Problem 58 - Calculus: Early Transcendentals

Question

Use series to approximate the definite integral to within the indicated accuracy. \( \int^1_0 \sin (x^4) dx \) \( \text { (four decimal places)} \)

Step-by-Step Solution

Answer

0.1872

Step 1: Understand the Function

The function we need to integrate is \( \sin(x^4) \). This function is not elementary, meaning it cannot be integrated directly using basic integration techniques. Therefore, we must use a series expansion to approximate the integral.

Step 2: Expand the Function into a Series

The sine function can be expanded as a Taylor series around 0: \( \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \). Applying this to \( \sin(x^4) \), we substitute \( x^4 \) into the series: \( x^4 - \frac{(x^4)^3}{3!} + \frac{(x^4)^5}{5!} - \cdots \). This simplifies to \( x^4 - \frac{x^{12}}{6} + \frac{x^{20}}{120} - \cdots \).

Step 3: Integrate the Series Term-by-Term

We approximate the integral \( \int^1_0 \sin(x^4) \, dx \) by integrating the series term by term. The integral becomes \( \int_0^1 \left( x^4 - \frac{x^{12}}{6} + \frac{x^{20}}{120} - \cdots \right) \, dx \). Calculate individual terms: \( \frac{x^5}{5} \big|_0^1 \), \( -\frac{x^{13}}{13 \times 6} \big|_0^1 \), and \( \frac{x^{21}}{21 \times 120} \big|_0^1 \). Only compute terms that contribute to four decimal place accuracy.

Step 4: Evaluate the Result

Calculate each integral term: the first term is \( \frac{1}{5} \), the second term is \( -\frac{1}{78} \), and the third term is approximately \( 0 \) when rounded to four decimal places. Therefore, we add these results to estimate the integral: \( \frac{1}{5} - \frac{1}{78} \).

Step 5: Estimate to Required Accuracy

We have: \( \frac{1}{5} = 0.2 \) and \( \frac{1}{78} \approx 0.0128 \). Subtracting these gives \( 0.2 - 0.0128 = 0.1872 \). Additional terms contribute less than 0.0001, so they can be ignored.

Key Concepts

Taylor Series

The Taylor Series is a mathematical tool that allows us to express functions as infinite sums of terms derived from the function's derivatives at a specific point. It is especially useful when the original function is difficult or impossible to integrate directly, as it can be approximated by polynomials.
For the function \( \sin(x) \), the Taylor Series expansion around zero is given by:

• \( \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \)

This series becomes valuable for approximation when the aim is to integrate a function like \( \sin(x^4) \). Here, by substituting \( x^4 \) into the series, we can convert the original trigonometric function into a polynomial, making integration feasible by evaluating polynomials instead.

Definite Integral

A definite integral is computed over a specified interval and it calculates the net area under the curve of a function between two points. For the integral \( \int^1_0 \sin(x^4) \, dx \), it evaluates the area under the curve of \( \sin(x^4) \) from 0 to 1.
Since \( \sin(x^4) \) is not straightforward to integrate using elementary methods, we use its series expansion to evaluate this integral indirectly. Definite integrals are vital in various fields such as calculus, engineering, and physics because they allow calculation of quantities that vary continuously.

Trigonometric Function

Trigonometric functions like sine, cosine, and tangent are fundamental in mathematics and describe relationships between angles and lengths in right triangles. In the context of integral approximation, we dealt with \( \sin(x^4) \). This is a composition of a simple power function and the sine function, making it non-elementary for direct integration.
To bypass this challenge, we took advantage of the function's periodic nature by using its Taylor Series expansion, transforming it into a polynomial. This transformation is essential, as polynomials are easier to integrate term-by-term—each term being a simple power function.

Series Expansion

Series Expansion involves expressing complex functions as sums of simpler polynomial terms. This concept is used extensively in calculus to approximate functions that are otherwise difficult to handle. It allows for an infinite series where each subsequent term becomes significantly smaller.
In our exercise, by expanding \( \sin(x^4) \) into its series form, we obtain:

• \( x^4 - \frac{x^{12}}{6} + \frac{x^{20}}{120} - \cdots \)

We then integrated each term independently over the interval \([0, 1]\). This method uses the properties of polynomial integration to estimate our integral's value to a high degree of accuracy, ignoring smaller terms that do not significantly affect the result within four decimal places.