Problem 47 - Essential Calculus Early Transcendentals

Question

Use series to approximate the definite integral to within the indicated accuracy. $$\int_{0}^{1} x \cos \left(x^{3}\right) d x \quad(\text { three decimal places })$$

Step-by-Step Solution

Answer

The integral evaluates to approximately 0.441.

Step 1: Understand the Problem

We need to approximate the integral \( \int_{0}^{1} x \cos\left(x^{3}\right) \, dx \) to an accuracy of three decimal places. The function inside the integral, \( x \cos\left(x^{3}\right) \), suggests that we can use the Taylor series for \( \cos(u) \), where \( u = x^3 \), to simplify the integration.

Step 2: Expand the Function Using Taylor Series

The cosine function can be expanded using the Taylor series: \( \cos(u) = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \cdots \). Setting \( u = x^3 \), the series becomes: \( \cos(x^3) = 1 - \frac{(x^3)^2}{2} + \frac{(x^3)^4}{4!} - \cdots \).

Step 3: Insert the Series into the Integral

Substituting the Taylor series into the integral gives: \( \int_{0}^{1} x \left( 1 - \frac{x^6}{2} + \frac{x^{12}}{4!} - \cdots \right) \, dx \). This simplifies to: \( \int_{0}^{1} \left( x - \frac{x^7}{2} + \frac{x^{13}}{4!} - \cdots \right) \, dx \).

Step 4: Compute the Integral Term-by-Term

Integrate each term separately:- \( \int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1}{2} \).- \( \int_{0}^{1} \frac{x^7}{2} \, dx = \left[ \frac{x^8}{16} \right]_{0}^{1} = \frac{1}{16} \).- \( \int_{0}^{1} \frac{x^{13}}{24} \, dx = \left[ \frac{x^{14}}{336} \right]_{0}^{1} = \frac{1}{336} \).

Step 5: Sum the Series to Approximated Value

Approximate the value using the first few terms:\[ \frac{1}{2} - \frac{1}{16} + \frac{1}{336} \approx 0.5000 - 0.0625 + 0.0030 \approx 0.4405 \]. Adding more terms if needed, check that further terms decrease the value negligibly for three decimal accuracy.

Key Concepts

Taylor Series

The Taylor Series is a powerful mathematical tool that allows us to approximate functions using an infinite sum of terms calculated from the values of the function's derivatives at a single point. In the context of this problem, we use the Taylor series to express the function \(\cos(u)\) where \(u = x^3\). By expanding \(\cos(u)\) as a Taylor series, we obtain:

• \( \cos(u) = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \cdots \).

For our integral, this becomes \(\cos(x^3) = 1 - \frac{(x^3)^2}{2} + \frac{(x^3)^4}{4!} - \cdots \).
Substituting this series into the integral allows each term to be integrated separately, providing an approximation of the integral's value.
The key advantage here is that we can choose the number of terms in the series based on the desired level of accuracy. The more terms we use, the closer we get to the true value, but even a few terms can offer a good approximation for many practical purposes.

Definite Integral

A Definite Integral represents the area under the curve of a function over a specific interval. It is represented by:

• \( \int_{a}^{b} f(x) \, dx \), where \(a\) and \(b\) are the limits of integration.

In this exercise, the definite integral is \( \int_{0}^{1} x \cos(x^3) \, dx \).
We are not looking for an exact symbolic answer here, but an approximation to three decimal places.
By using the Taylor series to simplify the function inside the integral, this problem becomes manageable. Each term in the series forms its own simpler integral, which can be calculated independently.
The result of each of these integrals will then be summed to approximate the value of the original integral.

Integration

Integration is the process of finding the integral of a function, which can be thought of as finding the total accumulation, or area under the curve, of that function. In a practical sense, integration undoes differentiation.

• When approaching integration with a series, such as in this problem, each term of the series is integrated separately.
• This term-by-term integration takes advantage of the linearity of the integral and simplifies complicated functions to series of basic polynomials.

Here, each term from the Taylor series expansion of \( \cos(x^3) \) is integrated separately over the interval from 0 to 1.
If we consider \( x - \frac{x^7}{2} + \frac{x^{13}}{24} - \cdots \), each component is essentially a polynomial function that can be integrated using known rules.
For example, integrating the first term gives \( \frac{x^2}{2} \), evaluated from 0 to 1. Each higher-degree polynomial term is similarly integrated and then substituted back into the original definite integral limits.
This approach drastically simplifies the integration and allows us to find an approximate numerical result to the desired level of accuracy.