taylor series expansion as the approximation is particularly useful because it can be integrated term by term, making the process manageable.In essence, to approximate \(\int_{0.1}^{0.3} \sqrt{1+x^{3}} dx\), we first express \(\sqrt{1+x^{3}}\) as its Taylor series, and then we calculate the integral of this series within the specified limits. This way, we transform the complex problem of integrating an irrational function into the easier one of integrating a polynomial. As we add more terms from the Taylor series, the approximation becomes more accurate, thus allowing us to achieve a predefined accuracy level, such as an error less than \(0.0001\).

The Binomial Theorem and Its Role in Expansions

The Binomial theorem provides us with a formula to expand expressions that are raised to a power, like \( (1+x)^{n} \). It's particularly useful when dealing with power series approximations, like in our exercise, where the function includes a square root—a fractional exponent.The Binomial theorem states that:\[ (1+x)^{n} = 1 + nx + \frac{n(n-1)}{2!}x^{2} + \frac{n(n-1)(n-2)}{3!}x^{3} + \cdots \]In the context of the exercise, we apply the Binomial theorem where \(n=\frac{1}{2}\) and \(x\) is cubed to expand the function \(\sqrt{1+x^{3}}\). The resulting series is a form of a Taylor series centered around \(a=0\). Each term in the series expansion contributes to a more precise approximation, which is crucial when calculating the integral with the desired accuracy.