Problem 6 - The Calculus with Analytic Geometry

Question

Compute the value of the given integral, accurate to four decimal places, by using series.\(\int_{0}^{1} h(x) d x\), where \(h(x)= \begin{cases}\frac{\sinh x}{x} & \text { if } x \neq 0 \\ 1 & \text { if } x=0\end{cases}\)

Step-by-Step Solution

Answer

The value of the integral is 0.9258.

Step 1: Break Down the Given Function

The given function is: \[ h(x)= \begin{cases}\frac{\sinh x}{x} & \text{ if } x eq 0 \ 1 & \text{ if } x = 0 \end{cases} \]

Step 2: Define the Series Expansion for \sinh x

Recall the Taylor series expansion for \sinh x: \[ \sinh x = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!} \]

Step 3: Divide the Series by x

Now, compute \frac{\sinh x}{x} using the series expansion: \[ \frac{\sinh x}{x} = \frac{1}{x} \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n+1)!} \]

Step 4: Integrate Term-by-Term

Integrate the series term-by-term from 0 to 1: \[ \int_{0}^{1} \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n+1)!} d x = \sum_{n=0}^{\infty} \int_{0}^{1} \frac{x^{2n}}{(2n+1)!} d x \]

Step 5: Compute Each Integral

Calculate the integrals: \[ \int_{0}^{1} x^{2n} d x = \frac{1}{2n+1} \] Therefore, \[ \int_{0}^{1} \frac{x^{2n}}{(2n+1)!} d x = \frac{1}{(2n+1)!} \cdot \frac{1}{2n+1} = \frac{1}{(2n+1)^{2}} \]

Step 6: Sum the Series

Combine the results to obtain \[ \int_{0}^{1} \frac{\sinh x}{x} d x = \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}} \] Summing up, we get \[ \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}} = 0.9258 \] (to four decimal places)

Key Concepts

Taylor Series

The Taylor series is a powerful tool in calculus that allows us to approximate more complex functions using polynomials. The series is represented as an infinite sum of terms calculated from the values of a function's derivatives at a single point. For example, the Taylor series expansion for the hyperbolic sine function \( \sinh x \) around zero is: \[ \sinh x = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!} \] This series expands the function into an infinite polynomial, making it easier to work with in various calculus problems.

The Taylor series is especially useful when working with integrals and limits, as it allows us to approximate functions that might otherwise be difficult to handle.

Hyperbolic Functions

Hyperbolic functions, like \( \sinh x \) (hyperbolic sine) and \( \cosh x \) (hyperbolic cosine), are analogs of the trigonometric functions but for a hyperbola instead of a circle. The hyperbolic sine function is defined as: \[ \sinh x = \frac{e^x - e^{-x}}{2} \] These functions arise in various areas of mathematics, including calculus and complex analysis.

In the given exercise, we used the Taylor series expansion of \( \sinh x \) to evaluate the integral of \( \frac{\sinh x}{x} \) over the interval from 0 to 1. By breaking down \( \sinh x \) into its Taylor series, we simplify the process of finding the integral.

Term-by-term Integration

Term-by-term integration is a method that allows us to integrate an infinite series by integrating each term in the series individually. In our example, after expanding and dividing \( \sinh x \) by \( x \), we obtained: \[ \frac{\sinh x}{x} = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n+1)!} \] By integrating each term from 0 to 1, we used the rule: \[ \int_{0}^{1} x^{2n} dx = \frac{1}{2n+1} \] This process simplifies the integration of complex functions by dealing with simpler polynomial terms instead of the original function.

Definite Integral

The definite integral represents the area under a curve within a specific interval. In our problem, we computed the definite integral of \( h(x) \) from 0 to 1. To solve: \[ \int_{0}^{1} \frac{\sinh x}{x} dx \] We used the infinite series expansion of \( \frac{\sinh x}{x} \), integrating term-by-term.

This technique allowed us to handle the complex function more easily. A definite integral calculation ensures our result is accurate over the specified interval.

Infinite Series

An infinite series sums infinitely many terms, which can help approximate functions that are otherwise difficult to handle. For instance, in our exercise, we rewrote \( \frac{\sinh x}{x} \) as an infinite series: \[ \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}} = 0.9258 \] Summing this series, we found an approximate value for the integral accurate to four decimal places.

Infinite series are fundamental in calculus because they link finite polynomials with more complex functions, allowing simpler calculations. Understanding how to manipulate and evaluate them is critical in advanced mathematics and various practical applications.